{"id":539888,"date":"2021-09-25T07:56:10","date_gmt":"2021-09-25T07:56:10","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=539888"},"modified":"2022-12-26T09:16:25","modified_gmt":"2022-12-26T09:16:25","slug":"rd-sharma-solutions-for-class-12-maths-chapter-2-functions","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-12-maths-chapter-2-functions\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Chapter 2\u2013Functions"},"content":{"rendered":"\n

Class 12: Maths Chapter 1 solutions. Complete Class 12 Maths Chapter 1 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 2\u2013Functions<\/h2>\n\n\n\n

RD Sharma 12th Maths Chapter 1, Class 12 Maths Chapter 1 solutions<\/p>\n\n\n\n

Page No 2.31:<\/h4>\n\n\n\n

Question 1:<\/h4>\n\n\n\n

Give an example of a function
(i) which is one-one but not onto
(ii) which is not one-one but onto
(iii) which is neither one-one nor onto<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(i) which is one-one but not onto.

f<\/em>: Z<\/em> \u2192 Z<\/em> given by f<\/em>(x<\/em>)=3x<\/em>+2

Injectivity:
Let x<\/em> and y<\/em> be any two elements in the domain (Z<\/em>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).
 f<\/em>(x<\/em>)=f<\/em>(y<\/em>)
\u21d23x<\/em> <\/em>+ 2 =3y<\/em> + 2
\u21d23x<\/em> = 3y<\/em>
\u21d2x<\/em> = y<\/em>
\u21d2f<\/em>(x) = f<\/em>(y<\/em>) \u21d2x<\/em> = y<\/em>
So, f<\/em> is one-one.

Surjectivity:
Let y<\/em> be any element in the co-domain (Z<\/em>), such that f<\/em>(x<\/em>) = y<\/em> for some element x<\/em> in Z<\/em> (domain).
f<\/em>(x<\/em>) = y<\/em>
\u21d23x<\/em> + 2 = y<\/em>
\u21d23x<\/em> = y<\/em> – 2

\u21d2x<\/em> = y<\/em>-23. It may not be in the domain (Z<\/em>) because if we take y = 3,x<\/em> = y<\/em>-23 = 3-23 = 13\u2209 domain Z<\/em>.
So, for every element in the co domain there need not be any element in the domain such that f<\/em>(x<\/em>) = y<\/em>.
Thus, f<\/em> is not onto.

(ii) which is not one-one but onto.
f<\/em>: Z<\/em> \u2192 N<\/em> \u222a {0} given by f<\/em>(x<\/em>) = |x<\/em>|

Injectivity:
Let x <\/em>and y<\/em> be any two elements in the domain (Z<\/em>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).
\u21d2|x<\/em>| = |y<\/em>|
\u21d2x<\/em>= \u00b1 y<\/em>
So, different elements of domain f<\/em> may give the same image.
So, f<\/em> is not one-one.

Surjectivity:
Let y<\/em> be any element in the co domain (Z<\/em>), such that f<\/em>(x<\/em>) = y<\/em> for some element x<\/em> in Z<\/em> (domain).
f<\/em>(x<\/em>) = y<\/em>
\u21d2|x<\/em>| = y<\/em>
\u21d2x<\/em> = \u00b1 y<\/em>, which is an element in Z<\/em> (domain).
So, for every element in the co-domain, there exists a pre-image in the domain.
Thus, f<\/em> is onto.

(iii) which is neither one-one nor onto.

f<\/em>: Z<\/em> \u2192 Z<\/em> given by f<\/em>(x<\/em>) = 2x<\/em>2<\/sup> + 1

Injectivity:
Let x<\/em> and y<\/em> be any two elements in the domain (Z<\/em>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).
f<\/em>(x<\/em>) = f<\/em>(y<\/em>)\u21d22x<\/em>2+1 = 2y<\/em>2+1\u21d22x<\/em>2 = 2y<\/em>2\u21d2x<\/em>2 = y<\/em>2\u21d2x<\/em> = \u00b1y<\/em>
So, different elements of domain f<\/em> may give the same image.
Thus, f<\/em> is not one-one.

Surjectivity:
Let y<\/em> be any element in the co-domain (Z<\/em>), such that f<\/em>(x<\/em>) = y<\/em> for some element x<\/em> in Z<\/em> (domain).
f<\/em>(x<\/em>) = y<\/em>
\u21d22x<\/em>2+1=y<\/em>\u21d22x<\/em>2=y<\/em>-1\u21d2x<\/em>2=y<\/em>-12\u21d2x<\/em>=\u00b1\u221ay<\/em>-12, \u2209 Z<\/em> always.For example, if we take, y<\/em> = 4,x<\/em>=\u00b1\u221ay<\/em>-12=\u00b1\u221a4-12=\u00b1\u221a32, \u2209 ZSo, x<\/em> may not be in Z<\/em> (domain).

Thus, f<\/em> is not onto.<\/p>\n\n\n\n

Page No 2.31:<\/h4>\n\n\n\n

Question 2:<\/h4>\n\n\n\n

Which of the following functions from A<\/em> to B<\/em> are one-one and onto?
(i) f<\/em>1<\/sub> = {(1, 3), (2, 5), (3, 7)} ; A<\/em> = {1, 2, 3}, B<\/em> = {3, 5, 7}
(ii) f<\/em>2<\/sub> = {(2, a<\/em>), (3, b<\/em>), (4, c<\/em>)} ; A<\/em> = {2, 3, 4}, B<\/em> = {a<\/em>, b<\/em>, c<\/em>}
(iii) f<\/em>3<\/sub> = {(a<\/em>, x<\/em>), (b<\/em>, x<\/em>), (c<\/em>, z<\/em>), (d<\/em>, z<\/em>)} ; A<\/em> = {a<\/em>, b<\/em>, c<\/em>, d<\/em>,}, B<\/em> = {x<\/em>, y<\/em>, z<\/em>}<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(i) f<\/em>1<\/sub> = {(1, 3), (2, 5), (3, 7)} ; A<\/em> = {1, 2, 3}, B<\/em> = {3, 5, 7}

Injectivity:
f<\/em>1<\/sub> (1) = 3
f<\/em>1<\/sub> <\/sub>(2) = 5
f<\/em>1<\/sub> (3) = 7
\u21d2Every element of A <\/em>has different images in B<\/em>.
So, f<\/em>1<\/sub> is one-one.

Surjectivity:
Co-domain of f<\/em>1<\/sub> = {3, 5, 7}
Range of f<\/em>1<\/sub> =set of images  =  {3, 5, 7}
\u21d2Co-domain = range
So, f<\/em>1<\/sub> is onto.

(ii) f<\/em>2<\/sub> = {(2, a<\/em>), (3, b<\/em>), (4, c<\/em>)} ; A<\/em> = {2, 3, 4}, B<\/em> = {a<\/em>, b<\/em>, c<\/em>}

Injectivity:
f<\/em>2<\/sub> <\/em>(2) = a<\/em>
f<\/em>2<\/sub> <\/sub><\/em>(3) = b<\/em>
f<\/em>2<\/sub> <\/em>(4) = c<\/em>
\u21d2Every element of A <\/em>has different images in B<\/em>.
So, f<\/em>2<\/sub> is one-one.

Surjectivity:
Co-domain of f<\/em>2<\/sub> = {a<\/em>, b<\/em>, c<\/em>}
Range of f<\/em>2<\/sub> = set of images = {a<\/em>, b<\/em>, c<\/em>}
\u21d2Co-domain = range
So, f<\/em>2<\/sub> is onto.

(iii) f<\/em>3<\/sub> = {(a<\/em>, x<\/em>), (b<\/em>, x<\/em>), (c<\/em>, z<\/em>), (d<\/em>, z<\/em>)} ; A<\/em> = {a<\/em>, b<\/em>, c<\/em>, d<\/em>,}, B<\/em> = {x<\/em>, y<\/em>, z<\/em>}

Injectivity:
f3<\/sub> (a) = x
f3<\/sub> <\/sub>(b) = x
f3<\/sub> (c) = z
f3<\/sub> (d) = z<\/em>
\u21d2a<\/em> and b <\/em>have the same image x<\/em>. (Also <\/em>c<\/em> and d<\/em> have the same image z<\/em>)
So, f<\/em>3<\/sub> is not one-one.

Surjectivity:
Co-domain of f<\/em>1<\/sub> ={x, y, z}<\/em>
Range of f<\/em>1<\/sub> =set of images  =  {x, z}<\/em>
So, the co-domain  is not same as the range.
So, f3<\/sub><\/em> is not onto.<\/p>\n\n\n\n

Page No 2.31:<\/h4>\n\n\n\n

Question 3:<\/h4>\n\n\n\n

Prove that the function f<\/em> : N<\/em> \u2192 N<\/em>, defined by f<\/em>(x<\/em>) = x<\/em>2<\/sup> + x<\/em> + 1, is one-one but not onto.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em> : N<\/em> \u2192 N<\/em>, defined by f<\/em>(x<\/em>) = x<\/em>2<\/sup> + x<\/em> + 1

Injectivity:
Let x<\/em> and y<\/em> be any two elements in the domain (N<\/em>), such that f(x) = f(y).<\/em>
\u21d2x2+x+1=y2+y+1\u21d2(x2-y2)+(x-y)=0\u21d2(x+y)(x-y)+(x-y)=0\u21d2(x-y)(x+y+1)=0\u21d2x-y=0     [ ( x+y+1) cannot be zero because x and y are natural numbers]\u21d2x=y<\/em>
 So, f <\/em>is one-one.

Surjectivity:
The minimum number in N is 1.When x=1,x2+x+1=1+1+1=3\u21d2x2+x+1\u22653, for every x in N.\u21d2f(x) will not assume the values 1 and 2.So, f is not onto.<\/em><\/p>\n\n\n\n

Page No 2.31:<\/h4>\n\n\n\n

Question 4:<\/h4>\n\n\n\n

Let A<\/em> = {\u22121, 0, 1} and f<\/em> = {(x<\/em>, x<\/em>2<\/sup>) : x<\/em> \u2208 A<\/em>}. Show that f<\/em> : A<\/em> \u2192 A<\/em> is neither one-one nor onto.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

A<\/em> = {\u22121, 0, 1} and f<\/em> = {(x<\/em>, x<\/em>2<\/sup>) : x<\/em> \u2208 A<\/em>}
Given, f(x) = x2<\/sup><\/em>

Injectivity:
f<\/em>(1) = 12<\/sup>=1 and
f<\/em>(-1)=(-1)2<\/sup>=1

\u21d21 and -1 have the same images.
So, f  <\/em>is not one-one.

Surjectivity:
Co-domain of  f  = <\/em>{-1, 0, 1}

f<\/em>(1) = 12<\/sup> = 1,
f<\/em>(-1) = (-1)2<\/sup> = 1 and
f<\/em>(0) = 0
\u21d2Range of f  <\/em>= {0, 1}
So, both are not same.
Hence, f  <\/em>is not onto.<\/p>\n\n\n\n

Page No 2.31:<\/h4>\n\n\n\n

Question 5:<\/h4>\n\n\n\n

Classify the following functions as injection, surjection or bijection :
(i) f<\/em> : N<\/strong> \u2192 N<\/strong> given by f<\/em>(x<\/em>) = x<\/em>2<\/sup>
(ii) f<\/em> : Z<\/strong> \u2192 Z<\/strong> given by f<\/em>(x<\/em>) = x<\/em>2<\/sup>
(iii) f<\/em> : N<\/strong> \u2192 N<\/strong> given by f<\/em>(x<\/em>) = x<\/em>3<\/sup>
(iv) f<\/em> : Z<\/strong> \u2192 Z<\/strong> given by f<\/em>(x<\/em>) = x<\/em>3<\/sup>
(v) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = |x<\/em>|
(vi) f<\/em> : Z<\/strong> \u2192 Z<\/strong>, defined by f<\/em>(x<\/em>) = x<\/em>2<\/sup> + x<\/em>
(vii) f<\/em> : Z<\/strong> \u2192 Z<\/strong>, defined by f<\/em>(x<\/em>) = x<\/em> \u2212 5
(viii) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = sinx<\/em>
(ix) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = x<\/em>3<\/sup> + 1
(x) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = x<\/em>3<\/sup> \u2212 x<\/em>
(xi) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = sin2<\/sup>x<\/em> + cos2<\/sup>x<\/em>
(xii) f<\/em> : Q<\/strong> \u2212 {3} \u2192 Q<\/strong>, defined by f<\/em>(x<\/em>)=2x<\/em>+3x<\/em>-3
(xiii) f<\/em> : Q<\/strong> \u2192 Q<\/strong>, defined by f<\/em>(x<\/em>) = x<\/em>3<\/sup> + 1
(xiv) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = 5x<\/em>3<\/sup> + 4
(xv) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = 3 \u2212 4x<\/em>
(xvi) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = 1 + x<\/em>2<\/sup>
(xvii) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = x<\/em>x<\/em>2+1                                                                                                                    [NCERT EXEMPLAR]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(i) f<\/em> : N<\/strong> \u2192 N<\/strong>,<\/em> given by f<\/em>(x<\/em>) = x<\/em>2<\/sup>

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (N<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).<\/em>

f<\/em>(x<\/em>)=f<\/em>(y<\/em>)

x<\/em>2=y<\/em>2x<\/em>=y<\/em>       (We do not get \u00b1 because x<\/em> and y<\/em> are in N<\/strong>)

So, f<\/em> is an injection .

Surjection test:

Let y<\/em> be any element in the co-domain (N<\/strong>),<\/em> such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in N<\/strong> (domain).

f<\/em>(x<\/em>) = y<\/em>

x<\/em>2=y<\/em>x<\/em>=\u221ay<\/em>, which may not be in N<\/strong>.For example, if y<\/em>=3,x<\/em>=\u221a3 is not in N<\/strong>.

So, f <\/em>is not a surjection.

So, f<\/em> is not a bijection.

(ii) f<\/em> : Z<\/strong> \u2192 Z<\/strong>,<\/em> given by f<\/em>(x<\/em>) = x<\/em>2<\/sup>

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (Z<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

x<\/em>2=y<\/em>2x<\/em>=\u00b1y<\/em>

So, f<\/em> is not an injection .

Surjection test:

Let y<\/em> be any element in the co-domain (Z<\/strong>),<\/em> such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in Z<\/strong> (domain).

f<\/em>(x<\/em>) = y<\/em>

x<\/em>2=y<\/em>x<\/em>=\u00b1\u221ay<\/em> which may not be in Z<\/strong>.For example, if y<\/em>=3,x<\/em>=\u00b1\u221a3 is not in Z<\/strong>.

So, f <\/em>is not a surjection.

So, f<\/em> is not a bijection.

(iii) f<\/em> : N<\/strong> \u2192 N<\/strong>, given by f<\/em>(x<\/em>) = x<\/em>3<\/sup>

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (N<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

x<\/em>3=y<\/em>3x<\/em>=y<\/em>

So, f<\/em> is an injection .

Surjection test:

Let y<\/em> be any element in the co-domain (N<\/strong>),<\/em> such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in N <\/strong>(domain).

f<\/em>(x<\/em>) = y<\/em>

x<\/em>3=y<\/em>x<\/em>=3\u221ay<\/em>which may not be in N<\/strong>.For example, if y<\/em>=3,x<\/em>=3\u221a3 is not in N<\/strong>.

So, f <\/em>is not a surjection and  f<\/em> is not a bijection.

(iv) f<\/em> : Z<\/strong> \u2192 Z<\/strong>,<\/em> given by f<\/em>(x<\/em>) = x<\/em>3<\/sup>

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (Z<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

x<\/em>3=y<\/em>3x<\/em>=y<\/em>

So, f<\/em> is an injection.

Surjection test:

Let y<\/em> be any element in the co-domain (Z<\/strong>),<\/em> such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in Z<\/strong> (domain).

f(x) = y<\/em>

x<\/em>3=y<\/em>x<\/em>=3\u221ay<\/em> which may not be in Z<\/strong>.For example, if y<\/em>=3,x<\/em>=3\u221a3 is not in Z<\/strong>.

So, f <\/em>is not a surjection and f<\/em> is not a bijection.

(v) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = |x<\/em>|

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (R<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

|x<\/em>|=|y<\/em>|x<\/em>=\u00b1y<\/em>

So, f<\/em> is not an injection .

Surjection test:

Let y<\/em> be any element in the co-domain (R<\/strong>), such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in R<\/strong> (domain).

f<\/em>(x<\/em>) = y<\/em>

|x<\/em>|=y<\/em>x<\/em>=\u00b1y<\/em>\u2208Z<\/strong>

So, f <\/em>is a surjection and  f<\/em> is not a bijection.

(vi) f<\/em> : Z<\/strong> \u2192 Z<\/strong>, defined by f<\/em>(x<\/em>) = x<\/em>2<\/sup> + x<\/em>

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (Z<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

x<\/em>2+x<\/em>=y<\/em>2+y<\/em>Here, we cannot say that x<\/em> = y<\/em>.For example, x = 2 and y = – 3Then, x<\/em>2+x<\/em>=22+2= 6y<\/em>2+y<\/em>=(-3)2-3= 6So, we have two numbers 2 and -3 in the domain Z<\/strong> whose image is same as 6.

So, f<\/em> is not an injection .

Surjection test:

Let y<\/em> be any element in the co-domain (Z<\/strong>), such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in Z<\/strong> (domain).

f<\/em>(x<\/em>) = y<\/em>

x<\/em>2+x<\/em>=y<\/em>Here, we cannot say x<\/em>\u2208Z<\/strong>.For example, y<\/em> =-4.x<\/em>2+x<\/em>=-4x<\/em>2+x<\/em>+4=0x<\/em>=-1\u00b1\u221a-152=-1\u00b1i<\/em>\u221a152 which is not in Z<\/strong>.

So, f <\/em>is not a surjection and  f<\/em> is not a bijection.

(vii) f<\/em> : Z<\/strong> \u2192 Z<\/strong>, defined by f<\/em>(x<\/em>) = x<\/em> \u2212 5

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (Z<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

x<\/em> – 5 = y<\/em> – 5

x = y<\/em>

So, f<\/em> is an injection .

Surjection test:

Let y<\/em> be any element in the co-domain (Z<\/strong>),<\/em> such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in Z<\/strong> (domain).

f<\/em>(x<\/em>) = y<\/em>

x <\/em>– 5 = y<\/em>

x = y <\/em>+ 5, which is in Z<\/strong>.

So, f <\/em>is a surjection and f<\/em> is a bijection.

(viii) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = sinx<\/em>

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (R<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

sinx<\/em>=siny<\/em>Here, x<\/em> may not be equal to y<\/em> because sin0=sin\u03c0.So, 0 and \u03c0 have the same image 0.

So, f<\/em> is not an injection .

Surjection test:

Range of f<\/em> = [-1, 1]

Co-domain of f <\/em>= R<\/strong>

Both are not same.

So, f<\/em> is not a surjection and f<\/em> is not a bijection.

(ix) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = x<\/em>3<\/sup> + 1

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (R<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

x<\/em>3+1=y<\/em>3+1x<\/em>3=y<\/em>3x<\/em>=y<\/em>

So, f<\/em> is an injection.

Surjection test:

Let y<\/em> be any element in the co-domain (R<\/strong>), <\/em>such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in R <\/strong>(domain).

f(x) = y<\/em>

x<\/em>3+1=y<\/em>x<\/em>=3\u221ay<\/em>-1\u2208R<\/strong>

So, f <\/em>is a surjection.

So, f<\/em> is a bijection.

(x) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = x<\/em>3<\/sup> \u2212 x<\/em>

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (R<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

x<\/em>3-x<\/em>=y<\/em>3-y<\/em>Here, we cannot say x<\/em>=y<\/em>.For example, x<\/em>=1 and y<\/em>=-1x<\/em>3-x<\/em>=1-1=0y<\/em>3-y<\/em>=(-1)3-(-1)-1+1=0So, 1 and -1 have the same image 0.

So, f<\/em> is not an injection.

Surjection test:

Let y<\/em> be any element in the co-domain (R<\/strong>),<\/em> such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in R <\/strong>(domain).

f<\/em>(x<\/em>) = y<\/em>

x<\/em>3-x<\/em>=y<\/em>By observation we can say that there exist some x<\/em> in R<\/strong>, such that x<\/em>3-x=y.

So, f <\/em>is a surjection and  f<\/em> is not a bijection.

(xi) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = sin2<\/sup>x<\/em> + cos2<\/sup>x<\/em>

f<\/em>(x<\/em>) = sin2<\/sup>x<\/em> + cos2<\/sup>x <\/em>= 1

So, f<\/em>(x<\/em>) = 1 for every x<\/em> in <\/em>R<\/strong>.

So, for all elements in the domain, the image is 1.

So, f<\/em> is not an injection.

Range of f <\/em>= {1}

Co-domain of f<\/em> = R<\/strong>

Both are not same.

So, f<\/em> is not a surjection and  f <\/em>is not a bijection.

(xii) f<\/em> : Q<\/strong> \u2212 {3} \u2192 Q<\/strong>, defined by f<\/em>(x<\/em>)=2x<\/em>+3x<\/em>-3

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (Q<\/strong> \u2212 {3}), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

2x<\/em>+3x<\/em>-3=2y<\/em>+3y<\/em>-3(2x<\/em>+3)(y<\/em>-3)=(2y<\/em>+3)(x<\/em>-3)2x<\/em>y<\/em>-6x<\/em>+3y<\/em>-9=2x<\/em>y<\/em>-6y<\/em>+3x<\/em>-99x<\/em>=9y<\/em>x<\/em>=y<\/em>

So, f<\/em> is an injection.

Surjection test:

Let y<\/em> be any element in the co-domain (Q<\/strong> \u2212 {3}),<\/em> such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in Q<\/strong> (domain).

f<\/em>(x<\/em>) = y<\/em>

2x<\/em>+3x<\/em>-3=y<\/em>2x<\/em>+3=x<\/em>y<\/em>-3y<\/em>2x<\/em>–x<\/em>y<\/em>=-3y<\/em>-3x<\/em>(2-y<\/em>)=-3(y<\/em>+1)x<\/em>=3(y<\/em>+1)y<\/em>-2, which is not defined at y<\/em>=2.

So, f<\/em> is not a surjection and f<\/em> is not a bijection.

(xiii) f<\/em> : Q<\/strong> \u2192 Q<\/strong>, defined by f<\/em>(x<\/em>) = x<\/em>3<\/sup> + 1

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (Q<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).<\/em>

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

x<\/em>3+1=y<\/em>3+1x<\/em>3=y<\/em>3x<\/em>=y<\/em>

So, f<\/em> is an injection .

Surjection test:

Let y<\/em> be any element in the co-domain (Q<\/strong>),<\/em> such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in Q<\/strong> (domain).

f<\/em>(x<\/em>) = y<\/em>

x<\/em>3+1=y<\/em>x<\/em>=3\u221ay<\/em>-1, which may not be in Q<\/strong>.For example, if y<\/em>= 8,x<\/em>3+1= 8x<\/em>3=7x<\/em>=3\u221a7, which is not in Q<\/strong>.

So, f<\/em> is not a surjection and f<\/em> is not a bijection.

So, f <\/em>is a surjection and f<\/em> is a bijection.

(xiv) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = 5x<\/em>3<\/sup> + 4

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (R<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

5x<\/em>3+4 = 5y<\/em>3+45x<\/em>3= 5y<\/em>3x<\/em>3= y<\/em>3x<\/em>=y<\/em>

So, f<\/em> is an injection .

Surjection test:

Let y<\/em> be any element in the co-domain (R<\/strong>),<\/em> such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in R <\/strong>(domain).

f(x) = y<\/em>

5x<\/em>3+4=y<\/em>5x<\/em>3=y<\/em>-4x<\/em>3=y<\/em>-45x<\/em>=3\u221ay<\/em>-45\u2208R<\/strong>

So, f <\/em>is a surjection and f<\/em> is a bijection.

(xv) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = 3 \u2212 4x<\/em>

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (R<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

3-4x<\/em>=3-4y<\/em>-4x<\/em>=-4y<\/em>x<\/em>= y<\/em>

So, f<\/em> is an injection .

Surjection test:

Let y<\/em> be any element in the co-domain (R<\/strong>),<\/em> such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in R <\/strong>(domain).

f(x) = y<\/em>

3-4x<\/em>=y<\/em>4x<\/em>=3-y<\/em>x<\/em>=3-y<\/em>4\u2208R<\/strong>

So, f <\/em>is a surjection and f<\/em> is a bijection.

(xvi) f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = 1 + x<\/em>2<\/sup>

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (R<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).<\/em>

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

1+x<\/em>2=1+y<\/em>2x<\/em>2=y<\/em>2x<\/em>= \u00b1y<\/em>

So, f<\/em> is not an injection.

Surjection test:

Let y<\/em> be any element in the co-domain (R<\/strong>),<\/em> such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in R <\/strong>(domain).

f<\/em>(x<\/em>) = y<\/em>

1+x<\/em>2=y<\/em>x<\/em>2=y<\/em>-1x<\/em>=\u00b1\u221ay<\/em>-1 which may not be in R<\/strong>For example, if y<\/em>=0,x<\/em>=\u00b1\u221a-1=\u00b1i<\/em> is not in R<\/strong>.

So, f <\/em>is not a surjection and f<\/em> is not a bijection.

(xvii)  f<\/em> : R<\/strong> \u2192 R<\/strong>, defined by f<\/em>(x<\/em>) = x<\/em>x<\/em>2+1

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (R<\/strong>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).<\/em>

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

x<\/em>x<\/em>2+1=y<\/em>y<\/em>2+1x<\/em>y<\/em>2+x<\/em>=x<\/em>2y<\/em>+y<\/em>x<\/em>y<\/em>2-x<\/em>2y<\/em>+x<\/em>–y<\/em>=0-x<\/em>y<\/em>(-y<\/em>+x<\/em>)+1(x<\/em>–y<\/em>)=0(x<\/em>–y<\/em>)(1-x<\/em>y<\/em>)=0x<\/em>=y<\/em> or x<\/em>=1y<\/em>

So, f<\/em> is not an injection.

Surjection test:

Let y<\/em> be any element in the co-domain (R<\/strong>),<\/em> such that f<\/em>(x<\/em>) = y <\/em>for some element x<\/em> in R <\/strong>(domain).

f<\/em>(x<\/em>) = y<\/em>

x<\/em>x<\/em>2+1=y<\/em>y<\/em>x<\/em>2-x<\/em>+y<\/em>=0x<\/em>=-(-1)\u00b1\u221a1-4y<\/em>22y<\/em>, if y<\/em>\u22600=1\u00b1\u221a1-4y<\/em>22y<\/em>, which may not be in R<\/strong>For example, if y<\/em>=1, thenx<\/em>=1\u00b1\u221a1-42=1\u00b1i\u221a32, which is not in R<\/strong>So, f<\/em> is not surjection and f<\/em> is not bijection.

So, f <\/em>is not a surjection and f<\/em> is not a bijection.<\/p>\n\n\n\n

Page No 2.31:<\/h4>\n\n\n\n

Question 6:<\/h4>\n\n\n\n

If f<\/em> : A<\/em> \u2192 B<\/em> is an injection, such that range of f<\/em> = {a<\/em>}, determine the number of elements in A<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Range of f<\/em> = {a<\/em>}
So, the number of images of  f <\/em>= 1
Since, f  <\/em>is an injection, there will be exactly one image for each element of f<\/em> .
So, number of elements in A<\/em> = 1.<\/p>\n\n\n\n

Page No 2.31:<\/h4>\n\n\n\n

Question 7:<\/h4>\n\n\n\n

Show that the function f<\/em> : R<\/em> \u2212 {3} \u2192 R<\/em> \u2212 {2} given by f<\/em>(x<\/em>)=x<\/em>-2x<\/em>-3 is a bijection.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em> : R<\/em> \u2212 {3} \u2192 R<\/em> \u2212 {2} given by
f<\/em>(x<\/em>)=x<\/em>-2x<\/em>-3
Injectivity:
Let x<\/em> and y<\/em> be any two elements in the domain (R<\/em> \u2212 {3}), such that f(x) = f(y)<\/em>.
 f(x) = f(y)
\u21d2x-2x-3=y-2y-3\u21d2(x-2)(y-3)=(y-2)(x-3)\u21d2xy-3x-2y+6=xy-3y-2x+6\u21d2x=y<\/em>
So, f<\/em> is one-one.

Surjectivity:
Let y<\/em> be any element in the co-domain (R<\/em> \u2212 {2}), such that f(x) = y <\/em>for some element x<\/em> in R<\/em> \u2212 {3} <\/em>(domain).
f(x) = y<\/em>
\u21d2x<\/em>-2x<\/em>-3=y<\/em>\u21d2x<\/em>-2=x<\/em>y<\/em>-3y<\/em>\u21d2x<\/em>y<\/em>–x<\/em>=3y<\/em>-2\u21d2x<\/em>(y<\/em>-1)=3y<\/em>-2\u21d2x<\/em>=3y<\/em>-2y<\/em>-1, which is in R<\/em>-{3}
So, for every element in the co-domain, there exists some pre-image in the domain.
\u21d2f  <\/em>is onto.
Since, f  <\/em>is both one-one and onto, it is a bijection.<\/p>\n\n\n\n

Page No 2.32:<\/h4>\n\n\n\n

Question 8:<\/h4>\n\n\n\n

Let A<\/em> = [-1, 1]. Then, discuss whether the following functions from A<\/em> to itself are one-one, onto or bijective:

(i) f<\/em>(x<\/em>) = x<\/em>2                                (ii) g<\/em>(x<\/em>) = |x<\/em>|                                (iii) h<\/em>(x<\/em>) = x<\/em>2<\/sup>                                                               [NCERT EXEMPLAR]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(i) f<\/em> : A<\/em> \u2192 A<\/em>, given by f<\/em>(x<\/em>) = x<\/em>2

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (A<\/em>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

x<\/em>2 = y<\/em>2

x<\/em> = y<\/em>

So, f<\/em> is one-one.

Surjection test:

Let y<\/em> be any element in the co-domain (A<\/em>), such that f<\/em>(x<\/em>) = y<\/em> for some element x<\/em> in A<\/em> (domain)

f<\/em>(x<\/em>) = y<\/em>

x<\/em>2 = y<\/em>

x<\/em> = 2y<\/em>, which may not be in A<\/em>.

For example, if y<\/em> = 1, then

x<\/em> = 2, which is not in A<\/em>.

So, f<\/em> is not onto.

So, f<\/em> is not bijective.

(ii) g<\/em>(x<\/em>) = |x<\/em>|

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (A<\/em>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

|x<\/em>| = |y<\/em>|

x<\/em> = \u00b1y<\/em>

So, f<\/em> is not one-one.

Surjection test:

For y<\/em> = -1, there is no value of x<\/em> in A<\/em>.

So, f<\/em> is not onto.

So, f<\/em> is not bijective.

(iii) h<\/em>(x<\/em>) = x<\/em>2<\/sup>

Injection test:

Let x<\/em> and y<\/em> be any two elements in the domain (A<\/em>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).

f<\/em>(x<\/em>) = f<\/em>(y<\/em>)

x<\/em>2<\/sup> = y<\/em>2<\/sup>

x<\/em> = \u00b1y<\/em>

So, f<\/em> is not one-one.

Surjection test:

For y<\/em> = -1, there is no value of x<\/em> in A<\/em>.

So, f<\/em> is not onto.

So, f<\/em> is not bijective.

 <\/p>\n\n\n\n

Page No 2.32:<\/h4>\n\n\n\n

Question 9:<\/h4>\n\n\n\n

Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:

(i) {(x<\/em>, y<\/em>) : x<\/em> is a person, y<\/em> is the mother of x<\/em>}
(ii) {(a<\/em>, b<\/em>) : a<\/em> is a person, b<\/em> is an ancestor of a<\/em>}                                                                                                               [NCERT EXEMPLAR]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(i) f<\/em> = {(x<\/em>, y<\/em>) : x<\/em> is a person, y<\/em> is the mother of x<\/em>}

As, for each element x<\/em> in domain set, there is a unique related element y<\/em> in co-domain set.

So, f<\/em> is the function.

Injection test:

As, y<\/em> can be mother of two or more persons

So, f<\/em> is not injective.

Surjection test:

For every mother y<\/em> defined by (x<\/em>, y<\/em>), there exists a person x<\/em> for whom y<\/em> is mother.

So, f<\/em> is surjective.

Therefore, f<\/em> is surjective function.

(ii) g<\/em> = {(a<\/em>, b<\/em>) : a<\/em> is a person, b<\/em> is an ancestor of a<\/em>}

Since, the ordered map (a<\/em>, b<\/em>) does not map ‘a<\/em>‘ – a person to a living person.
So, g<\/em> is not a function.<\/p>\n\n\n\n

Page No 2.32:<\/h4>\n\n\n\n

Question 10:<\/h4>\n\n\n\n

Let A<\/em> = {1, 2, 3}. Write all one-one from A<\/em> to itself.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

A =<\/em>{1, 2, 3}
Number of elements in  A <\/em>= 3
Number of one-one functions = number of ways of arranging 3 elements = 3! = 6

(i) {(1, 1), (2, 2), (3, 3)}
(ii) {(1, 1), (2, 3), (3, 2)}
(iii) {(1, 2 ), (2, 2), (3, 3 )}
(iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1, 3), (2, 2), (3, 1)}
(vi) {(1, 3), (2, 1), (3,2 )}<\/p>\n\n\n\n

Page No 2.32:<\/h4>\n\n\n\n

Question 11:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> be the function defined by f<\/em>(x<\/em>) = 4x<\/em>3<\/sup> + 7, show that f<\/em> is a bijection.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity:
Let x<\/em> and y<\/em> be any two elements in the domain (R<\/em>), such that f(x) = f(y)<\/em>
\u21d24x<\/em>3+7=4y<\/em>3+7\u21d24x<\/em>3=4y<\/em>3\u21d2x<\/em>3=y<\/em>3\u21d2x<\/em>=y<\/em>
So, f <\/em>is one-one.

Surjectivity:
Let y<\/em> be any element in the co-domain (R),<\/em> such that f(x) = y <\/em>for some element x<\/em> in R <\/em>(domain).
f(x) = y<\/em>
\u21d24x<\/em>3+7=y<\/em>\u21d24x<\/em>3=y<\/em>-7\u21d2x<\/em>3=y<\/em>-74\u21d2x<\/em>=3\u221ay<\/em>-74\u2208R<\/em>
So, for every element in the co-domain, there exists some pre-image in the domain.
\u21d2f <\/em>is onto.
Since, f<\/em> is both one-to-one and onto, it is a bijection.<\/p>\n\n\n\n

Page No 2.32:<\/h4>\n\n\n\n

Question 12:<\/h4>\n\n\n\n

Show that the exponential function f<\/em> : R<\/em> \u2192 R<\/em>, given by f<\/em>(x<\/em>) = ex<\/sup><\/em>, is one-one but not onto. What happens if the co-domain is replaced by R<\/em>+0 (set of all positive real numbers)?<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em> : R<\/em> \u2192 R<\/em>, given by f<\/em>(x<\/em>) = ex<\/sup>

Injectivity:
Let x<\/em> and y<\/em> be any two elements in the domain (R<\/em>), such that f(x) = f(y)<\/em>
 f(x)=f(y)
\u21d2ex=ey\u21d2x=y<\/em>
So, f <\/em>is one-one.

Surjectivity:
We know that range of ex<\/sup><\/em> is (0, \u221e) = R+<\/sup><\/em>
Co-domain = R<\/em>
Both are not same.
So, f<\/em> is not onto.

If the co-domain is replaced by R+<\/sup><\/em>, then the co-domain and range become the same and in that case, f<\/em> is onto and hence, it is a bijection.<\/p>\n\n\n\n

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Question 13:<\/h4>\n\n\n\n

Show that the logarithmic function f<\/em> : R<\/em>+0\u2192R<\/em> given by f<\/em>(x<\/em>)=loga<\/em> x<\/em>, a<\/em>>0 is a bijection.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em>:R<\/em>+<\/sup>\u2192R<\/em> given by f<\/em>(x<\/em>)= loga<\/em> x<\/em>, a<\/em>>0
Injectivity:
Let x<\/em> and y<\/em> be any two elements in the domain (N<\/em>), such that f<\/em>(x<\/em>) = f<\/em>(y<\/em>).
 f(x) = f(y)<\/em>
loga<\/em> x<\/em>=loga<\/em> y<\/em>\u21d2x<\/em>=y<\/em>
So, f<\/em> is one-one.

Surjectivity:
Let y<\/em> be any element in the co-domain (R),<\/em> such that f(x) = y <\/em>for some element x<\/em> in R+<\/sup> <\/em>(domain).
f(x) = y<\/em>
loga<\/em> x<\/em>=y<\/em>\u21d2x<\/em>=a<\/em>y<\/em> \u2208R<\/em>+
So, for every element in the co-domain, there exists some pre-image in the domain.
\u21d2f<\/em> is onto.
Since f<\/em> is one-one and onto, it is a bijection.<\/p>\n\n\n\n

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Question 14:<\/h4>\n\n\n\n

If A<\/em> = {1, 2, 3}, show that a one-one function f<\/em> : A<\/em> \u2192 A<\/em> must be onto.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

A =<\/em>{1, 2, 3}
Number of elements in  A <\/em>= 3
Number of one – one functions = number of ways of arranging 3 elements = 3! = 6
So, the possible one -one functions can be the following:

(i) {(1, 1), (2, 2), (3, 3)}
(ii) {(1, 1), (2, 3), (3, 2)}
(iii) {(1, 2 ), (2, 2), (3, 3 )}
(iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1, 3), (2, 2), (3, 1)}
(vi) {(1, 3), (2, 1), (3,2 )}
Here, in each function, range = {1, 2, 3}, which is same as the co-domain.
So, all the functions are onto.<\/p>\n\n\n\n

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Question 15:<\/h4>\n\n\n\n

If A<\/em> = {1, 2, 3}, show that a onto function f<\/em> : A<\/em> \u2192 A<\/em> must be one-one.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

A =<\/em>{1, 2, 3}
Possible onto functions from A<\/em> to A<\/em> can be the following:

(i) {(1, 1), (2, 2), (3, 3)}
(ii) {(1, 1), (2, 3), (3, 2)}
(iii) {(1, 2 ), (2, 2), (3, 3 )}
(iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1, 3), (2, 2), (3, 1)}
(vi) {(1, 3), (2, 1), (3,2 )}

Here, in each function, different elements of the domain have different images.
So, all the functions are one-one.<\/p>\n\n\n\n

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Question 16:<\/h4>\n\n\n\n

Find the number of all onto functions from the set A<\/em> = {1, 2, 3, …, n<\/em>} to itself.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We know that every onto function from A<\/em> to itself is one-one.
So, the number of one-one functions = number of bijections = n<\/em>!<\/p>\n\n\n\n

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Question 17:<\/h4>\n\n\n\n

Give examples of two one-one functions f<\/em>1<\/sub> and f<\/em>2<\/sub> from R<\/em> to R,<\/em> such that f<\/em>1<\/sub> + f<\/em>2<\/sub> : R<\/em> \u2192 R<\/em>. defined by (f<\/em>1<\/sub> + f<\/em>2<\/sub>) (x<\/em>) = f<\/em>1<\/sub> (x<\/em>) + f<\/em>2<\/sub> (x<\/em>) is not one-one.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We know that f<\/em>1<\/sub>: R<\/em> \u2192 R,<\/em> given by f<\/em>1<\/sub>(x)=x, <\/em>and <\/em>f<\/em>2<\/sub>(x)=-x<\/em> are one-one.
Proving f<\/em>1<\/sub> <\/sub>is one-one:
Let f<\/em>1(x<\/em>)=f<\/em>1(y<\/em>)\u21d2x<\/em>=y<\/em>
So, f<\/em>1 <\/sub>is one-one.

Proving f<\/em>2<\/sub> <\/sub>is one-one:
Let f<\/em>2(x<\/em>)=f<\/em>2(y<\/em>)\u21d2-x<\/em>=-y<\/em>\u21d2x<\/em>=y<\/em>
So, f<\/em>2<\/sub>  <\/sub><\/em>is one-one.

Proving (f<\/em>1<\/sub> + f<\/em>2<\/sub>) is not one-one:
Given:
(f<\/em>1<\/sub> + f<\/em>2<\/sub>) (x<\/em>) = f<\/em>1<\/sub> (x<\/em>) + f<\/em>2<\/sub> (x<\/em>)= x<\/em> + (-x<\/em>) =0
So, for every real number x<\/em>, (f<\/em>1<\/sub> + f<\/em>2<\/sub>) (x<\/em>)=0
So, the image of ever number in the domain is same as 0.
Thus, (f<\/em>1<\/sub> + f<\/em>2<\/sub>) is not one-one.<\/p>\n\n\n\n

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Question 18:<\/h4>\n\n\n\n

Give examples of two surjective functions f<\/em>1<\/sub> and f<\/em>2<\/sub> from Z<\/em> to Z<\/em> such that f<\/em>1<\/sub> + f<\/em>2<\/sub> is not surjective.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We know that f<\/em>1<\/sub>: R<\/em> \u2192 R,<\/em> given by f<\/em>1<\/sub>(x) = x, <\/em>and <\/em>f<\/em>2<\/sub>(x) = -x<\/em> are surjective functions.
Proving f<\/em>1<\/sub> <\/sub>is surjective :
Let y<\/em> be an element in the co-domain (R<\/em>), such that <\/em>f<\/em>1<\/sub>(x) = y<\/em>.
f<\/em>1<\/sub>(x) = y<\/em>
\u21d2x<\/em> = y<\/em>, which is in R<\/em>.
So, <\/em>for every element in the co-domain, there exists some pre-image in the domain.1(x<\/em>)=f<\/em>1(y<\/em>)x<\/em>=y<\/em>
So, f<\/em>1<\/sub>is surjective .

Proving f<\/em>2<\/sub> is surjective :L<\/em>e<\/em>t<\/em> f<\/em>2(x<\/em>)=f<\/em>2(y<\/em>)\u2212x<\/em>=\u2212y<\/em>x<\/em>=y<\/em>
Let y<\/em> be an element in the co domain (R<\/em>) such that <\/em>f<\/em>2<\/sub>(x) = y<\/em>.
 <\/em>f<\/em>2<\/sub>(x) = y<\/em>
\u21d2x<\/em> = y<\/em>, which is in R<\/em>.
So, <\/em>for every element in the co-domain, there exists some pre-image in the domain.1(x<\/em>)=f<\/em>1(y<\/em>)x<\/em>=y<\/em>
So, f<\/em>2<\/sub> <\/sub>is surjective .

Proving (f<\/em>1<\/sub> + f<\/em>2<\/sub>) is not surjective :
Given:
(f<\/em>1<\/sub> + f<\/em>2<\/sub>) (x<\/em>) = f<\/em>1<\/sub> (x<\/em>) + f<\/em>2<\/sub> (x<\/em>)= x + (-x) =0
So, for every real number x<\/em>, (f<\/em>1<\/sub> + f<\/em>2<\/sub>) (x<\/em>)=0
So, <\/em>the image of every number in the domain is same as 0.
\u21d2<\/em>Range = {0}
Co-domain = R
So, both are not same.
So, f<\/em>1<\/sub> + f<\/em>2<\/sub> <\/sub>is not surjective.<\/p>\n\n\n\n

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Question 19:<\/h4>\n\n\n\n

Show that if f<\/em>1<\/sub> and f<\/em>2<\/sub> are one-one maps from R<\/em> to R<\/em>, then the product f<\/em>1\u00d7f<\/em>2 : R<\/em>\u2192R<\/em> defined by (f<\/em>1\u00d7f<\/em>2) (x<\/em>)=f<\/em>1 (x<\/em>) f<\/em>2 (x<\/em>) need not be one-one.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We know that f<\/em>1<\/sub>: R<\/em> \u2192 R<\/em>, given by f<\/em>1<\/sub>(x) = x, <\/em>and <\/em>f<\/em>2<\/sub>(x) = x<\/em> are one-one.
Proving f<\/em>1<\/sub> <\/sub>is one-one:
Let x<\/em> and y<\/em> be two elements in the domain R,<\/em> such that
f<\/em>1<\/sub>(x) = <\/em>f<\/em>1<\/sub>(y)
\u21d2x = y<\/em>e<\/em>t<\/em> f<\/em>1(x<\/em>)=f<\/em>1(y<\/em>)x<\/em>=y<\/em>
So, f1<\/sub> <\/sub><\/em>is one-one.

Proving f<\/em>2<\/sub> <\/sub>is one-one:
Let x<\/em> and y<\/em> be two elements in the domain R,<\/em> such that
f<\/em>2<\/sub>(x) = <\/em>f<\/em>2<\/sub>(y)
\u21d2x = y<\/em>e<\/em>t<\/em> f<\/em>1(x<\/em>)=f<\/em>1(y<\/em>)x<\/em>=y<\/em>
So, f<\/em>2<\/sub> <\/sub><\/em>is one-one.

Proving f<\/em>1 \u00d7 f<\/em>2 is not one-one:
Given:
(f<\/em>1 \u00d7 f<\/em>2)(x<\/em>)=f<\/em>1 (x<\/em>) \u00d7 f<\/em>2 (x<\/em>)=x<\/em> \u00d7 x<\/em>=x<\/em>2Let x<\/em> and y<\/em> be two elements in the domain R<\/em>, such that(f<\/em>1 \u00d7 f<\/em>2)(x<\/em>)=(f<\/em>1 \u00d7 f<\/em>2)(y<\/em>)\u21d2x<\/em>2 = y<\/em>2\u21d2x<\/em>=\u00b1 y<\/em>So, (f<\/em>1 \u00d7 f<\/em>2) is not one-one.f<\/em>1\u00d7f<\/em>2<\/p>\n\n\n\n

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Question 20:<\/h4>\n\n\n\n

Suppose f<\/em>1<\/sub> and f<\/em>2<\/sub> are non-zero one-one functions from R<\/em> to R<\/em>. Is f<\/em>1f<\/em>2 necessarily one-one? Justify your answer. Here, f<\/em>1f<\/em>2:R<\/em>\u2192R<\/em> is given by (f<\/em>1f<\/em>2) (x<\/em>)=f<\/em>1 (x<\/em>)f<\/em>2 (x<\/em>) for all x<\/em>\u2208R<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We know that f<\/em>1<\/sub>: R<\/em> \u2192 R,<\/em> given by f<\/em>1<\/sub>(x)=x3<\/sup> <\/em>and <\/em>f<\/em>2<\/sub>(x)=x<\/em> are one-one.
Injectivity of f<\/em>1<\/sub>:
Let x<\/em> and y<\/em> be two elements in the domain R,<\/em> such that
f<\/em>1(x<\/em>)=f<\/em>2(y<\/em>)\u21d2x<\/em>3=y<\/em>\u21d2x<\/em>=3\u221ay<\/em>\u2208R<\/em>L<\/em>e<\/em>t<\/em> f<\/em>1(x<\/em>)=f<\/em>1(y<\/em>)x<\/em>=y<\/em>
So, f1<\/sub> <\/sub><\/em>is one-one.

Injectivity of f<\/em>2<\/sub>:
Let x<\/em> and y<\/em> be two elements in the domain R,<\/em> such that
f<\/em>2(x<\/em>)=f<\/em>2(y<\/em>)\u21d2x<\/em>=y<\/em> \u21d2x<\/em>\u2208R<\/em>.L<\/em>e<\/em>t<\/em> f<\/em>2(x<\/em>)=f<\/em>2(y<\/em>)\u2212x<\/em>=\u2212y<\/em>x<\/em>=y<\/em>
So, f2<\/sub>  <\/sub><\/em>is one-one.

Proving f<\/em>1f<\/em>2is not one-one:
Given that f<\/em>1f<\/em>2(x<\/em>)=f<\/em>1(x<\/em>)f<\/em>2(x<\/em>)=x<\/em>3x<\/em>=x<\/em>2
Let x<\/em> and y<\/em> be two elements in the domain R, such that

f<\/em>1f<\/em>2(x<\/em>)=f<\/em>1f<\/em>2(y<\/em>)\u21d2x<\/em>2=y<\/em>2\u21d2x<\/em>=\u00b1y<\/em>
So, f<\/em>1f<\/em>2 is not one-one.<\/p>\n\n\n\n

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Question 21:<\/h4>\n\n\n\n

Given A<\/em> = {2, 3, 4}, B<\/em> = {2, 5, 6, 7}. Construct an example of each of the following:

(i) an injective map from A<\/em> to B<\/em>
(ii) a mapping from A<\/em> to B<\/em> which is not injective
(iii) a mapping from A<\/em> to B<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(i) {(2, 7), (3, 6), (4, 5)}

(ii) {(2, 2), (3, 2), (4, 5)}

(iii) {(2, 5), (3, 6), (4, 7)}

Disclaimer: There are many more possibilities of each case.<\/p>\n\n\n\n

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Question 22:<\/h4>\n\n\n\n

Show that f<\/em> : R<\/em> \u2192 R<\/em>, given by f<\/em>(x<\/em>) = x<\/em> – [x<\/em>], is neither one-one nor onto.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have, f<\/em>(x<\/em>) = x<\/em> – [x<\/em>]

Injection test:

f<\/em>(x<\/em>) = 0 for all x<\/em> \u2208 Z<\/strong>

So, f<\/em> is a many-one function.

Surjection test:

Range (f<\/em>) = [0, 1) \u2260 R<\/strong>.

So, f<\/em> is an into function.

Therefore, f<\/em> is neither one-one nor onto.<\/p>\n\n\n\n

Page No 2.32:<\/h4>\n\n\n\n

Question 23:<\/h4>\n\n\n\n

Let f<\/em> : N<\/strong> \u2192 N<\/strong> be defined by

f<\/em>(n<\/em>)={n<\/em>+1, if n<\/em> is oddn<\/em>-1, if n<\/em> is even

Show that f<\/em> is a bijection.                                                                                                                                                   [CBSE 2012, NCERT]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,f<\/em>(n<\/em>)={n<\/em>+1, if n<\/em> is oddn<\/em>-1, if n<\/em> is evenInjection test:Case I: If n<\/em> is odd,Let x<\/em>, y<\/em>\u2208N<\/strong> <\/strong>such that f<\/em>(x<\/em>)=f<\/em>(y<\/em>)As, f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>+1=y<\/em>+1\u21d2x<\/em>=y<\/em>Case II: If n<\/em> is even,Let x<\/em>, y<\/em>\u2208N<\/strong> <\/strong>such that f<\/em>(x<\/em>)=f<\/em>(y<\/em>)As, f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>-1=y<\/em>-1\u21d2x<\/em>=y<\/em>So, f<\/em> is injective.Surjection test:Case I: If n<\/em> is odd,As, for every n<\/em>\u2208N<\/strong>, there exists y<\/em>=n<\/em>-1 in N<\/strong> such thatf(y<\/em>)=f<\/em>(n<\/em>-1)=n<\/em>-1+1=n<\/em>Case II: If n<\/em> is even,As, for every n<\/em>\u2208N<\/strong>, there exists y<\/em>=n<\/em>+1 in N<\/strong> such thatf(y<\/em>)=f<\/em>(n<\/em>+1)=n<\/em>+1-1=n<\/em>So, f<\/em> is surjective.So, f<\/em> is a bijection.<\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 1:<\/h4>\n\n\n\n

Find gof<\/em> and fog<\/em> when f<\/em> : R<\/em> \u2192 R<\/em> and g<\/em> : R<\/em> \u2192 R<\/em> are defined by
(i) f<\/em>(x<\/em>) = 2x<\/em> + 3                and       g<\/em>(x<\/em>) = x<\/em>2<\/sup> + 5
(ii) f<\/em>(x<\/em>) = 2x<\/em> + x<\/em>2<\/sup>             and       g<\/em>(x<\/em>) = x<\/em>3<\/sup>
(iii) f<\/em>(x<\/em>) = x<\/em>2<\/sup> + 8              and       g<\/em>(x<\/em>) = 3x<\/em>3<\/sup> + 1
(iv) f<\/em>(x<\/em>) = x<\/em>                       and       g<\/em>(x<\/em>) = |x<\/em>|
(v) f<\/em>(x<\/em>) = x<\/em>2<\/sup> + 2x<\/em> \u2212 3       and       g<\/em>(x<\/em>) = 3x<\/em> \u2212 4
(vi) f<\/em>(x<\/em>) = 8x<\/em>3<\/sup>                   and       g<\/em>(x<\/em>) = x<\/em>1<\/sup>\/3<\/sup><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given, f<\/em> : R<\/em> \u2192 R<\/em> and g<\/em> : R<\/em> \u2192 R<\/em>
So, gof <\/em>: R<\/em> \u2192 R<\/em>  and fog<\/em> : R<\/em> \u2192 R<\/em>

(i) f<\/em>(x<\/em>) = 2x<\/em> + 3  and g<\/em>(x<\/em>) = x<\/em>2<\/sup> + 5
Now, (gof) (x)
= g (f (x))
= g <\/em>(2x<\/em> +3)
= <\/em>(2x<\/em> + 3)2<\/sup> + 5
= <\/em>4x<\/em>2<\/sup>+ 9 + 12x<\/em> +5
=4x2<\/sup><\/em>+  12x<\/em> <\/em>+ 14

(fog) (x)
=f (g (x))
= f (x2<\/sup> <\/em>+ 5)
=<\/em> 2 (x2<\/sup> <\/em>+ 5) +3
= <\/em>2 x2<\/sup><\/em>+ 10 + 3
=<\/em> 2x2<\/sup><\/em> <\/em>+ 13

(ii) f<\/em>(x<\/em>) = 2x<\/em> + x<\/em>2<\/sup> and g<\/em>(x<\/em>) = x<\/em>3<\/sup>
(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (2x<\/em>+x<\/em>2)=(2x<\/em>+x<\/em>2)3(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (x<\/em>3)=2 (x<\/em>3)+(x<\/em>3)2=2x<\/em>3+x<\/em>6<\/sup>

(iii) f<\/em>(x<\/em>) = x<\/em>2<\/sup> + 8  and g<\/em>(x<\/em>) = 3x<\/em>3<\/sup> + 1
(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em>(x<\/em>))=g<\/em> (x<\/em>2+8)=3 (x<\/em>2+8)3+1(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (3x<\/em>3+1)=(3x<\/em>3+1)2+8=9x<\/em>6+6x<\/em>3+1+8=9x<\/em>6+6x<\/em>3+9

(iv) f<\/em>(x<\/em>) = x<\/em> and g<\/em>(x<\/em>) = |x<\/em>|
(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em>(x<\/em>))=g<\/em> (x<\/em>)=|x<\/em>|(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (|x<\/em>|)=|x<\/em>|

(v) f<\/em>(x<\/em>) = x<\/em>2<\/sup> + 2x<\/em> \u2212 3 and g<\/em>(x<\/em>) = 3x<\/em> \u2212 4
(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em>(x<\/em>))=g<\/em> (x<\/em>2+2x<\/em>-3)=3 (x<\/em>2+2x<\/em>-3)-4=3x<\/em>2+6x<\/em>-9-4=3x<\/em>2+6x<\/em>-13(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (3x<\/em>-4)=(3x<\/em>-4)2+2 (3x<\/em>-4)-3=9x<\/em>2+16-24x<\/em>+6x<\/em>-8-3=9x<\/em>2-18x<\/em>+5

(vi) f<\/em>(x<\/em>) = 8x<\/em>3<\/sup> and g<\/em>(x<\/em>) = x<\/em>1<\/sup>\/3<\/sup>
(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (8x<\/em>3)=(8x<\/em>3)13=[(2x<\/em>)3]13=2x<\/em>(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (x<\/em>13)=8 (x<\/em>13)3=8x<\/em><\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 2:<\/h4>\n\n\n\n

Let f<\/em> = {(3, 1), (9, 3), (12, 4)} and g<\/em> = {(1, 3), (3, 3) (4, 9) (5, 9)}. Show that gof<\/em> and fog<\/em> are both defined. Also, find fog<\/em> and gof<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em> = {(3, 1), (9, 3), (12, 4)} and g<\/em> = {(1, 3), (3, 3) (4, 9) (5, 9)}

f <\/em>: {3, 9, 12} \u2192 {1, 3,4} and g <\/em>: {1, 3, 4, 5} \u2192 {3, 9}

Co-domain of f<\/em> is a subset of the domain of g.<\/em>
So, gof<\/em> exists and gof <\/em>: {3, 9, 12} \u2192 {3, 9}
(g<\/em>o<\/em>f<\/em>) (3)=g<\/em> (f<\/em> (3))=g<\/em> (1)=3(g<\/em>o<\/em>f<\/em>) (9)=g<\/em> (f<\/em> (9))=g<\/em> (3)=3(g<\/em>o<\/em>f<\/em>) (12)=g<\/em> (f<\/em> (12))=g<\/em> (4)=9\u21d2g<\/em>o<\/em>f<\/em> ={(3, 3), (9, 3), (12, 9)}
Co-domain of g<\/em> is a subset of the domain of f.<\/em>
So, fog<\/em> exists and fog <\/em>: {1, 3, 4, 5} \u2192 {3, 9, 12}
(f<\/em>o<\/em>g<\/em>) (1)=f<\/em> (g<\/em> (1))=f<\/em> (3)=1(f<\/em>o<\/em>g<\/em>) (3)=f<\/em> (g<\/em> (3))=f<\/em> (3)=1(f<\/em>o<\/em>g<\/em>) (4)=f<\/em> (g<\/em> (4))=f<\/em> (9)=3(f<\/em>o<\/em>g<\/em>) (5)=f<\/em> (g<\/em> (5))=f<\/em> (9)=3\u21d2f<\/em>o<\/em>g<\/em>={(1, 1), (3, 1), (4, 3), (5, 3)}<\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 3:<\/h4>\n\n\n\n

Let f<\/em> = {(1, \u22121), (4, \u22122), (9, \u22123), (16, 4)} and g<\/em> = {(\u22121, \u22122), (\u22122, \u22124), (\u22123, \u22126), (4, 8)}. Show that gof<\/em> is defined while fog<\/em> is not defined. Also, find gof<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em> = {(1, \u22121), (4, \u22122), (9, \u22123), (16, 4)} and g<\/em> = {(\u22121, \u22122), (\u22122, \u22124), (\u22123, \u22126), (4, 8)}
f <\/em>: {1, 4, 9, 16} \u2192 {-1, -2, -3, 4} and g<\/em> : {-1, -2, -3, 4} \u2192 {-2, -4, -6, 8}
Co-domain of f<\/em> = domain of g<\/em>
So, gof <\/em>exists and gof <\/em>: {1, 4, 9, 16} \u2192 {-2, -4, -6, 8}
(g<\/em>o<\/em>f<\/em>) (1)=g<\/em> (f<\/em> (1))=g<\/em> (-1)=-2(g<\/em>o<\/em>f<\/em>) (4)=g<\/em> (f<\/em> (4))=g<\/em> (-2)=-4(g<\/em>o<\/em>f<\/em>) (9)=g<\/em> (f<\/em> (9))=g<\/em> (-3)=-6(g<\/em>o<\/em>f<\/em>) (16)=g<\/em> (f<\/em> (16))=g<\/em> (4)=8So, g<\/em>o<\/em>f<\/em>={(1, -2), (4, -4), (9, -6), (16, 8)}

But the co-domain of g<\/em> is not same as the domain of f<\/em>.
So, fog<\/em> does not exist.<\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 4:<\/h4>\n\n\n\n

Let A<\/em> = {a<\/em>, b<\/em>, c<\/em>}, B<\/em> = {u<\/em> v<\/em>, w<\/em>} and let f<\/em> and g<\/em> be two functions from A<\/em> to B<\/em> and from B<\/em> to A,<\/em> respectively, defined as :

    f<\/em> = {(a<\/em>, v<\/em>), (b<\/em>, u<\/em>), (c<\/em>, w<\/em>)}, g<\/em> = {(u<\/em>, b<\/em>), (v<\/em>, a<\/em>), (w<\/em>, c<\/em>)}.

Show that f<\/em> and g<\/em> both are bijections and find fog<\/em> and gof<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Proving f<\/em> is a bijection:
f<\/em> = {(a<\/em>, v<\/em>), (b<\/em>, u<\/em>), (c<\/em>, w<\/em>)} and f <\/em>: A \u2192 B<\/em>
Injectivity of f<\/em>: No two elements of A <\/em>have the same image in B.
So, f<\/em> is one-one.
Surjectivity of f<\/em>: Co-domain of f<\/em> = {u<\/em> v<\/em>, w<\/em>}
Range of f <\/em>= {u<\/em> v<\/em>, w<\/em>}
Both are same.
So,  f<\/em> is onto.
Hence, f<\/em> is a bijection.

Proving g <\/em>is a bijection:
g<\/em> = {(u<\/em>, b<\/em>), (v<\/em>, a<\/em>), (w<\/em>, c<\/em>)} and g <\/em>: B \u2192 A<\/em>
Injectivity of g<\/em>: No two elements of B  <\/em>have the same image in A.
So, g<\/em> is one-one.
Surjectivity of g: Co-domain of g<\/em> = {a<\/em>, b<\/em>, c<\/em>}
Range of g <\/em>= {a<\/em>, b<\/em>, c<\/em>}
Both are the same.
So, g<\/em> is onto.
Hence, g<\/em> is a bijection.

Finding  fog<\/em>:
Co-domain of <\/em>g<\/em> is same as the domain of f<\/em>.
So, fog <\/em>exists and fog <\/em>: {u<\/em> v<\/em>, w<\/em>} \u2192 <\/em>{u<\/em> v<\/em>, w<\/em>}
(f<\/em>o<\/em>g<\/em>) (u<\/em>)=f<\/em> (g<\/em> (u<\/em>))=f<\/em> (b<\/em>)=u<\/em>(f<\/em>o<\/em>g<\/em>) (v<\/em>)=f<\/em> (g<\/em> (v<\/em>))=f<\/em> (a<\/em>)=v<\/em>(f<\/em>o<\/em>g<\/em>) (w<\/em>)=f<\/em> (g<\/em> (w<\/em>))=f<\/em> (c<\/em>)=w<\/em>So, f<\/em>o<\/em>g<\/em> ={(u<\/em>, u<\/em>), (v<\/em>, v<\/em>), (w<\/em>, w<\/em>)}

Finding gof<\/em>:
Co-domain of <\/em>f<\/em> is same as the domain of g<\/em>.
So, fog <\/em>exists and gof <\/em>: {a<\/em>, b<\/em>, c<\/em>} \u2192 <\/em>{a<\/em>, b<\/em>, c<\/em>}
(g<\/em>o<\/em>f<\/em>) (a<\/em>)=g<\/em> (f<\/em> (a<\/em>))=g<\/em> (v<\/em>)=a<\/em>(g<\/em>o<\/em>f<\/em>) (b<\/em>)=g<\/em> (f<\/em> (b<\/em>))=g<\/em> (u<\/em>)=b<\/em>(g<\/em>o<\/em>f<\/em>) (c<\/em>)=g<\/em> (f<\/em> (c<\/em>))=g<\/em> (w<\/em>)=c<\/em>So, g<\/em>o<\/em>f<\/em>={(a<\/em>, a<\/em>), (b<\/em>, b<\/em>), (c<\/em>, c<\/em>)}<\/p>\n\n\n\n

\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 5:<\/h4>\n\n\n\n

Find  fog<\/em> (2) and gof<\/em> (1) when : f<\/em> : R<\/em> \u2192 R<\/em> ; f<\/em>(x<\/em>) = x<\/em>2<\/sup> + 8 and g<\/em> : R<\/em> \u2192 R<\/em>; g<\/em>(x<\/em>) = 3x<\/em>3<\/sup> + 1.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(f<\/em>o<\/em>g<\/em>) (2)=f<\/em> (g<\/em> (2))=f<\/em>(3\u00d723+1)=f<\/em>(25)=252+8=633(g<\/em>o<\/em>f<\/em>) (1)=g<\/em> (f<\/em> (1))=g<\/em> (12+8)=g<\/em> (9)=3\u00d793+1=2188<\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 6:<\/h4>\n\n\n\n

Let R<\/em>+<\/sup> be the set of all non-negative real numbers. If f<\/em> : R<\/em>+<\/sup> \u2192 R<\/em>+<\/sup> and g<\/em> : R<\/em>+<\/sup> \u2192 R<\/em>+<\/sup> are defined as f<\/em>(x<\/em>)=x<\/em>2 and g<\/em>(x<\/em>)=+\u221ax<\/em>, find fog<\/em> and gof<\/em>. Are they equal functions?<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given,  f<\/em> : R<\/em>+<\/sup> \u2192 R<\/em>+<\/sup> and g<\/em> : R<\/em>+<\/sup> \u2192 R<\/em>+<\/sup>
So,  fog <\/em>: R<\/em>+<\/sup> \u2192 R<\/em>+<\/sup>  and gof <\/em>: R<\/em>+<\/sup> \u2192 R<\/em>+<\/sup>
Domains of fog  <\/em>and gof  <\/em>are the same.
(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (\u221ax<\/em>)=(\u221ax<\/em>)2=x<\/em>(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (x<\/em>2)=\u221ax<\/em>2=x<\/em>So, (f<\/em>o<\/em>g<\/em>) (x<\/em>)=(g<\/em>o<\/em>f<\/em>) (x<\/em>),\u2200x<\/em>\u2208R<\/em>+
Hence, fog <\/em>= gof<\/em><\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 7:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em> \u2192 R<\/em> and g<\/em> : R<\/em> \u2192 R<\/em> be defined by f<\/em>(x<\/em>) = x<\/em>2<\/sup> and g<\/em>(x<\/em>) = x<\/em> + 1. Show that fog<\/em> \u2260 gof<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given,  f<\/em> : R<\/em> \u2192 R<\/em> and g<\/em> : R<\/em> \u2192 R.
So, the domains of f<\/em> and g<\/em> are the same.

(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (x<\/em>+1)=(x<\/em>+1)2=x<\/em>2+1+2x<\/em>(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (x<\/em>2)=x<\/em>2+1
So,  fog \u2260 gof<\/em><\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 8:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em> \u2192 R<\/em> and g<\/em> : R<\/em> \u2192 R<\/em> be defined by f<\/em>(x<\/em>) = x <\/em>+ 1 and g<\/em>(x<\/em>) = x<\/em> \u2212 1. Show that fog<\/em> = gof = IR<\/sub><\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given,  f<\/em> : R<\/em> \u2192 R<\/em> and g<\/em> : R<\/em> \u2192 R
\u21d2fog <\/em>:  R<\/em> \u2192 R and gof <\/em>: R<\/em> \u2192 R (Also, we know that IR<\/sub> : <\/em>R<\/em> \u2192 R)
So, the domains of all fog, gof <\/em>and IR<\/sub> <\/sub><\/em>are the same.
(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (x<\/em>-1)=x<\/em>-1+1=x<\/em>=I<\/em>R<\/em> (x<\/em>)      … (1)(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (x<\/em>+1)=x<\/em>+1-1=x<\/em>=I<\/em>R<\/em> (x<\/em>)      … (2)From (1) and (2), (f<\/em>o<\/em>g<\/em>) (x<\/em>)=(g<\/em>o<\/em>f<\/em>) (x<\/em>)=I<\/em>R<\/em> (x<\/em>), \u2200x<\/em>\u2208R<\/em>Hence, f<\/em>o<\/em>g<\/em>=g<\/em>o<\/em>f<\/em>=I<\/em>R<\/em><\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 9:<\/h4>\n\n\n\n

Verify associativity for the following three mappings : f<\/em> : N<\/em> \u2192 Z0<\/sub> (the set of non-zero integers), g<\/em> : Z0<\/sub> \u2192 Q<\/em> and h<\/em> : Q<\/em> \u2192 R<\/em> given by f<\/em>(x<\/em>) = 2x<\/em>, g<\/em>(x<\/em>) = 1\/x<\/em> and h<\/em>(x<\/em>) = ex<\/sup><\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given that f<\/em> : N<\/em> \u2192 Z0<\/sub> , g<\/em> : Z0<\/sub> \u2192 Q<\/em> and h<\/em> : Q<\/em> \u2192 R<\/em> .
gof <\/em>: N <\/em>\u2192 Q<\/em>  <\/sub>and hog <\/em>: Z0<\/sub> \u2192 R<\/em>
\u21d2h o (gof ) <\/em>: N <\/em>\u2192 R <\/em>and (hog<\/em>) o f<\/em>: N <\/em>\u2192 R
So, both have the same domains.
(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (2x<\/em>)=12x<\/em>         …(1)(h<\/em>o<\/em>g<\/em>) (x<\/em>)=h<\/em> (g<\/em> (x<\/em>))=h<\/em> (1x<\/em>)=e<\/em>1x<\/em>       …(2)Now,(h<\/em> o<\/em>(g<\/em>o<\/em>f<\/em>)) (x<\/em>)=h<\/em>((g<\/em>o<\/em>f<\/em>) (x<\/em>))=h<\/em> (12x<\/em>)=e<\/em>12x<\/em>           [from (1)]((h<\/em>o<\/em>g<\/em>) o<\/em> f<\/em>)(x<\/em>)=(h<\/em>o<\/em>g<\/em>) (f<\/em> (x<\/em>))= (h<\/em>o<\/em>g<\/em>) (2x<\/em>)=e<\/em>12x<\/em>    [from (2)]\u21d2(h<\/em> o<\/em>(g<\/em>o<\/em>f<\/em>)) (x<\/em>)=((h<\/em>o<\/em>g<\/em>) o<\/em> f<\/em>)(x<\/em>), \u2200x<\/em>\u2208N<\/em>So, h<\/em> o<\/em>(g<\/em>o<\/em>f<\/em>)=(h<\/em>o<\/em>g<\/em>) o<\/em> f<\/em>
Hence, the associative property has been verified.<\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 10:<\/h4>\n\n\n\n

Consider f<\/em> : N<\/em> \u2192 N<\/em>, g<\/em> : N<\/em> \u2192 N<\/em> and h<\/em> : N<\/em> \u2192 R<\/em> defined as f<\/em>(x<\/em>) = 2x<\/em>, g<\/em>(y<\/em>) = 3y<\/em> + 4 and h<\/em>(z<\/em>) = sin z<\/em> for all x<\/em>, y<\/em>, z<\/em> \u2208 N<\/em>. Show that ho<\/em> (gof<\/em>) = (hog<\/em>) of<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given, f<\/em> : N<\/em> \u2192 N<\/em>, g<\/em> : N<\/em> \u2192 N<\/em> and h<\/em> : N<\/em> \u2192 R<\/em>
\u21d2gof <\/em>: N<\/em> \u2192 N <\/em>and hog <\/em>: N<\/em> \u2192 R<\/em>
\u21d2ho<\/em> (gof<\/em>) : N<\/em> \u2192 R<\/em> and (hog<\/em>) of <\/em>: N<\/em> \u2192 R<\/em>
So, both have the same domains.
(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (2x<\/em>)=3 (2x<\/em>)+4=6x<\/em>+4  …(1)(h<\/em>o<\/em>g<\/em>) (x<\/em>)=h<\/em>(g<\/em> (x<\/em>))=h<\/em> (3x<\/em>+4)=sin (3x<\/em>+4)        … (2)Now,(h<\/em> o<\/em> (g<\/em>o<\/em>f<\/em>)) (x<\/em>)=h<\/em> ((g<\/em>o<\/em>f<\/em>) (x<\/em>))=h<\/em>(6x<\/em>+4) = sin (6x<\/em>+4)    [from (1)]((h<\/em>o<\/em>g<\/em>) o<\/em> f<\/em>) (x<\/em>)=(h<\/em>o<\/em>g<\/em>) (f<\/em> (x<\/em>))=(h<\/em>o<\/em>g<\/em>) (2x<\/em>)=sin (6x<\/em>+4)    [from (2)]So, (h<\/em> o<\/em> (g<\/em>o<\/em>f<\/em>)) (x<\/em>)=((h<\/em>o<\/em>g<\/em>) o<\/em> f<\/em>) (x<\/em>), \u2200x<\/em>\u2208N<\/em>Hence, h<\/em> o<\/em> (g<\/em>o<\/em>f<\/em>)=(h<\/em>o<\/em>g<\/em>) o<\/em> f<\/em><\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 11:<\/h4>\n\n\n\n

Give examples of two functions f<\/em> : N<\/em> \u2192 N<\/em> and g<\/em> : N<\/em> \u2192 N<\/em>, such that gof<\/em> is onto but f<\/em> is not onto.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let us consider a function f<\/em> : N<\/em> \u2192 N <\/em>given by f(x) = x +1 ,<\/em> which is not onto.
[This not onto because if we take 0 in N<\/em> (co-domain), then,
0=x<\/em>+1
\u21d2x<\/em>=-1\u2209N<\/em>]

Let us consider g<\/em> : N<\/em> \u2192 N  <\/em>given by
g<\/em> (x<\/em>)={x<\/em>-1, if x<\/em>>11, if x<\/em>=1Now, let us find (g<\/em>o<\/em>f<\/em>) (x<\/em>)Case 1: x<\/em>>1(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (x<\/em>+1)=x<\/em>+1-1=x<\/em>Case 2: x<\/em>=1(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (x<\/em>+1)=1From case-1 and case-2, (g<\/em>o<\/em>f<\/em>) (x<\/em>)=x<\/em>, \u2200x<\/em>\u2208N<\/em>, which is an identity function and, hence, it is onto.<\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 12:<\/h4>\n\n\n\n

Give examples of two functions f<\/em> : N<\/em> \u2192 Z<\/em> and g<\/em> : Z<\/em> \u2192 Z, such that gof<\/em> is injective but g<\/em> is not injective.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em> : N<\/em> \u2192 Z <\/em>be given by f (x) = x, <\/em>which is injective.
(If we take f(x) = f(y)<\/em>, then it gives x = y<\/em>)

Let g<\/em> : Z \u2192 Z<\/em> be given by g (x) = |x|, <\/em>which is not injective.
If we take f(x) = f(y), <\/em>we get:
|x<\/em>| = |y<\/em>|
\u21d2x<\/em> = \u00b1 y<\/em>

Now, gof <\/em>: N<\/em> \u2192 Z.
(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (x<\/em>)=|x<\/em>|
Let us take two elements x<\/em> and y<\/em> in the domain of gof , <\/em>such that
(g<\/em>o<\/em>f<\/em>) (x<\/em>)=(g<\/em>o<\/em>f<\/em>) (y<\/em>)\u21d2|x<\/em>|=|y<\/em>|\u21d2x<\/em>=y<\/em> (We don’t get \u00b1 here because x, y \u2208N)
So, gof <\/em>is injective.
 <\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 13:<\/h4>\n\n\n\n

If f<\/em> : A<\/em> \u2192 B<\/em> and g<\/em> : B<\/em> \u2192 C<\/em> are one-one functions, show that gof<\/em> is a one-one function.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given,  f<\/em> : A<\/em> \u2192 B<\/em> and g<\/em> : B<\/em> \u2192 C are one – one.
Then, gof : <\/em>A<\/em> \u2192 B<\/em>
Let us take two elements x<\/em> and y<\/em> from A,<\/em> such that
(g<\/em>o<\/em>f<\/em>) (x<\/em>)=(g<\/em>o<\/em>f<\/em>) (y<\/em>)\u21d2g<\/em> (f<\/em> (x<\/em>))=g<\/em> (f<\/em> (y<\/em>))\u21d2f<\/em> (x<\/em>)=f<\/em> (y<\/em>) (As, g<\/em> is one-one)\u21d2x<\/em>=y<\/em> (As, f<\/em> is one-one)
Hence, gof <\/em>is one-one.<\/p>\n\n\n\n

Page No 2.46:<\/h4>\n\n\n\n

Question 14:<\/h4>\n\n\n\n

If f<\/em> : A<\/em> \u2192 B<\/em> and g<\/em> : B<\/em> \u2192 C<\/em> are onto functions, show that gof<\/em> is a onto function.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given,  f<\/em> : A<\/em> \u2192 B<\/em> and g<\/em> : B<\/em> \u2192 C are onto.
Then, gof : <\/em>A<\/em> \u2192 C<\/em>
Let us take an element z<\/em> in the co-domain (C<\/em>).
Now, z<\/em> is in C<\/em> and g<\/em> : B<\/em> \u2192 C is onto.
So, there exists some element y<\/em> in B,<\/em> such that g (y) = z<\/em> … (1)
Now, y<\/em> is in B<\/em> and f<\/em> : A<\/em> \u2192 B <\/em>is onto.
So, there exists some x<\/em> in A<\/em>, such that f (x)<\/em> = y<\/em> … (2)
From (1) and (2),
z = g (y) = g (f (x)) = (gof) (x)<\/em>
So, z = (gof) (x)<\/em>, where x<\/em> is in A<\/em>.
Hence, gof <\/em>is onto.<\/p>\n\n\n\n

Page No 2.54:<\/h4>\n\n\n\n

Question 1:<\/h4>\n\n\n\n

Find fog<\/em> and gof<\/em>  if
(i) f<\/em>(x<\/em>)=e<\/em>x<\/em>, g<\/em>(x<\/em>)=loge<\/em> x<\/em>
(ii) f<\/em>(x<\/em>)=x<\/em>2, g<\/em>(x<\/em>)=cos x<\/em>
(iii) f<\/em>(x<\/em>)=|x<\/em>|, g<\/em> (x<\/em>)=sin x<\/em>
(iv) f<\/em>(x<\/em>)=x<\/em>+1, g<\/em>(x<\/em>)=e<\/em>x<\/em>
(v) f<\/em>(x<\/em>)=sin-1 x<\/em>, g<\/em>(x<\/em>)=x<\/em>2
(vi) f<\/em>(x<\/em>)=x<\/em>+1, g<\/em>(x<\/em>)=sin x<\/em>
(vii) f<\/em>(x<\/em>)=x<\/em>+1, g<\/em>(x<\/em>)=2x<\/em>+3
(viii) f<\/em>(x<\/em>)=c<\/em>, c<\/em> \u2208 R<\/em>, g<\/em>(x<\/em>)=sin x<\/em>2
(ix) f<\/em>(x<\/em>)=x<\/em>2+2, g<\/em>(x<\/em>)=1-11-x<\/em><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(i<\/em>) f<\/em> (x<\/em>)=e<\/em>x<\/em>, g<\/em>(x<\/em>)=loge<\/em> x<\/em>f<\/em>:R<\/em>\u2192(0,\u221e); g<\/em>:(0,\u221e)\u2192R<\/em>Computing f<\/em>o<\/em>g<\/em>:Clearly, the range of g<\/em> is a subset of the domain of f<\/em>.f<\/em>o<\/em>g<\/em> : (0,\u221e)\u2192R<\/em>(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (loge<\/em> x<\/em>)=loge<\/em> e<\/em>x<\/em>=x<\/em>Computing g<\/em>o<\/em>f<\/em>:Clearly, the range of f<\/em> is a subset of the domain of g<\/em>.\u21d2f<\/em>o<\/em>g<\/em> : R<\/em>\u2192R<\/em>(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (e<\/em>x<\/em>)=loge<\/em> e<\/em>x<\/em>=x<\/em>

(i<\/em>i<\/em>) f<\/em> (x<\/em>)=x<\/em>2, g<\/em>(x<\/em>)=cos x<\/em>f<\/em>:R<\/em>\u2192[0, \u221e) ; g<\/em>:R<\/em>\u2192[-1, 1]Computing fog:Clearly, the range of g<\/em> is not a subset of the domain of f<\/em>.\u21d2Domain (f<\/em>o<\/em>g<\/em>)={x<\/em>: x<\/em>\u2208domain of g<\/em> and g<\/em>(x<\/em>)\u2208domain of f<\/em>}\u21d2Domain (f<\/em>o<\/em>g<\/em>)=x<\/em>: x<\/em>\u2208R<\/em> and cos x<\/em> \u2208R<\/em>}\u21d2Domain of (f<\/em>o<\/em>g<\/em>)=R<\/em>f<\/em>o<\/em>g<\/em>: R<\/em>\u2192R<\/em>(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (cos x<\/em>)=cos2x<\/em>Computing g<\/em>o<\/em>f<\/em>:Clearly, the range of f<\/em> is a subset of the domain of g<\/em>.\u21d2f<\/em>o<\/em>g<\/em> : R<\/em>\u2192R<\/em>(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (x<\/em>2)=cos (x<\/em>2)


(i<\/em>i<\/em>i<\/em>) f<\/em> (x<\/em>)=|x<\/em>|, g<\/em>(x<\/em>)=sin x<\/em>f<\/em>:R<\/em>\u2192(0, \u221e); g<\/em>:R<\/em>\u2192[-1, 1]Computing f<\/em>o<\/em>g<\/em>:Clearly, the range of g<\/em> is a subset of the domain of f<\/em>.\u21d2f<\/em>o<\/em>g<\/em> : R<\/em>\u2192R<\/em>(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (sin x<\/em>)=|sin x<\/em>|Computing g<\/em>o<\/em>f<\/em>:Clearly, the range of f<\/em> is a subset of the domain of g<\/em>.\u21d2f<\/em>o<\/em>g<\/em> : R<\/em>\u2192R<\/em>(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (|x<\/em>|)=sin |x<\/em>|

(i<\/em>v<\/em>) f<\/em> (x<\/em>)=x<\/em>+1, g<\/em>(x<\/em>)=e<\/em>x<\/em>f<\/em>:R<\/em>\u2192R<\/em>; g<\/em>:R<\/em>\u2192[1, \u221e)Computing f<\/em>o<\/em>g<\/em>:Clearly, range of g<\/em> is a subset of domain of f<\/em>.\u21d2f<\/em>o<\/em>g<\/em> : R<\/em>\u2192R<\/em>(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (e<\/em>x<\/em>)=e<\/em>x<\/em>+1Computing g<\/em>o<\/em>f<\/em>:Clearly, range of f<\/em> is a subset of domain of g<\/em>.\u21d2f<\/em>o<\/em>g<\/em> : R<\/em>\u2192R<\/em>(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (x<\/em>+1)=e<\/em>x<\/em>+1

(v<\/em>) f<\/em> (x<\/em>)=sin-1x<\/em>, g<\/em>(x<\/em>)=x<\/em>2f<\/em>:[-1,1]\u2192[-\u03c02,\u03c02] ; g<\/em>:R<\/em>\u2192[0, \u221e)Computing <\/strong>f<\/em><\/strong>o<\/em><\/strong>g<\/em><\/strong>:<\/strong>Clearly, the range of g<\/em> is not a subset of the domain of f<\/em>.Domain (f<\/em>o<\/em>g<\/em>)={x<\/em>: x<\/em>\u2208domain of g<\/em> and g<\/em>(x<\/em>)\u2208domain of f<\/em>}Domain (f<\/em>o<\/em>g<\/em>)={x<\/em>: x<\/em>\u2208R<\/em> and x<\/em>2\u2208[-1,1]}Domain (f<\/em>o<\/em>g<\/em>)={x<\/em>: x<\/em>\u2208R<\/em> and x<\/em>\u2208[-1,1]}Domain of (f<\/em>o<\/em>g<\/em>)=[-1,1]f<\/em>o<\/em>g<\/em>: [-1,1]\u2192R<\/em>(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (x<\/em>2)=sin-1 (x<\/em>2)Computing <\/strong>g<\/em><\/strong>o<\/em><\/strong>f<\/em><\/strong>:<\/strong>Clearly, the range of f<\/em> is a subset of the domain of g<\/em>.f<\/em>o<\/em>g<\/em> : [-1,1]\u2192R<\/em>(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (sin-1x<\/em>)=(sin-1 x<\/em>)2

(vi) f<\/em>(x<\/em>)=x<\/em>+1, g<\/em>(x<\/em>)=sin x<\/em>f<\/em>:R<\/em>\u2192R<\/em> ; g<\/em>:R<\/em>\u2192[-1, 1]Computing f<\/em>o<\/em>g<\/em>:Clearly, the range of g<\/em> is a subset of the domain of f<\/em>.\u21d2f<\/em>o<\/em>g<\/em>: R<\/em>\u2192R<\/em>(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (sin x<\/em>)=sin x<\/em>+1Computing g<\/em>o<\/em>f<\/em>:Clearly, the range of f<\/em> is a subset of the domain of g<\/em>.\u21d2f<\/em>o<\/em>g<\/em> : R<\/em>\u2192R<\/em>(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (x<\/em>+1)=sin (x<\/em>+1)

(v<\/em>i<\/em>i<\/em>) f<\/em> (x<\/em>)=x<\/em>+1, g<\/em>(x<\/em>)=2x<\/em>+3f<\/em>:R<\/em>\u2192R<\/em> ; g<\/em>:R<\/em>\u2192R<\/em>Computing f<\/em>o<\/em>g<\/em>:Clearly, the range of g<\/em> is a subset of the domain of f<\/em>.\u21d2f<\/em>o<\/em>g<\/em>: R<\/em>\u2192R<\/em>(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (2x<\/em>+3)=2x<\/em>+3+1=2x<\/em>+4Computing g<\/em>o<\/em>f<\/em>:Clearly, the range of f<\/em> is a subset of the domain of g<\/em>.\u21d2f<\/em>o<\/em>g<\/em> : R<\/em>\u2192R<\/em>(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (x<\/em>+1)=2 (x<\/em>+1)+3=2x<\/em>+5

(v<\/em>i<\/em>i<\/em>i<\/em>) f<\/em> (x<\/em>)=c<\/em>, g<\/em>(x<\/em>)=sin x<\/em>2f<\/em>:R<\/em>\u2192{c<\/em>} ; g<\/em>:R<\/em>\u2192[0, 1]Computing f<\/em>o<\/em>g<\/em>:Clearly, the range of g<\/em> is a subset of the domain of f<\/em>.f<\/em>o<\/em>g<\/em>: R<\/em>\u2192R<\/em>(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (sin x<\/em>2)=c<\/em>Computing g<\/em>o<\/em>f<\/em>:Clearly, the range of f<\/em> is a subset of the domain of g<\/em>.\u21d2f<\/em>o<\/em>g<\/em> : R<\/em>\u2192R<\/em>(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (c<\/em>)=sin c<\/em>2

(i<\/em>x<\/em>) f<\/em>(x<\/em>)=x<\/em>2+2f<\/em>:R<\/em>\u2192[2,\u221e)  g<\/em>(x<\/em>)=1-11-x<\/em>For domain of g<\/em>: 1-x<\/em>\u22600 \u21d2x<\/em>\u22601\u21d2Domain of g=R<\/em>-{1}g<\/em>(x<\/em>)=1-11-x<\/em>=1-x<\/em>-11-x<\/em>=-x<\/em>1-x<\/em>For range of g<\/em>:y<\/em>=-x<\/em>1-x<\/em>\u21d2y<\/em>–x<\/em>y<\/em>=-x<\/em>\u21d2y<\/em>=x<\/em>y<\/em>–x<\/em>\u21d2y<\/em>=x<\/em>(y<\/em>-1)\u21d2x<\/em>=y<\/em>y<\/em>-1Range of g =R<\/em>-{1}So, g: R<\/em>-{1}\u2192R<\/em>-{1}Computing f<\/em>o<\/em>g<\/em>:Clearly, the range of g<\/em> is a subset of the domain of f<\/em>.\u21d2f<\/em>o<\/em>g<\/em>: R<\/em>-{1}\u2192R<\/em>(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (-x<\/em>x<\/em>-1)=(-x<\/em>x<\/em>-1)2+2=x<\/em>2+2x<\/em>2+2-4x<\/em>(1-x<\/em>)2=3x<\/em>2-4x<\/em>+2(1-x<\/em>)2Computing g<\/em>o<\/em>f<\/em>:Clearly, the range of f<\/em> is a subset of the domain of g<\/em>.\u21d2g<\/em>o<\/em>f<\/em> : R<\/em>\u2192R<\/em>(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (x<\/em>2+2)=1-11-(x<\/em>2+2)=1-1-(x<\/em>2+1)=x<\/em>2+2x<\/em>2+1<\/p>\n\n\n\n

Page No 2.54:<\/h4>\n\n\n\n

Question 2:<\/h4>\n\n\n\n

Let f<\/em>(x<\/em>) = x<\/em>2<\/sup> + x<\/em> + 1 and g<\/em>(x<\/em>) = sin x<\/em>. Show that fog<\/em> \u2260 gof<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em>(sin x<\/em>)=sin2x<\/em>+sin x<\/em>+1and (g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (x<\/em>2+x<\/em>+1)= sin (x<\/em>2+x<\/em>+1)So, f<\/em>o<\/em>g<\/em>\u2260g<\/em>o<\/em>f<\/em>.<\/p>\n\n\n\n

Page No 2.54:<\/h4>\n\n\n\n

Question 3:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>) = |x<\/em>|, prove that fof<\/em> = f<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Domains of  f<\/em> and fof<\/em> are same as R<\/em>.
(f<\/em>o<\/em>f<\/em>) (x<\/em>)=f<\/em> (f<\/em> (x<\/em>))=f<\/em> (|x<\/em>|)=| |x<\/em>| |=|x<\/em>|=f<\/em> (x<\/em>)So,(f<\/em>o<\/em>f<\/em>) (x<\/em>)=f<\/em> (x<\/em>), \u2200x<\/em>\u2208R<\/em>Hence, f<\/em>o<\/em>f<\/em>=f<\/em><\/p>\n\n\n\n

Page No 2.54:<\/h4>\n\n\n\n

Question 4:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>) = 2x<\/em> + 5 and g<\/em>(x<\/em>) = x<\/em>2<\/sup> + 1 be two real functions, then describe each of the following functions:
(i) fog<\/em>
(ii) gof<\/em>
(iii) fof<\/em>
(iv) f<\/em>2<\/sup>

Also, show that fof <\/em>\u2260 f<\/em>2<\/sup><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f(x) <\/em>and g(x) <\/em>are polynomials.
\u21d2f <\/em>: R <\/em>\u2192 R <\/em>and g <\/em>: R <\/em>\u2192 R<\/em>.
So, fog <\/em>: R <\/em>\u2192 R  <\/em>and gof <\/em>: R <\/em>\u2192 R.
(i) (fog) (x)=f (g (x))=f (x2+1)=2 (x2+1)+5=2×2+2+5=2×2+7

(ii) (gof) (x)=g (f (x))=g (2x+5)=(2x+5)2+1=4×2+20x+26

(iii) (fof) (x)=f (f (x))=f (2x+5)=2 (2x+5)+5=4x+10+5=4x+15

(iv) f2 (x)=f(x)\u00d7f(x)=(2x+5)(2x+5)=(2x+5)2=4×2+20x+25<\/em>


\u2192\u2192  \u2192<\/p>\n\n\n\n

Page No 2.54:<\/h4>\n\n\n\n

Question 5:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>) = sin x<\/em> and g<\/em>(x<\/em>) = 2x<\/em> be two real functions, then describe gof <\/em>and fog. <\/em>Are these equal functions?
 <\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We know thatf<\/em>:R<\/em>\u2192[-1, 1] and g<\/em>: R<\/em>\u2192R<\/em>Clearly, the range of f<\/em> is a subset of the domain of g<\/em>.g<\/em>o<\/em>f<\/em>:R<\/em>\u2192R<\/em>(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (sin x<\/em>)=2sin x<\/em>

Clearly, the range of g<\/em> is a subset of the domain of f<\/em>.f<\/em>o<\/em>g<\/em>:R<\/em>\u2192R<\/em>So, (f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (2x<\/em>)=sin (2x<\/em>)

Clearly, fog\u2260gof<\/em><\/p>\n\n\n\n

Page No 2.54:<\/h4>\n\n\n\n

Question 6:<\/h4>\n\n\n\n

Let f<\/em>, g<\/em>, h<\/em> be real functions given by f<\/em>(x<\/em>) = sin x<\/em>, g<\/em> (x<\/em>) = 2x<\/em> and h<\/em> (x<\/em>) = cos x<\/em>. Prove that fog<\/em> = go<\/em> (fh<\/em>).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We know that f<\/em>:R<\/em>\u2192[-1, 1] and g<\/em>:R<\/em>\u2192R<\/em>Clearly, the range of g<\/em> is a subset of the domain of f<\/em>.f<\/em>o<\/em>g<\/em>:R<\/em>\u2192R<\/em>Now, (f<\/em>h<\/em>) (x<\/em>)=f<\/em>(x<\/em>)h<\/em>(x<\/em>)=(sin x<\/em>) (cos x<\/em>)=12 sin (2x<\/em>)Domain of f<\/em>h<\/em> is R<\/em>.Since range of sin x<\/em> is [-1,1],-1\u2264sin 2x<\/em>\u22641\u21d2-12\u2264sin x<\/em>2\u226412Range of f<\/em>h<\/em>  =[-12, 12]So, (f<\/em>h<\/em>):R<\/em>\u2192[-12, 12]Clearly, range of f<\/em>h<\/em> is a subset of g<\/em>.\u21d2g<\/em>o<\/em>(f<\/em>h<\/em>):R<\/em>\u2192R<\/em>\u21d2domains of f<\/em>o<\/em>g<\/em> and g<\/em>o<\/em>(f<\/em>h<\/em>) are the same.So, (f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (2x<\/em>)=sin (2x<\/em>)and (g<\/em>o<\/em>(f<\/em>h<\/em>))(x<\/em>)= g<\/em> ((f<\/em>h<\/em>) (x<\/em>))=g<\/em> (sinx<\/em> cos x<\/em>)=2sin x<\/em> cos x<\/em>=sin (2x<\/em>)\u21d2(f<\/em>o<\/em>g<\/em>) (x<\/em>)=( g<\/em>o<\/em>(f<\/em>h<\/em>))(x<\/em>), \u2200x<\/em>\u2208R<\/em>Hence, f<\/em>o<\/em>g<\/em> = g<\/em>o<\/em>(f<\/em>h<\/em>)<\/p>\n\n\n\n

Page No 2.54:<\/h4>\n\n\n\n

Question 7:<\/h4>\n\n\n\n

Let  f<\/em>  be any real function and let g<\/em> be a function given by g<\/em>(x<\/em>) = 2x<\/em>. Prove that gof<\/em> = f<\/em> + f<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given, f<\/em>:R<\/em>\u2192R<\/em>Since g(x<\/em>)=2x<\/em> is a polynomial, g<\/em>:R<\/em>\u2192R<\/em>Clearly, g<\/em>o<\/em>f<\/em>:R<\/em>\u2192R<\/em> a<\/em>n<\/em>d<\/em> f<\/em>+f<\/em>:R<\/em>\u2192R<\/em>So, domains of gof and f+f are the same.(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=2 f<\/em>(x<\/em>)(f<\/em>+f<\/em>) (x<\/em>)=f<\/em>(x<\/em>)+f<\/em>(x<\/em>)=2 f<\/em>(x<\/em>)\u21d2(g<\/em>o<\/em>f<\/em>) (x<\/em>)=(f<\/em>+f<\/em>) (x<\/em>), \u2200x<\/em>\u2208R<\/em>Hence, g<\/em>o<\/em>f<\/em> =f<\/em>+f<\/em><\/p>\n\n\n\n

Page No 2.54:<\/h4>\n\n\n\n

Question 8:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>)=\u221a1-x<\/em> and g<\/em>(x<\/em>)=loge<\/em> x<\/em> are two real functions, then describe functions fog<\/em> and gof<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

 f<\/em>(x<\/em>)=\u221a1-x<\/em>For domain, 1-x\u22650\u21d2x<\/em>\u22641\u21d2domain of f<\/em> =(-\u221e, 1]\u21d2f<\/em>:(-\u221e, 1]\u2192(0,\u221e) g<\/em>(x<\/em>)=loge<\/em> x<\/em>Clearly, g<\/em> : (0, \u221e)\u2192R<\/em>Computation of f<\/em>o<\/em>g<\/em>:Clearly, the range of g<\/em> is not a subset of the domain of f<\/em>.So,we need to compute the domain of f<\/em>o<\/em>g<\/em>.\u21d2Domain (f<\/em>o<\/em>g<\/em>)={x<\/em> : x<\/em>\u2208Domain (g<\/em>) and g<\/em>(x<\/em>)\u2208Domain of f}\u21d2Domain (f<\/em>o<\/em>g<\/em>)={x<\/em>: x<\/em>\u2208(0, \u221e) and loge<\/em> x \u2208 (-\u221e, 1]}\u21d2Domain (f<\/em>o<\/em>g<\/em>)={x<\/em>:x<\/em>\u2208(0, \u221e) and x<\/em>\u2208 (0, e<\/em>]}\u21d2Domain (f<\/em>o<\/em>g<\/em>)={x<\/em>: x<\/em> \u2208(0, e<\/em>]}\u21d2Domain (f<\/em>o<\/em>g<\/em>)=(0, e]\u21d2f<\/em>o<\/em>g<\/em>: (0, e<\/em>)\u2192R<\/em>So, (f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em> (g<\/em> (x<\/em>))=f<\/em> (loge<\/em> x<\/em>)=\u221a1-loge<\/em> x<\/em> Computation of g<\/em>o<\/em>f<\/em>:Clearly, the range of f<\/em> is  a subset of the domain of g<\/em>.\u21d2g<\/em>o<\/em>f<\/em>:(-\u221e,1]\u2192R<\/em>\u21d2(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (\u221a1-x<\/em>)=loge<\/em>\u221a1-x<\/em>=loge<\/em> (1-x<\/em>)12=12loge<\/em> (1-x<\/em>)<\/p>\n\n\n\n

Page No 2.54:<\/h4>\n\n\n\n

Question 9:<\/h4>\n\n\n\n

If f<\/em>:(-\u03c0<\/em>2,\u03c0<\/em>2)\u2192R<\/em> and g<\/em>:[-1, 1]\u2192R<\/em> be defined as
f<\/em>(x<\/em>)=tan x<\/em> and g<\/em>(x<\/em>)=\u221a1-x<\/em>2 respectively, describe fog<\/em> and gof<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

g<\/em> (x<\/em>)=\u221a1-x<\/em>2\u21d2x<\/em>2\u22650, \u2200x<\/em>\u2208[-1, 1]\u21d2-x<\/em>2\u22640, \u2200x<\/em>\u2208[-1, 1]\u21d21-x<\/em>2\u22641, \u2200x<\/em>\u2208[-1, 1]We know that 1-x<\/em>2\u22650\u21d20\u22641-x<\/em>2\u22641\u21d2Range of g<\/em>(x<\/em>)=[0, 1]So, f<\/em>:<\/em>(–<\/em>\u03c0<\/em>2<\/em>,<\/em> <\/em>\u03c0<\/em>2<\/em>)\u2192<\/em>R<\/em> <\/em>and <\/em>g<\/em>:<\/em>[–<\/em>1<\/em>,<\/em> <\/em>1<\/em>]\u2192<\/em> <\/em>[0<\/em>,<\/em> <\/em>1<\/em>]Computation of fog:<\/em>Clearly, the range of g<\/em> <\/em>is a subset of the domain of f.<\/em>So, fog: <\/em>[-1, 1]\u2192R<\/em>(f<\/em>o<\/em>g<\/em>) <\/em>(x<\/em>)=f<\/em> <\/em>(g<\/em> <\/em>(x<\/em>))=f<\/em> <\/em>(\u221a1-x2)=tan<\/em> <\/em>\u221a1-x2Computation of gof:<\/em>Clearly, the range of f<\/em> is not a subset of the domain of g<\/em>.<\/em>\u21d2Domain (g<\/em>o<\/em>f<\/em>)={x<\/em>\u2208domain of f<\/em> <\/em>and f(x)\u2208domain of g<\/em>}\u21d2Domain (gof)={x<\/em>\u2208(–<\/em>\u03c02<\/em>,<\/em> <\/em>\u03c02<\/em>) and tan x<\/em> <\/em>\u2208<\/em>[–<\/em>1<\/em>,<\/em>1<\/em>]}\u21d2Domain (gof)={x\u2208(–<\/em>\u03c02<\/em>,<\/em> <\/em>\u03c02<\/em>) and x <\/em>\u2208<\/em>(–<\/em>\u03c0<\/em>4<\/em>,<\/em> <\/em>\u03c0<\/em>4<\/em>)}\u21d2Domain (gof)={x\u2208<\/em>(–<\/em>\u03c04<\/em>,<\/em> <\/em>\u03c04<\/em>)}Now, g<\/em>o<\/em>f<\/em>:<\/em>(–<\/em>\u03c0<\/em>4<\/em>,<\/em> <\/em>\u03c0<\/em>4<\/em>)\u2192<\/em>R<\/em>So, (g<\/em>o<\/em>f<\/em>) <\/em>(x<\/em>)=g<\/em> <\/em>(f<\/em> <\/em>(x<\/em>))=<\/em>g<\/em> <\/em>(tan <\/em>x<\/em>)=<\/em>\u221a1-tan2x<\/em> <\/em><\/p>\n\n\n\n

Page No 2.54:<\/h4>\n\n\n\n

Question 10:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>)=\u221ax<\/em>+3 and g<\/em>(x<\/em>)=x<\/em>2+1 be two real functions, then find fog<\/em> and gof<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em>(x<\/em>)=\u221ax<\/em>+3For domain,x<\/em>+3\u22650\u21d2x<\/em>\u2265-3Domain of f<\/em> =[-3, \u221e)Since f<\/em> is a square root function, range of f<\/em> =[0, \u221e)f<\/em>: [-3, \u221e)\u2192[0, \u221e)g<\/em>(x<\/em>)=x<\/em>2+1 is a polynomial.\u21d2g<\/em>:R<\/em>\u2192R<\/em>Computation of f<\/em>o<\/em>g<\/em>:Range of g<\/em>  is not a subset of the domain of f<\/em>.and domain (f<\/em>o<\/em>g<\/em>)={x<\/em>: x<\/em>\u2208domain of g<\/em> and g<\/em>(x<\/em>)\u2208domain of f<\/em>(x<\/em>)}\u21d2Domain (f<\/em>o<\/em>g<\/em>)={x<\/em>:x<\/em>\u2208R<\/em> and  x<\/em>2+1\u2208[-3, \u221e)}\u21d2Domain (f<\/em>o<\/em>g<\/em>)={x<\/em>:x<\/em>\u2208R<\/em> and  x<\/em>2+1\u2265-3}\u21d2Domain (f<\/em>o<\/em>g<\/em>)={x<\/em>:x<\/em>\u2208R<\/em> and  x<\/em>2+4\u22650}\u21d2Domain (f<\/em>o<\/em>g<\/em>)={x<\/em>:x<\/em>\u2208R<\/em> and x<\/em>\u2208R<\/em>}\u21d2Domain (f<\/em>o<\/em>g<\/em>)=Rf<\/em>o<\/em>g<\/em>:R<\/em>\u2192R<\/em>(f<\/em>o<\/em>g<\/em>) (x<\/em>)=f<\/em>(g<\/em> (x<\/em>))=f<\/em> (x<\/em>2+1)=\u221ax<\/em>2+1+3=\u221ax<\/em>2+4Computation of g<\/em>o<\/em>f<\/em>:Range of f<\/em>  is a subset of the domain of g<\/em>.g<\/em>o<\/em>f<\/em>: [-3, \u221e)\u2192R<\/em>\u21d2(g<\/em>o<\/em>f<\/em>) (x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (\u221ax<\/em>+3)=(\u221ax<\/em>+3)2+1=x<\/em>+3+1=x<\/em>+4<\/p>\n\n\n\n

Page No 2.54:<\/h4>\n\n\n\n

Question 11:<\/h4>\n\n\n\n

Let f<\/em> be a real function given by f<\/em>(x<\/em>)=\u221ax<\/em>-2.
Find each of the following:
(i) fof<\/em>
(ii) fofof<\/em>
(iii) (fofof<\/em>) (38)
(iv) f<\/em>2<\/sup>

Also, show that fof<\/em> \u2260 f<\/em>2<\/sup> .<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em>(x<\/em>)=\u221ax<\/em>-2For domain,x<\/em>-2\u22650\u21d2x<\/em>\u22652Domain of f<\/em>=[2,\u221e)Since f<\/em>is a square-root function, range of f<\/em>=(0,\u221e)So, f<\/em>:[2,\u221e)\u2192(0,\u221e)(i) f<\/em>o<\/em>f<\/em>Range of f<\/em> is not a subset of the domain of f<\/em>.\u21d2Domain(f<\/em>o<\/em>f<\/em>)={x<\/em>: x<\/em> \u2208domain of f<\/em>and f<\/em>(x<\/em>)\u2208domain of f}\u21d2Domain(f<\/em>o<\/em>f<\/em>)={x<\/em>: x<\/em> \u2208[2,\u221e) and \u221ax<\/em>-2\u2208[2,\u221e)}\u21d2Domain(f<\/em>o<\/em>f<\/em>)={x<\/em>: x<\/em> \u2208[2,\u221e) and \u221ax<\/em>-2\u22652}\u21d2Domain(f<\/em>o<\/em>f<\/em>)={x<\/em>: x<\/em> \u2208[2,\u221e) and  x<\/em>-2\u22654}\u21d2Domain(f<\/em>o<\/em>f<\/em>)={x<\/em>: x<\/em> \u2208[2,\u221e) and  x<\/em>\u22656}\u21d2Domain(f<\/em>o<\/em>f<\/em>)={x<\/em>: x<\/em>\u22656}\u21d2Domain(f<\/em>o<\/em>f<\/em>)=[6, \u221e)f<\/em>o<\/em>f<\/em> :[6, \u221e)\u2192R<\/em>(f<\/em>o<\/em>f<\/em>) (x<\/em>)=f<\/em> (f<\/em> (x<\/em>))=f<\/em> (\u221ax<\/em>-2)=\u221a\u221ax<\/em>-2-2

(ii) f<\/em>o<\/em>f<\/em>o<\/em>f<\/em>= (f<\/em>o<\/em>f<\/em>) o<\/em>f<\/em>We have, f<\/em>:[2,\u221e)\u2192(0,\u221e) and f<\/em>o<\/em>f<\/em> : [6, \u221e)\u2192R<\/em>\u21d2Range of f<\/em> is not a subset of the domain of f<\/em>o<\/em>f<\/em>.Then, domain((f<\/em>o<\/em>f<\/em>)o<\/em>f<\/em>)={x<\/em>: x<\/em> \u2208domain of f<\/em>and f<\/em>(x<\/em>)\u2208domain of f<\/em>o<\/em>f<\/em>}\u21d2Domain((f<\/em>o<\/em>f<\/em>)o<\/em>f<\/em>)={x<\/em>: x<\/em> \u2208[2,\u221e) and \u221ax<\/em>-2\u2208[6,\u221e)}\u21d2Domain((f<\/em>o<\/em>f<\/em>)o<\/em>f<\/em>)={x<\/em>: x<\/em> \u2208[2,\u221e) and \u221ax<\/em>-2\u22656}\u21d2Domain((f<\/em>o<\/em>f<\/em>)o<\/em>f<\/em>)={x<\/em>: x<\/em> \u2208[2,\u221e) and  x<\/em>-2\u226536}\u21d2Domain((f<\/em>o<\/em>f<\/em>)o<\/em>f<\/em>)={x<\/em>: x<\/em> \u2208[2,\u221e) and  x<\/em>\u226538}\u21d2Domain((f<\/em>o<\/em>f<\/em>)o<\/em>f<\/em>)={x<\/em>: x<\/em>\u226538}\u21d2Domain((f<\/em>o<\/em>f<\/em>)o<\/em>f<\/em>)=[38, \u221e)f<\/em>o<\/em>f<\/em> :[38,\u221e)\u2192R<\/em>So, ((f<\/em>o<\/em>f<\/em>)o<\/em>f<\/em>) (x<\/em>)=(f<\/em>o<\/em>f<\/em>) (f<\/em> (x<\/em>))=(f<\/em>o<\/em>f<\/em>) (\u221ax<\/em>-2)=\u221a\u221a\u221ax<\/em>-2-2-2

(i<\/em>i<\/em>i<\/em>) We have, (f<\/em>o<\/em>f<\/em>o<\/em>f<\/em>) (x<\/em>)=\u221a\u221a\u221ax<\/em>-2-2-2So, (f<\/em>o<\/em>f<\/em>o<\/em>f<\/em>) (38)=\u221a\u221a\u221a38-2-2-2=\u221a\u221a\u221a36-2-2=\u221a\u221a6-2-2=\u221a2-2=0

(iv) We have, f<\/em>o<\/em>f<\/em>=\u221a\u221ax<\/em>-2-2\u21d2f<\/em>2(x<\/em>)=f<\/em>(x<\/em>)\u00d7f<\/em>(x<\/em>)=\u221ax<\/em>-2\u00d7\u221ax<\/em>-2=x<\/em>-2So, f<\/em>o<\/em>f<\/em> \u2260 f<\/em>2<\/p>\n\n\n\n

Page No 2.55:<\/h4>\n\n\n\n

Question 12:<\/h4>\n\n\n\n

Let f<\/em>(x<\/em>)={1+x<\/em>,0\u2264x<\/em>\u226423-x<\/em>,2<x<\/em>\u22643. Find fof<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em>(x<\/em>)={1+x<\/em>,0\u2264x<\/em>\u226423-x<\/em>,2<x<\/em>\u22643It can be written as,f<\/em>(x<\/em>)={ 1+x<\/em>,0\u2264x<\/em>\u226411+x<\/em>,1<x<\/em>\u226423-x<\/em>,2<x<\/em>\u22643  When,0\u2264x<\/em>\u22641Then, f<\/em>(x<\/em>)=1+x<\/em>Now when ,0\u2264x<\/em>\u22641 then ,1\u2264x<\/em>+1\u22642Then, f<\/em>(f<\/em>(x<\/em>))=1+(1+x<\/em>)=2+x<\/em>   [\u22351\u2264f<\/em>(x<\/em>)<2]When ,1<x<\/em>\u22642Then, f<\/em>(x<\/em>)=1+x<\/em>Now when ,1<x<\/em>\u22642 then,2<x<\/em>+1\u22643Then, f<\/em>(f<\/em>(x<\/em>))=3-(1+x<\/em>)=2-x<\/em>  [\u22352\u2264f<\/em>(x<\/em>)<3]When ,2<x<\/em>\u22643Then, f<\/em>(x<\/em>)=3-x<\/em>Now when ,2<x<\/em>\u22643 then ,0\u22643-x<\/em><1Then, f<\/em>(f<\/em>(x<\/em>))=1+(3-x<\/em>)=4-x<\/em>     [\u22350\u2264f<\/em>(x<\/em>)<1]f<\/em>(f<\/em>(x<\/em>))={ 2+x<\/em>,0\u2264x<\/em>\u226412-x<\/em>,1<x<\/em>\u226424-x<\/em>,2<x<\/em>\u22643  <\/p>\n\n\n\n

Page No 2.55:<\/h4>\n\n\n\n

Question 13:<\/h4>\n\n\n\n

If f<\/em>, g : R \u2192 R be two functions defined as f<\/em>(x<\/em>) = |x<\/em>| + x<\/em> and g(x<\/em>) = |x<\/em>| \u2013x<\/em>,\u2200x<\/em>\u2208R. Then find fog and gof. Hence find fog(\u20133), fog(5) and gof (\u20132).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = |x<\/em>| + x<\/em> 
and g<\/em>(x<\/em>) = |x<\/em>| \u2013 x<\/em>,\u2200x<\/em>\u2208R

f<\/em>o<\/em>g<\/em> = f<\/em>(g<\/em>(x<\/em>)) = |g<\/em>(x<\/em>)| + g<\/em>(x<\/em>)                           =||x<\/em>| – x<\/em>|+(|x<\/em>| – x<\/em>)

Therefore,
 f<\/em>(g<\/em>(x<\/em>)) = {0                x<\/em>\u226504x<\/em>             x<\/em><0f<\/em>o<\/em>g<\/em> = {4x<\/em>       x<\/em><00          x<\/em>\u22650

g<\/em>o<\/em>f<\/em> = g<\/em>(f<\/em>(x<\/em>))=|f<\/em>(x<\/em>)|-f<\/em>(x<\/em>)                          =||x<\/em>|+x<\/em>|-(|x<\/em>|+x<\/em>)g<\/em>(f<\/em>(x<\/em>))={0         x<\/em>\u226500         x<\/em><0
Therefore, g<\/em>(f<\/em>(x<\/em>)) = gof<\/em> = 0

Now, fog<\/em>(\u22123) =(4)(\u22123) = \u221212                 (since, fog<\/em> = 4x<\/em> for x<\/em> < 0)

fog<\/em>(5) = 0                                                (since, fog<\/em> = 0 for x<\/em> \u2265 0)

gof<\/em>(\u22122) = 0                                               (since, gof<\/em> = 0 for x<\/em> < 0)<\/p>\n\n\n\n

Page No 2.68:<\/h4>\n\n\n\n

Question 1:<\/h4>\n\n\n\n

State with reasons whether the following functions have inverse:
(i) f<\/em> : {1, 2, 3, 4} \u2192 {10} with f<\/em> = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g<\/em> : {5, 6, 7, 8} \u2192 {1, 2, 3, 4} with g<\/em> = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h<\/em> : {2, 3, 4, 5} \u2192 {7, 9, 11, 13} with h<\/em> = {(2, 7), (3, 9), (4, 11), (5, 13)}<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(i) f<\/em> : {1, 2, 3, 4} \u2192 {10} with f<\/em> = {(1, 10), (2, 10), (3, 10), (4, 10)}
We have:
f <\/em>(1) = f <\/em>(2) = f <\/em>(3) = f <\/em>(4) = 10
\u21d2f <\/em>is not one-one.
\u21d2f <\/em>is not a bijection.
So, f <\/em>does not have an inverse.

(ii) g<\/em> : {5, 6, 7, 8} \u2192 {1, 2, 3, 4} with g<\/em> = {(5, 4), (6, 3), (7, 4), (8, 2)}
g <\/em>(5) = g <\/em>(7) = 4
\u21d2f <\/em>is not one-one.
\u21d2f <\/em>is not a bijection.
So, f <\/em>does not have an inverse.

(iii) h<\/em> : {2, 3, 4, 5} \u2192 {7, 9, 11, 13} with h<\/em> = {(2, 7), (3, 9), (4, 11), (5, 13)}
Here, different elements of the domain have different images in the co-domain.
\u21d2h <\/em>is one-one.
Also, each element in the co-domain has a pre-image in the domain.
\u21d2h <\/em>is onto.
\u21d2h <\/em>is a bijection.
\u21d2h <\/em>has an inverse and it is given by
h-1<\/sup><\/em>={(7, 2), (9, 3), (11, 4), (13, 5)}<\/p>\n\n\n\n

Page No 2.68:<\/h4>\n\n\n\n

Question 2:<\/h4>\n\n\n\n

Find f<\/em> \u22121<\/sup> if it exists : f<\/em> : A<\/em> \u2192 B,<\/em> where
(i) A<\/em> = {0, \u22121, \u22123, 2}; B<\/em> = {\u22129, \u22123, 0, 6} and f<\/em>(x<\/em>) = 3 x<\/em>.
(ii) A<\/em> = {1, 3, 5, 7, 9}; B<\/em> = {0, 1, 9, 25, 49, 81} and f<\/em>(x<\/em>) = x<\/em>2<\/sup><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(i) A<\/em> = {0, \u22121, \u22123, 2}; B<\/em> = {\u22129, \u22123, 0, 6} and f<\/em>(x<\/em>) = 3 x<\/em>.
Given: f<\/em>(x<\/em>) = 3 x<\/em>
So,  f = <\/em>{(0, 0), (-1, -3), (-3, -9), (2, 6)}
Clearly, this is one-one.
Range of f = <\/em>Range of f <\/em>=B
So, f <\/em>is a bijection and, thus, f -1<\/sup> <\/em>exists.  
Hence, f -1<\/sup>=<\/em> {(0, 0), (-3, -1), (-9, -3), (6, 2)}

(ii) A<\/em> = {1, 3, 5, 7, 9}; B<\/em> = {0, 1, 9, 25, 49, 81} and f<\/em>(x<\/em>) = x<\/em>2<\/sup>
Given: f<\/em>(x<\/em>) = x2<\/sup><\/em>
So, f = <\/em>{(1, 1), (3, 9), (5, 25), (7,49), (9, 81)}
Clearly, f <\/em>is one-one.
But this is not onto because the element 0 in the co-domain (B<\/em>) has no pre-image in the domain (A) <\/em>.
\u21d2f <\/em>is not a bijection.
So, f <\/sup><\/em>-1<\/sup><\/em>does not exist.<\/p>\n\n\n\n

Page No 2.68:<\/h4>\n\n\n\n

Question 3:<\/h4>\n\n\n\n

Consider f<\/em> : {1, 2, 3} \u2192 {a<\/em>, b<\/em>, c<\/em>} and g<\/em> : {a<\/em>, b<\/em>, c<\/em>} \u2192 {apple, ball, cat} defined as f<\/em> (1) = a<\/em>, f<\/em> (2) = b<\/em>, f<\/em> (3) = c<\/em>, g<\/em> (a) = apple, g<\/em> (b) = ball and g<\/em> (c) =  cat. Show that f<\/em>, g<\/em> and gof<\/em> are invertible. Find f<\/em>\u22121<\/sup>, g<\/em>\u22121<\/sup> and gof<\/em>\u22121<\/sup> and show that (gof<\/em>)\u22121<\/sup> = f<\/em> \u2212<\/sup>1<\/sup>o<\/em> g<\/em>\u22121<\/sup>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f={(1, a), (2, b), (3, c)} and g={(a, apple), (b, ball), (c, cat)}Clearly, f and g are bijections.So, f and g are invertible.Now,f-1={(a, 1), (b, 2), (c, 3)} and g-1={(apple, a), (ball, b), (cat, c)}So, f-1o g-1={(apple, 1), (ball, 2), (cat, 3)}         …(1)f:{1, 2, 3}\u2192{a, b, c} and g:{a, b, c}\u2192{apple, ball, cat}So, gof:{1, 2, 3}\u2192{apple, ball, cat}\u21d2(gof) (1)=g (f (1))=g (a)=apple(gof) (2)=g (f (2))=g (b)=ball,and (gof) (3)=g (f (3))=g (c)=cat\u2234gof ={(1, apple), (2, ball), (3, cat)}Clearly, gofis a bijection.So, gof is invertible.(gof)-1={(apple, 1), (ball, 2), (cat, 3)}          …(2)From (1) and (2), we get:(gof)-1=f-1o g-1<\/em><\/p>\n\n\n\n

Page No 2.68:<\/h4>\n\n\n\n

Question 4:<\/h4>\n\n\n\n

Let A<\/em> = {1, 2, 3, 4}; B<\/em> = {3, 5, 7, 9}; C<\/em> = {7, 23, 47, 79} and f<\/em> : A<\/em> \u2192 B<\/em>, g<\/em> : B<\/em> \u2192 C<\/em> be defined as f<\/em>(x<\/em>) = 2x<\/em> + 1 and g<\/em>(x<\/em>) = x<\/em>2<\/sup> \u2212 2. Express (gof<\/em>)\u22121<\/sup> and f<\/em>\u22121<\/sup> og<\/em>\u22121<\/sup> as the sets of ordered pairs and verify that (gof<\/em>)\u22121<\/sup> = f<\/em>\u22121<\/sup> og<\/em>\u22121<\/sup>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

 f<\/em>(x<\/em>)=2x<\/em>+1\u21d2f<\/em>={(1, 2(1)+1), (2, 2(2)+1), (3, 2(3)+1), (4, 2(4)+1)}={(1, 3), (2, 5), (3, 7), (4, 9)}g<\/em>(x<\/em>)=x<\/em>2-2\u21d2g<\/em>={(3, 32-2), (5, 52-2), (7, 72-2), (9, 92-2)}={(3, 7), (5, 23), (7, 47), (9, 79)}Clearly f <\/em>and g<\/em> are bijections and, hence, f<\/em>-1:B<\/em>\u2192A<\/em> and g<\/em>-1: C<\/em>\u2192B<\/em> exist.So, f<\/em>-1={(3, 1), (5, 2), (7, 3), (9, 4)} and g<\/em>-1={(7, 3), (23, 5), (47, 7), (79, 9)}Now, (f<\/em>-1 o<\/em> g<\/em>-1):C<\/em>\u2192A<\/em>f<\/em>-1 o<\/em> g<\/em>-1={(7, 1), (23, 2), (47, 3), (79, 4)}        …(1)Also, f<\/em>:A<\/em>\u2192B<\/em> and g<\/em>:B<\/em>\u2192C<\/em>,\u21d2g<\/em>o<\/em>f<\/em>:A<\/em>\u2192C<\/em>, (g<\/em>o<\/em>f<\/em>)-1:C<\/em>\u2192A<\/em>So, f<\/em>-1 o<\/em> g<\/em>-1and (g<\/em>o<\/em>f<\/em>)-1 have same domains.(g<\/em>o<\/em>f<\/em>)(x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (2x<\/em>+1)=(2x<\/em>+1)2-2\u21d2 (g<\/em>o<\/em>f<\/em>)(x<\/em>)=4x<\/em>2+4x<\/em>+1-2\u21d2 (g<\/em>o<\/em>f<\/em>)(x<\/em>)=4x<\/em>2+4x<\/em>-1Then, (g<\/em>o<\/em>f<\/em>)(1)=g<\/em> (f<\/em> (1))=4+4-1=7,(g<\/em>o<\/em>f<\/em>)(2)=g<\/em> (f<\/em> (2))=4+4-1=23,(g<\/em>o<\/em>f<\/em>)(3)=g<\/em> (f<\/em> (3))=4+4-1=47 and (g<\/em>o<\/em>f<\/em>)(4)=g<\/em> (f<\/em> (4))=4+4-1=79So, g<\/em>o<\/em>f<\/em>={(1, 7), (2, 23), (3, 47), (4, 79)}\u21d2(g<\/em>o<\/em>f<\/em>)-1={(7, 1), (23, 2), (47, 3), (79, 4)}     …(2)From (1) and (2), we get: (g<\/em>o<\/em>f<\/em>)-1=f<\/em>-1 o<\/em> g<\/em>-1<\/p>\n\n\n\n

Page No 2.68:<\/h4>\n\n\n\n

Question 5:<\/h4>\n\n\n\n

Show that the function f<\/em> : Q<\/em> \u2192 Q,<\/em> defined by f<\/em>(x<\/em>) = 3x<\/em> + 5, is invertible. Also, find f<\/em>\u22121<\/sup><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity of f<\/em>:
Let x <\/em>and y <\/em>be two elements of the domain (Q<\/em>), such that
f(x)=f(y)<\/em>
\u21d23x<\/em> + 5 =3y<\/em> + 5
\u21d23x<\/em> = 3y<\/em>
\u21d2x = y<\/em>
So, f <\/em>is one-one.

Surjectivity of f<\/em>:
Let y<\/em> be in the co-domain (Q<\/em>), such that f(x) = y

\u21d23x+5=y\u21d23x=y-5\u21d2x=y-53\u2208Q (domain)<\/em>

\u21d2f<\/em> is onto.
So, f<\/em> is a bijection and, hence, it is invertible.

Finding f  –<\/sup><\/em>1<\/sup>:
Let f<\/em>-1(x<\/em>)=y<\/em>                    …(1)\u21d2x<\/em>=f<\/em>(y<\/em>)\u21d2x<\/em>=3y<\/em>+5\u21d2x<\/em>-5=3y<\/em>\u21d2y<\/em>=x<\/em>-53So, f<\/em>-1(x<\/em>)=x<\/em>-53       [from (1)]<\/p>\n\n\n\n

Page No 2.68:<\/h4>\n\n\n\n

Question 6:<\/h4>\n\n\n\n

Consider f<\/em> : R<\/em> \u2192 R<\/em> given by f<\/em>(x<\/em>) = 4x<\/em> + 3. Show that f<\/em> is invertible. Find the inverse of f<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity of f<\/em> :
Let x <\/em>and y <\/em>be two elements of domain (R<\/em>), such that
f(x) = f(y)<\/em>
\u21d24x<\/em> + 3 = 4y<\/em> + 3
\u21d24x<\/em> = 4y<\/em>
\u21d2x = y<\/em>
So, f <\/em>is one-one.

Surjectivity of f<\/em> :
Let y<\/em> be in the co-domain (R<\/em>), such that f(x) = y.

\u21d24x+3=y\u21d24x=y- 3\u21d2x=y- 34\u2208R(Domain)<\/em>

\u21d2f<\/em> is onto.
So, f<\/em> is a bijection and, hence, is invertible.

Finding f  –<\/sup><\/em>1<\/sup>:
Let f<\/em>-1(x<\/em>)=y<\/em>                   …(1)\u21d2x<\/em>=f<\/em>(y<\/em>)\u21d2x<\/em>=4y<\/em>+3\u21d2x<\/em>-3=4y<\/em>\u21d2y<\/em>=x<\/em>-34So, f<\/em>-1(x<\/em>)=x<\/em>-34       [from (1)]<\/p>\n\n\n\n

Page No 2.68:<\/h4>\n\n\n\n

Question 7:<\/h4>\n\n\n\n

Consider f<\/em> : R<\/em> \u2192 R<\/em>+<\/sub> \u2192 [4, \u221e) given by f<\/em>(x<\/em>) = x<\/em>2<\/sup> + 4. Show that f<\/em> is invertible with inverse f<\/em>\u22121<\/sup> of f<\/em> given by f\u22121<\/sup> (x<\/em>)=\u221ax<\/em>-4, where R<\/em>+<\/sup> is the set of all non-negative real numbers.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity of f<\/em> :
Let x <\/em>and y <\/em>be two elements of the domain (Q<\/em>), such that
f(x)=f(y)<\/em>
\u21d2x<\/em>2+4=y<\/em>2+4\u21d2x<\/em>2=y<\/em>2\u21d2x<\/em>=y<\/em>   (as co-domain as R<\/em>+)
So, f <\/em>is one-one.

Surjectivity of f<\/em> :
Let y<\/em> be in the co-domain (Q<\/em>), such that f(x) = y

\u21d2x2+4=y\u21d2x2=y-4\u21d2x=\u221ay-4\u2208R<\/em>

\u21d2f<\/em> is onto.
So, f<\/em> is a bijection and, hence, it is invertible.

Finding f  –<\/sup><\/em>1<\/sup>:
Let f<\/em>-1(x<\/em>)=y<\/em>                    …(1)\u21d2x<\/em>=f<\/em>(y<\/em>)\u21d2x<\/em>=y<\/em>2+4\u21d2x<\/em>-4=y<\/em>2\u21d2y<\/em>=\u221ax<\/em>-4So, f<\/em>-1(x<\/em>)=\u221ax<\/em>-4          [from (1)]<\/p>\n\n\n\n

Page No 2.68:<\/h4>\n\n\n\n

Question 8:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>)=4x<\/em>+36x<\/em>-4, x<\/em>\u226023, show that fof<\/em>(x<\/em>) = x<\/em> for all x<\/em>\u226023. What is the inverse of f<\/em>?<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(f<\/em>o<\/em>f<\/em>)(x<\/em>)=f<\/em>(f<\/em>(x<\/em>))=f<\/em> (4x<\/em>+36x<\/em>-4)=4(4x<\/em>+36x<\/em>-4)+36(4x<\/em>+36x<\/em>-4)-4=16x<\/em>+12+18x<\/em>-1224x<\/em>+18-24x<\/em>+16=34x<\/em>34=x<\/em>\u21d2(f<\/em>o<\/em>f<\/em>)(x<\/em>)=x<\/em>=I<\/em>X<\/em>, where I<\/em> is an identity function.So, f<\/em>=f<\/em>-1 Hence, f<\/em>-1=4x<\/em>+36x<\/em>-4<\/p>\n\n\n\n

Page No 2.68:<\/h4>\n\n\n\n

Question 9:<\/h4>\n\n\n\n

Consider f<\/em> : R<\/em>+<\/sub> \u2192 [\u22125, \u221e) given by f<\/em>(x<\/em>) = 9x<\/em>2<\/sup> + 6x<\/em> \u2212 5. Show that f<\/em> is invertible with f<\/em>-1(x<\/em>)=\u221ax<\/em>+6-13.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity of f <\/em>:
Let x <\/em>and y <\/em>be two elements of domain (R+<\/sup><\/em>), such that
f(x)=f(y)<\/em>
\u21d29x<\/em>2+6x<\/em>-5=9y<\/em>2+6y<\/em>-5\u21d29x<\/em>2+6x<\/em>=9y<\/em>2+6y<\/em>\u21d2x<\/em>=y<\/em> (As, x<\/em>, y<\/em>\u2208R<\/em>+)
So, f <\/em>is one-one.

Surjectivity of f<\/em>:
Let y<\/em> is in the co domain (Q<\/em>) such that f(x) = y

\u21d29×2+6x-5=y\u21d29×2+6x=y+5\u21d29×2+6x+1=y+6 (Adding 1 on both sides)\u21d2(3x+1)2=y+6\u21d23x+1=\u221ay+6\u21d23x=\u221ay+6-1\u21d2x=\u221ay+6-13\u2208R+(domain)<\/em>

\u21d2f<\/em> is onto.
So, f<\/em> is a bijection and hence, it is invertible.

Finding f  –<\/sup><\/em>1<\/sup>:
Let f<\/em>-1(x<\/em>)=y<\/em>                          …(1)\u21d2x<\/em>=f<\/em>(y<\/em>)\u21d2x<\/em>=9y<\/em>2+6y<\/em>-5\u21d2x<\/em>+5=9y<\/em>2+6y<\/em>\u21d2x<\/em>+6=9y<\/em>2+6y<\/em>+1       (adding 1 on both sides)\u21d2x<\/em>+6=(3y<\/em>+1)2\u21d23y<\/em>+1=\u221ax<\/em>+6\u21d23y<\/em>=\u221ax<\/em>+6-1\u21d2y<\/em>=\u221ax<\/em>+6-13So, f<\/em>-1(x<\/em>)=\u221ax<\/em>+6-13     [from (1)]<\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 10:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> be defined by f<\/em>(x<\/em>) = x<\/em>3<\/sup> \u22123, then prove that f<\/em>\u22121<\/sup> exists and find a formula for f<\/em>\u22121<\/sup>. Hence, find f<\/em>\u22121<\/sup> (24) and f<\/em>\u22121<\/sup> (5).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity of f <\/em>:
Let x <\/em>and y <\/em>be two elements in domain (R<\/em>),
 
such that,  x<\/em>3-3=y<\/em>3-3            \u21d2x<\/em>3=y<\/em>3            \u21d2x<\/em>=y<\/em>
So, f <\/em>is one-one.

Surjectivity of f<\/em> :
Let y<\/em> be in the co-domain (R<\/em>) such that f(x) = y

\u21d2x3-3=y\u21d2x3=y+3\u21d2x=3\u221ay+3\u2208R<\/em>

\u21d2f<\/em> is onto.
So, f<\/em> is a bijection and, hence, it is invertible.

Finding f  –<\/sup><\/em>1<\/sup>:
Let f<\/em>-1(x<\/em>)=y<\/em>                         …(1)\u21d2x<\/em>=f<\/em>(y<\/em>)\u21d2x<\/em>=y<\/em>3-3\u21d2x<\/em>+3=y<\/em>3\u21d2y<\/em>=3\u221ax<\/em>+3 = f<\/em>-1(x<\/em>)       [from (1)]So, f<\/em>-1(x<\/em>)=3\u221ax<\/em>+3 Now, f<\/em>-1(24)=3\u221a24+3=3\u221a27=3\u221a33=3 and f<\/em>-1(5)=3\u221a5+3=3\u221a8=3\u221a23=2 <\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 11:<\/h4>\n\n\n\n

A function f<\/em> : R<\/em> \u2192 R<\/em> is defined as f<\/em>(x<\/em>) = x<\/em>3<\/sup> + 4. Is it a bijection or not? In case it is a bijection, find f<\/em>\u22121<\/sup> (3).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity of f<\/em>:
Let x <\/em>and y <\/em>be two elements of domain (R<\/em>), such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>3+4=y<\/em>3+4\u21d2x<\/em>3=y<\/em>3\u21d2x<\/em>=y<\/em>
So, f <\/em>is one-one.

Surjectivity of f<\/em>:
Let y<\/em> be in the co-domain (R<\/em>), such that f(x) = y.

\u21d2x3+4=y\u21d2x3=y-4\u21d2x=3\u221ay-4\u2208R (domain)<\/em>

\u21d2 f<\/em> is onto.
So, f<\/em> is a bijection and, hence, is invertible.

Finding f  –<\/sup><\/em>1<\/sup>:
Let f<\/em>-1(x<\/em>)=y<\/em>              …(1)\u21d2x<\/em>=f<\/em>(y<\/em>)\u21d2x<\/em>=y<\/em>3+4\u21d2x<\/em>-4=y<\/em>3\u21d2y<\/em>=3\u221ax<\/em>-4So, f<\/em>-1(x<\/em>)=3\u221ax<\/em>-4       [from (1)]f<\/em>-1(3)=3\u221a3-4 =3\u221a-1=-1<\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 12:<\/h4>\n\n\n\n

If f<\/em> : Q<\/em> \u2192 Q<\/em>, g<\/em> : Q<\/em> \u2192 Q<\/em> are two functions defined by f<\/em>(x<\/em>) = 2 x<\/em> and g<\/em>(x<\/em>) = x<\/em> + 2, show that f<\/em> and g<\/em> are bijective maps. Verify that (gof<\/em>)\u22121<\/sup> = f<\/em>\u22121<\/sup> og<\/em> \u22121<\/sup>.<\/em><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity of f<\/em>:
Let x <\/em>and y <\/em>be two elements of domain (Q<\/em>), such that
f(x) = f(y)<\/em>
\u21d22x<\/em> =<\/em> 2y<\/em>
\u21d2<\/em>x = y<\/em>
So, f <\/em>is one-one.

Surjectivity of f<\/em>:
Let y<\/em> be in the co-domain (Q<\/em>), such that f(x) = y.

\u21d22x= y\u21d2x= y2\u2208Q   (domain)<\/em>

\u21d2 f<\/em> is onto.
So, f<\/em> is a bijection and, hence, it is invertible.

Finding f  –<\/sup><\/em>1<\/sup>:
Let f<\/em>-1(x<\/em>)=y<\/em>             …(1)\u21d2x<\/em>=f<\/em>(y<\/em>)\u21d2x<\/em>=2y<\/em>\u21d2y<\/em>=x<\/em>2  So, f<\/em>-1(x<\/em>)=x<\/em>2        (from (1))

Injectivity of g<\/em>:
Let x <\/em>and y <\/em>be two elements of domain (Q<\/em>), such that
g(x) = g(y)<\/em>
\u21d2x + <\/em>2 = y<\/em> + 2
\u21d2x = y<\/em>
So, g <\/em>is one-one.

Surjectivity of g<\/em>:
Let y<\/em> be in the co domain (Q<\/em>), such that g(x) = y.

\u21d2x+2=y\u21d2x= 2-y\u2208Q   (domain)<\/em>

\u21d2 g<\/em> is onto.
So, g<\/em> is a bijection and, hence, it is invertible.

Finding g –<\/sup><\/em>1<\/sup>:
Let g<\/em>-1(x<\/em>)=y<\/em>             …(2)\u21d2x<\/em>=g<\/em>(y<\/em>)\u21d2x<\/em>=y<\/em>+2\u21d2y<\/em>=x<\/em>-2So, g<\/em>-1(x<\/em>)=x<\/em>-2     (From (2))

Verification of (gof<\/em>)\u22121<\/sup> = f<\/em>\u22121<\/sup> og<\/em> \u22121<\/sup>:

f<\/em>(x<\/em>)=2x<\/em>; g<\/em>(x<\/em>)=x<\/em>+2and f<\/em>-1(x<\/em>)=x<\/em>2; g<\/em>-1(x<\/em>)=x<\/em>-2Now, (f<\/em>-1o<\/em> g<\/em>-1)(x<\/em>)=f<\/em>-1(g<\/em>-1(x<\/em>))\u21d2(f<\/em>-1o<\/em> g<\/em>-1)(x<\/em>)=f<\/em>-1(x<\/em>-2)\u21d2(f<\/em>-1o<\/em> g<\/em>-1)(x<\/em>)=x<\/em>-22      …(3)(g<\/em>o<\/em>f<\/em>)(x<\/em>)=g<\/em> (f<\/em> (x<\/em>))=g<\/em> (2x<\/em>)=2x<\/em>+2Let (g<\/em>o<\/em>f<\/em>)-1(x<\/em>)=y  ….   (4)x<\/em>=(g<\/em>o<\/em>f<\/em>)(y<\/em>)\u21d2x<\/em>=2y<\/em>+2\u21d22y<\/em>=x<\/em>-2\u21d2y<\/em>=x<\/em>-22 \u21d2(g<\/em>o<\/em>f<\/em>)-1(x<\/em>)=x<\/em>-22   [from (4)   … (5)]From (3) and (5), (g<\/em>o<\/em>f<\/em>)-1=f<\/em>-1o<\/em> g<\/em>-1<\/strong><\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 13:<\/h4>\n\n\n\n

Let A<\/em> = R<\/em> – {3} and B<\/em> = R<\/em> – {1}. Consider the function f<\/em> : A<\/em> \u2192 B<\/em> defined by f<\/em>(x<\/em>) = x<\/em>-2x<\/em>-3. Show that f<\/em> is one-one and onto and
hence find f<\/em>-1<\/sup>.                                                                                                                                                                        [CBSE 2012, 2014]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,

A<\/em> = R<\/em> – {3} and B<\/em> = R<\/em> – {1}
The function f<\/em> : A<\/em> \u2192 B<\/em> defined by f<\/em>(x<\/em>) = x<\/em>-2x<\/em>-3

Let x<\/em>,y<\/em>\u2208A<\/em> such that f<\/em>(x<\/em>)=f<\/em>(y<\/em>). Then,x<\/em>-2x<\/em>-3=y<\/em>-2y<\/em>-3\u21d2x<\/em>y<\/em>-3x<\/em>-2y<\/em>+6=x<\/em>y<\/em>-2x<\/em>-3y<\/em>+6\u21d2-x<\/em>=-y<\/em>\u21d2x<\/em>=y<\/em>\u2234 f<\/em> is one-one.Let y<\/em>\u2208B<\/em>. Then, y<\/em>\u22601.The function f<\/em> is onto if there exists x<\/em>\u2208A<\/em> such that f<\/em>(x<\/em>)=y<\/em>.Now,f<\/em>(x<\/em>)=y<\/em>\u21d2x<\/em>-2x<\/em>-3=y<\/em>\u21d2x<\/em>-2=x<\/em>y<\/em>-3y<\/em>\u21d2x<\/em>–x<\/em>y<\/em>=2-3y<\/em>\u21d2x<\/em>(1-y<\/em>)=2-3y<\/em>\u21d2x<\/em>=2-3y<\/em>1-y<\/em>\u2208A<\/em>    [y<\/em>\u22601]Thus, for any y<\/em>\u2208B<\/em>, there exists 2-3y<\/em>1-y<\/em>\u2208A<\/em> such thatf<\/em>(2-3y<\/em>1-y<\/em>)=(2-3y<\/em>1-y<\/em>)-2(2-3y<\/em>1-y<\/em>)-3=2-3y<\/em>-2+2y<\/em>2-3y<\/em>-3+3y<\/em>=-y<\/em>-1=y<\/em>\u2234 f<\/em> is onto.So, f<\/em> is one-one and onto fucntion.Now,As, x<\/em>=(2-3y<\/em>1-y<\/em>)So, f<\/em>-1(x<\/em>)=(2-3x<\/em>1-x<\/em>)=3x<\/em>-2x<\/em>-1<\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 14:<\/h4>\n\n\n\n

Consider the function f<\/em> : R<\/strong>+<\/sup> \u2192[-9,\u221e] given by f<\/em>(x<\/em>) = 5x<\/em>2<\/sup> + 6x<\/em> – 9. Prove that f<\/em> is invertible with f<\/em>-1<\/sup>(y<\/em>) = \u221a54+5y<\/em>-35.      [CBSE 2015]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,f<\/em>(x<\/em>)=5x<\/em>2+6x<\/em>-9Let y<\/em>=5x<\/em>2+6x<\/em>-9=5(x<\/em>2+65x<\/em>-95)=5(x<\/em>2+2\u00d7x<\/em>\u00d735+925-925-95)=5((x<\/em>+35)2-925-95)=5(x<\/em>+35)2-95-9=5(x<\/em>+35)2-545\u21d2y<\/em>+545=5(x<\/em>+35)2\u21d25y<\/em>+5425=(x<\/em>+35)2\u21d2\u221a5y<\/em>+5425=x<\/em>+35\u21d2x<\/em>=\u221a5y<\/em>+545-35\u21d2x<\/em>=\u221a5y<\/em>+54-35Let g<\/em>(y<\/em>)=\u221a5y<\/em>+54-35Now,f<\/em>o<\/em>g<\/em>(y<\/em>)=f<\/em>(g<\/em>(y<\/em>))=f<\/em>(\u221a5y<\/em>+54-35)=5(\u221a5y<\/em>+54-35)2+6(\u221a5y<\/em>+54-35)-9=5(5y<\/em>+54+9-6\u221a5y<\/em>+5425)+(6\u221a5y<\/em>+54-185)-9=5y<\/em>+63-6\u221a5y<\/em>+545+6\u221a5y<\/em>+54-185-9=5y<\/em>+63-18-455=y<\/em>=I<\/em>Y<\/em>, Identity functionAlso, g<\/em>o<\/em>f<\/em>(x<\/em>)=g<\/em>(f<\/em>(x<\/em>))=g<\/em>(5x<\/em>2+6x<\/em>-9)=\u221a5(5x<\/em>2+6x<\/em>-9)+54-35=\u221a25x<\/em>2+30x<\/em>-45+54-35=\u221a25x<\/em>2+30x<\/em>+9-35=\u221a(5x<\/em>+3)2-35=5x<\/em>+3-35=x<\/em>=I<\/em>X<\/em>, Identity functionSo, f<\/em> is invertible.Also, f<\/em>-1(y<\/em>)=g<\/em>(y<\/em>)=\u221a5y<\/em>+54-35<\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 15:<\/h4>\n\n\n\n

 Let f<\/em> : N\u2192N <\/em>be a function defined as f (x)=9×2+6x-5. <\/em>Show that f <\/em>: N\u2192S, <\/em>where S <\/em> is the range of f, <\/em>is invertible. find the inverse of f <\/em>and hence find f <\/em>-1<\/sup>(43) and f <\/em>-1<\/sup>(163).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,

f<\/em> : N\u2192N <\/em>is a function defined as f <\/em>(x<\/em>) = 9x<\/em>2<\/sup> + 6x <\/em>– 5.

Let y<\/em> = f <\/em>(x<\/em>) = 9x<\/em>2<\/sup> + 6x <\/em>– 5

\u21d2y<\/em>=9x<\/em>2+6x<\/em>-5\u21d2y<\/em>=9x<\/em>2+6x<\/em>+1-1-5\u21d2y<\/em>=(9x<\/em>2+6x<\/em>+1)-6\u21d2y<\/em>=(3x<\/em>+1)2-6\u21d2y<\/em>+6=(3x<\/em>+1)2

\u21d2\u221ay<\/em>+6=3x<\/em>+1             (\u2235y<\/em>\u2208N<\/em>)\u21d2\u221ay<\/em>+6-1=3x<\/em>\u21d2x<\/em>=\u221ay<\/em>+6-13\u21d2g<\/em>(y<\/em>)=\u221ay<\/em>+6-13         [Let x<\/em>=g<\/em>(y<\/em>)]

Now,

f<\/em>o<\/em>g<\/em>(y<\/em>)=f<\/em>[g<\/em>(y<\/em>)]=f<\/em>(\u221ay<\/em>+6-13)=9(\u221ay<\/em>+6-13)2+6(\u221ay<\/em>+6-13)-5=9(y<\/em>+6-2\u221ay<\/em>+6+19)+2(\u221ay<\/em>+6-1)-5=y<\/em>+6-2\u221ay<\/em>+6+1+2\u221ay<\/em>+6-2-5=y<\/em>=I<\/em>Y<\/em>, Identity function

g<\/em>o<\/em>f<\/em>(x<\/em>)=g<\/em>[f<\/em>(x<\/em>)]=g<\/em>(9x<\/em>2+6x<\/em>-5)=\u221a(9x<\/em>2+6x<\/em>-5)+6-13=\u221a(9x<\/em>2+6x<\/em>+1)-13=\u221a(3x<\/em>+1)2-13=(3x<\/em>+1)-13=3x<\/em>3=x<\/em>=I<\/em>X<\/em>, Identity function

Since, fog<\/em>(y<\/em>) and gof<\/em>(x<\/em>) are identity function.

Thus, f<\/em> is invertible.

So, f<\/em>-1(x<\/em>)=g<\/em>(x<\/em>)=\u221ax<\/em>+6-13.

Now,

f <\/em>-1<\/sup>(43) = \u221a43+6-13=\u221a49-13=7-13=63=2

And f <\/em>-1<\/sup>(163) = \u221a163+6-13=\u221a169-13=13-13=123=4<\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 16:<\/h4>\n\n\n\n

Let f  <\/em>: R <\/em>-{-43} \u2192R<\/em> be a function defined as f(x)<\/em> <\/em>=4x<\/em>3x<\/em>+4 . Show that
     f : R <\/em>-{-43} \u2192 Rang (f<\/em>) is one-one and onto. Hence, find f <\/em>-1<\/sup>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

The function f<\/em>:R<\/strong>-{-43}\u2192R<\/strong>-{43} is given by f<\/em>(x<\/em>)=4x<\/em>3x<\/em>+4.
Injectivity: Let x<\/em>, y<\/em>\u2208R<\/strong>-{-43} be such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d24x<\/em>3x<\/em>+4=4y<\/em>3y<\/em>+4\u21d24x<\/em>(3y<\/em>+4)=4y<\/em>(3x<\/em>+4)\u21d212x<\/em>y<\/em>+16x<\/em>=12x<\/em>y<\/em>+16y<\/em>\u21d216x<\/em>=16y<\/em>\u21d2x<\/em>=y<\/em>
Hence, f <\/em>is one-one function.
Surjectivity: Let y<\/em> be an arbitrary element of R<\/strong>-{43}. Then,
f<\/em>(x<\/em>) = y<\/em>
\u21d24x<\/em>3x<\/em>+4=y<\/em>\u21d24x<\/em>=3x<\/em>y<\/em>+4y<\/em>\u21d24x<\/em>-3x<\/em>y<\/em>=4y<\/em>\u21d2x<\/em>=4y<\/em>4-3y<\/em>
As y<\/em>\u2208R<\/strong>-{43}, 4y<\/em>4-3y<\/em>\u2208R<\/strong>.
Also, 4y<\/em>4-3y<\/em>\u2260-43 because 4y<\/em>4-3y<\/em>=-43\u21d212y<\/em>=-16+12y<\/em>\u21d20=-16, which is not possible.
Thus,
x<\/em>=4y<\/em>4-3y<\/em>\u2208R<\/strong>-{-43} such that
f<\/em>(x<\/em>)=f<\/em>(4x<\/em>3x<\/em>+4)=4(4y<\/em>4-3y<\/em>)3(4y<\/em>4-3y<\/em>)+4=16y<\/em>12y<\/em>+16-12y<\/em>=16y<\/em>16=y<\/em>, so every element in R<\/strong>-{43} has pre-image in R<\/strong>-{-43}.
Hence, f<\/em> is onto.
Now,
x<\/em>=4y<\/em>4-3y<\/em>
Replacing x<\/em> by f<\/em>-1(x<\/em>) and y<\/em> by x<\/em>, we have
 f<\/em>-1(x<\/em>)=4x<\/em>4-3x<\/em>   <\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 17:<\/h4>\n\n\n\n

Let A<\/em> = R<\/em> \u2013 {2} and B<\/em> = R<\/em> \u2013 {1}. If f<\/em> : A<\/em> \u2192 B<\/em> is a function defined by f<\/em>(x<\/em>)=x<\/em>-1x<\/em>-2, show that f<\/em> is one-one and onto. Find f\u2013<\/sup><\/em>1<\/sup>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>)=x<\/em>-1x<\/em>-2

To show f<\/em> is one-one:

Let f<\/em>(x<\/em>1)=f<\/em>(x<\/em>2)\u21d2x<\/em>1-1x<\/em>1-2=x<\/em>2-1x<\/em>2-2\u21d2(x<\/em>1-1)(x<\/em>2-2)=(x<\/em>2-1)(x<\/em>1-2)\u21d2x<\/em>1x<\/em>2-2x<\/em>1-x<\/em>2+2=x<\/em>1x<\/em>2-2x<\/em>2-x<\/em>1+2\u21d2-2x<\/em>1-x<\/em>2=-2x<\/em>2-x<\/em>1\u21d2-2x<\/em>1+x<\/em>1=-2x<\/em>2+x<\/em>2\u21d2-x<\/em>1=-x<\/em>2\u21d2x<\/em>1=x<\/em>2Hence, f<\/em> is one-one.To show <\/em>f<\/em> is onto: Let y<\/em>\u2208B<\/em>\u2234 y<\/em>=f<\/em>(x<\/em>)\u21d2y<\/em>=x<\/em>-1x<\/em>-2\u21d2y<\/em>(x<\/em>-2)=x<\/em>-1\u21d2x<\/em>y<\/em>-2y<\/em>=x<\/em>-1\u21d2x<\/em>y<\/em>–x<\/em>=2y<\/em>-1\u21d2x<\/em>(y<\/em>-1)=2y<\/em>-1\u21d2x<\/em>=2y<\/em>-1y<\/em>-1Thus, for every value of y<\/em> in R<\/em>-{1}, there exists a pre-image x<\/em>=2y<\/em>-1y<\/em>-1 in R<\/em>-{2}.Hence, f<\/em> is onto.

Since, f <\/em>is one-one and onto
Therefore, f <\/em>is invertible with f<\/em>-1(y<\/em>)=2y<\/em>-1y<\/em>-1.

Hence, f<\/em>-1(x<\/em>)=2x<\/em>-1x<\/em>-1.

 <\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 18:<\/h4>\n\n\n\n

Show that the function f <\/em>: N<\/em> \u2192 N<\/em> defined by f<\/em>(x<\/em>) = x<\/em>2<\/sup> + x<\/em> + 1 is one-one but not onto. Find the inverse of f <\/em>: N<\/em> \u2192 S<\/em>, where S<\/em> is range of f<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: The function f <\/em>: N<\/em> \u2192 N<\/em> defined by f<\/em>(x<\/em>) = x<\/em>2<\/sup> + x<\/em> + 1

To show f<\/em> is one-one:

Let f<\/em>(x<\/em>1)=f<\/em>(x<\/em>2)\u21d2x<\/em>21+x<\/em>1+1=x<\/em>22+x<\/em>2+1\u21d2x<\/em>21+x<\/em>1=x<\/em>22+x<\/em>2\u21d2x<\/em>21+x<\/em>1-x<\/em>22-x<\/em>2=0\u21d2x<\/em>21-x<\/em>22+x<\/em>1-x<\/em>2=0\u21d2(x<\/em>1-x<\/em>2)(x<\/em>1+x<\/em>2)+(x<\/em>1-x<\/em>2)=0\u21d2(x<\/em>1-x<\/em>2)(x<\/em>1+x<\/em>2+1)=0\u21d2x<\/em>1-x<\/em>2=0 or x<\/em>1+x<\/em>2+1=0\u21d2x<\/em>1=x<\/em>2 or x<\/em>1=-(x<\/em>2+1)\u21d2x<\/em>1=x<\/em>2         (\u2235 x<\/em>1, x<\/em>2\u2208N<\/em>)Hence, f<\/em> is one-one.To show <\/em>f<\/em> is not onto: Since f<\/em>(x<\/em>)=x<\/em>2+x<\/em>+1\u2234 f<\/em>(1)=3f<\/em>(2)=7f<\/em>(3)=13and so onThus, Range of f<\/em>={3, 7, 13, …}\u2260N<\/em>Hence, f<\/em> is not onto.Now, Let f<\/em>:N<\/em>\u2192Range of f<\/em>y<\/em>=x<\/em>2+x<\/em>+1\u21d2x<\/em>2+x<\/em>+1-y<\/em>=0\u21d2x<\/em>2+x<\/em>+(1-y<\/em>)=0\u21d2x<\/em>=-1\u00b1\u221a12-4(1)(1-y<\/em>)2(1)\u21d2x<\/em>=-1\u00b1\u221a1-4+4y<\/em>2\u21d2x<\/em>=-1\u00b1\u221a4y<\/em>-32\u21d2x<\/em>=-1+\u221a4y<\/em>-32 or x<\/em>=-1-\u221a4y<\/em>-32\u21d2x<\/em>=-1+\u221a4y<\/em>-32        (\u2235 x<\/em>\u2208N<\/em>)Hence, f<\/em>-1(x<\/em>)=-1+\u221a4x<\/em>-32.


 <\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 19:<\/h4>\n\n\n\n

Let f<\/em> : [\u22121, \u221e) \u2192 [\u22121, \u221e) be given by f<\/em>(x<\/em>) = (x<\/em> + 1)2<\/sup> \u2212 1, x<\/em> \u2265 \u22121. Show that f<\/em> is invertible. Also, find the set S<\/em> = {x<\/em> : f<\/em>(x<\/em>) = f<\/em>\u22121<\/sup> (x<\/em>)}.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity: Let x and y \u2208[-1, \u221e), such that f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2(x<\/em>+1)2-1=(y<\/em>+1)2-1\u21d2(x<\/em>+1)2=(y<\/em>+1)2\u21d2(x<\/em>+1)=(y<\/em>+1)\u21d2x<\/em>=y<\/em>So, f<\/em> is a injection.Surjectivity: Let y \u2208[-1, \u221e). Then, f(x)=y\u21d2(x+1)2-1=y\u21d2x+1=\u221ay+1\u21d2x=\u221ay+1-1Clearly, x=\u221ay+1-1 is real for all y\u2265-1.Thus, every element y \u2208[-1, \u221e) has its pre-image x\u2208[-1, \u221e) given by x=\u221ay+1-1.\u21d2f<\/em> is a surjection.So, f<\/em> is a bijection.Hence, f<\/em> is invertible.Let f<\/em>-1(x<\/em>)=y<\/em>                          …(1)\u21d2f<\/em>(y<\/em>)=x<\/em>\u21d2(y<\/em>+1)2-1=x<\/em>\u21d2(y<\/em>+1)2=x<\/em>+1\u21d2y<\/em>+1=\u221ax<\/em>+1\u21d2y<\/em>=\u00b1\u221ax<\/em>+1-1\u21d2f<\/em>-1(x<\/em>)=\u00b1\u221ax<\/em>+1-1   [from (1)]f<\/em>(x<\/em>)=f<\/em>-1(x<\/em>)\u21d2(x<\/em>+1)2-1=\u00b1\u221ax<\/em>+1-1\u21d2(x<\/em>+1)2=\u00b1\u221ax<\/em>+1\u21d2(x<\/em>+1)4=x<\/em>+1\u21d2(x<\/em>+1)[(x<\/em>+1)3-1]=0\u21d2x<\/em>+1=0 or (x<\/em>+1)3-=0\u21d2x<\/em>=-1 or (x<\/em>+1)3=1\u21d2x<\/em>=-1 or x<\/em>+1=1\u21d2x<\/em>=-1 or x=0\u21d2S<\/em>={0, -1}<\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 20:<\/h4>\n\n\n\n

Let A<\/em> = {x<\/em> &epsis; R<\/em> | \u22121 \u2264 x<\/em> \u2264 1} and let f<\/em> : A<\/em> \u2192 A<\/em>, g<\/em> : A<\/em> \u2192 A<\/em> be two functions defined by f<\/em>(x<\/em>) = x<\/em>2<\/sup> and g<\/em>(x<\/em>) = sin (\u03c0<\/em> x<\/em>\/2). Show that g<\/em>\u22121<\/sup> exists but f<\/em>\u22121<\/sup> does not exist. Also, find g<\/em>\u22121.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

 f<\/em> is not one-one because
f<\/em>(-1)=(-1)2=1and f<\/em>(1)=12=1
\u21d2 -1 and 1 have the same image under f<\/em>.
\u21d2 f<\/em> is not a bijection.
So, f<\/em> -1 <\/sup>does not exist.

Injectivity of g:
Let x<\/em> and y<\/em> be any two elements in the domain (A<\/em>), such that
g<\/em>(x<\/em>)=g<\/em>(y<\/em>)\u21d2sin (\u03c0x2)=sin (\u03c0y2)\u21d2(\u03c0x2)=(\u03c0y2)\u21d2x<\/em>=y<\/em>
So, g <\/em>is one-one.

Surjectivity of g:
Range of g<\/em> = [sin (\u03c0(-1)2), sin (\u03c0(1)2)] = [sin (-\u03c02), sin (\u03c02)] = [-1, 1] = A <\/em>(co-domain of g<\/em>)
\u21d2g<\/em> is onto.
\u21d2g<\/em> is a bijection.
So, g<\/em>-1<\/sup> exists.

Also,
let g<\/em>-1(x<\/em>)=y<\/em>                      …(1)\u21d2g<\/em>(y<\/em>)=x<\/em>\u21d2sin(\u03c0y2)=x<\/em>\u21d2(\u03c0y2)=sin-1 x<\/em>\u21d2y<\/em>=2\u03c0sin-1 x<\/em>\u21d2g<\/em>-1(x<\/em>)=2\u03c0sin-1 x<\/em>        [from (1)]<\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 21:<\/h4>\n\n\n\n

Let f<\/em> be a function from R<\/em> to R,<\/em> such that f<\/em>(x<\/em>) = cos (x<\/em> + 2). Is f<\/em> invertible? Justify your answer.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity:
Let x <\/em>and y<\/em> be two elements in the domain (R<\/em>), such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2cos(x<\/em>+2)=cos(y<\/em>+2)\u21d2x<\/em>+2=y<\/em>+2 or x<\/em>+2=2\u03c0-(y<\/em>+2)\u21d2x<\/em>=y <\/em>or x<\/em>+2=2\u03c0-y<\/em>-2\u21d2x<\/em>=y <\/em>or x<\/em>=2\u03c0-y<\/em>-4So, we cannot say that x=yFor example,cos\u03c02=cos 3\u03c02=0So,\u03c02 and 3\u03c02  have the same image 0.
\u21d2 f <\/em>is not one-one.
\u21d2 f<\/em> is not a bijection.
Thus, f<\/em>  is not invertible.<\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 22:<\/h4>\n\n\n\n

If A<\/em> = {1, 2, 3, 4} and B<\/em> = {a<\/em>, b<\/em>, c<\/em>, d<\/em>}, define any four bijections from A<\/em> to B<\/em>. Also give their inverse functions.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em>1={(1, a<\/em>), (2, b<\/em>), (3, c<\/em>), (4, d<\/em>)}\u21d2f<\/em>1-1={(a<\/em>, 1), (b<\/em>, 2), (c<\/em>, 3), (d<\/em>, 4)}f<\/em>2={(1, b<\/em>), (2, a<\/em>), (3, c<\/em>), (4, d<\/em>)}\u21d2f<\/em>2-1={(b<\/em>, 1), (a<\/em>, 2), (c<\/em>, 3), (d<\/em>, 4)}f<\/em>3={(1, a<\/em>), (2, b<\/em>), (4, c<\/em>), (3, d<\/em>)}\u21d2f<\/em>3-1={(a<\/em>, 1), (b<\/em>, 2), (c<\/em>, 4), (d<\/em>, 3)}f<\/em>4={(1, b<\/em>), (2, a<\/em>), (4, c<\/em>), (3, d<\/em>)}\u21d2f<\/em>4-1={(b<\/em>, 1), (a<\/em>, 2), (c<\/em>, 4), (d<\/em>, 3)}
Clearly, all these are bijections because they are one-one and onto.<\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 23:<\/h4>\n\n\n\n

Let A<\/em> and B<\/em> be two sets, each with a finite number of elements. Assume that there is an injective map from A<\/em> to B<\/em> and that there is an injective map from B<\/em> to A<\/em>. Prove that there is a bijection from A<\/em> to B<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

 A<\/em> and B<\/em> are two non empty sets. Let f<\/em> be a function from A<\/em> to B<\/em> .It is given that there is injective map from A<\/em> to B<\/em>. That means f<\/em> is one-one function .It is also given that there is injective map from  B<\/em> to A<\/em> .That means every element of set B<\/em> has its image in set A<\/em>.\u21d2f<\/em> is onto function or surjective.\u2234 f<\/em> is bijective.(If a function is both injective and surjective, then the function is bijective.)     <\/p>\n\n\n\n

Page No 2.69:<\/h4>\n\n\n\n

Question 24:<\/h4>\n\n\n\n

If f<\/em> : A<\/em> \u2192 A<\/em>, g<\/em> : A<\/em> \u2192 A<\/em> are two bijections, then prove that
(i) fog<\/em> is an injection
(ii) fog<\/em> is a surjection<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: A<\/em> \u2192 A<\/em>, g<\/em> : A<\/em> \u2192 A<\/em> are two bijections.
Then,  fog<\/em> : A<\/em> \u2192 A<\/em>

(i) Injectivity of fog<\/em>:
Let x<\/em> and y<\/em> be two elements of the domain (A<\/em>), such that
(f<\/em>o<\/em>g<\/em>)(x<\/em>)=(f<\/em>o<\/em>g<\/em>)(y<\/em>)\u21d2f<\/em>(g<\/em>(x<\/em>))=f<\/em>(g<\/em>(y<\/em>))\u21d2g<\/em>(x<\/em>)=g<\/em>(y<\/em>) (As, f<\/em> is one-one)\u21d2x<\/em>=y<\/em>          (As, g<\/em> is one-one)
So,  fog <\/em>is an injection.

(ii) Surjectivity of fog<\/em>:
Let z<\/em> be an element in the co-domain of fog<\/em> (A<\/em>).
Now, z\u2208A <\/em>(co-domain of f<\/em>) and f<\/em> is a surjection.So, z<\/em>=<\/em>f<\/em>(y<\/em>),<\/em> <\/em>where y\u2208A <\/em>(domain of <\/em>f<\/em>) <\/em>.<\/em>.<\/em>.<\/em>(1)Now, y\u2208A <\/em>(co-domain of g<\/em>) and g<\/em> is a surjection.So, <\/em>y<\/em>=<\/em>g<\/em>(x<\/em>),<\/em> where x\u2208A <\/em>(domain of <\/em>g<\/em>) <\/em>.<\/em>.<\/em>.<\/em>(2)From (1) and (2),z<\/em>=<\/em>f<\/em>(y<\/em>)=<\/em>f<\/em>(g<\/em>(x<\/em>))=<\/em>(f<\/em>o<\/em>g<\/em>)(x<\/em>),<\/em> <\/em>where x<\/em>\u2208<\/em>A<\/em>(domain of f<\/em>o<\/em>g<\/em>)
So,  fog<\/em> is a surjection.<\/p>\n\n\n\n

Page No 2.72:<\/h4>\n\n\n\n

Question 1:<\/h4>\n\n\n\n

Let A<\/em>={x<\/em> \u2208 R<\/em> : -1 \u2264x<\/em>\u22641}=B<\/em> and C<\/em>={x<\/em> \u2208 R<\/em> : x<\/em>\u22650} and
let S<\/em>={(x<\/em>, y<\/em>) \u2208 A<\/em>\u00d7B<\/em> : x<\/em>2+y<\/em>2=1} and S<\/em>0={(x<\/em>, y<\/em>) \u2208 A<\/em>\u00d7C<\/em> : x<\/em>2+y<\/em>2=1}. Then,
(a) S<\/em> defines a function from A<\/em> to B<\/em>
(b) S<\/em>0<\/sub> defines a function from A<\/em> to C<\/em>
(c) S<\/em>0<\/sub> defines a function from A<\/em> to B<\/em>
(d) S<\/em> defines a function from A<\/em> to C<\/em><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(a) S<\/em> defines a function from A<\/em> to B<\/em>

Let x<\/em>\u2208A<\/em>\u21d2-1\u2264x<\/em>\u22641Now, x<\/em>2+y<\/em>2=1\u21d2y<\/em>2=1-x<\/em>2\u21d2y<\/em>=\u00b1\u221a1-x<\/em>2\u21d2-1\u2264y<\/em>\u22641\u2234 y<\/em>\u2208B<\/em>Thus, S<\/em> defines a function from A<\/em> to B<\/em>.<\/p>\n\n\n\n

Page No 2.72:<\/h4>\n\n\n\n

Question 2:<\/h4>\n\n\n\n

f<\/em> : R<\/em>\u2192R<\/em> given by f<\/em>(x<\/em>)=x<\/em>+\u221ax<\/em>2 is
(a) injective
(b) surjective
(c) bijective
(d) None of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em>(x<\/em>)=x<\/em>+\u221ax<\/em>2=x<\/em>\u00b1x<\/em>=0 or 2x<\/em>\u21d2 Each element of the domain has 2 images.
\u21d2f<\/em> is not a function.
So, the answer is (d).<\/p>\n\n\n\n

Page No 2.72:<\/h4>\n\n\n\n

Question 3:<\/h4>\n\n\n\n

If f<\/em> : A<\/em>\u2192B<\/em> given by 3f<\/em>(x<\/em>)+2-x<\/em>=4 is a bijection, then
(a) A<\/em>={x<\/em> \u2208 R<\/em> : -1 < x<\/em> < \u221e}, B<\/em>={x<\/em> \u2208 R<\/em> : 2 < x<\/em> < 4}
(b) A<\/em>={x<\/em> \u2208 R<\/em> : -3 < x<\/em> < \u221e}, B<\/em>={x<\/em> \u2208 R<\/em> : 2 < x<\/em> < 4}
(c) A<\/em>={x<\/em> \u2208 R<\/em> : -2 < x<\/em> < \u221e}, B<\/em>={x<\/em> \u2208 R<\/em> : 2 < x<\/em> < 4}
(d) None of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(d) None of these

f<\/em>:A<\/em>\u2192B<\/em>3f<\/em>(x<\/em>)+2-x<\/em>=4 \u21d23f<\/em>(x<\/em>)=4-2-x<\/em>Taking log on both the sides , f<\/em>(x<\/em>) log 3=log (4-2-x<\/em>)\u21d2f<\/em>(x<\/em>)=log (4-2-x<\/em>)log 3Logaritmic function will only be defined if 4-2-x<\/em>>0\u21d24>2-x<\/em>\u21d222>2-x<\/em>\u21d22>-x<\/em>\u21d2-2<x<\/em>\u21d2x<\/em>\u2208 (-2,\u221e)That means  A<\/em>={x<\/em>\u2208R<\/em>:-2<x<\/em><\u221e}As we know that, f<\/em>(x<\/em>)=log (4-2-x<\/em>)log 3We take x<\/em>=0\u2208 (-2,\u221e)\u21d2f<\/em>(x<\/em>)=1 which does not belong to any of the options .<\/p>\n\n\n\n

Page No 2.72:<\/h4>\n\n\n\n

Question 4:<\/h4>\n\n\n\n

The function f<\/em> : R<\/em> \u2192 R<\/em> defined by f<\/em>(x<\/em>)=2x<\/em>+2|x<\/em>| is
(a) one-one and onto
(b) many-one and onto
(c) one-one and into
(d) many-one and into<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(d) many-one and into

Graph for the given function is as follows.

\"\"
A line parallel to X axis is cutting the graph at two different values.

Therefore, for two different values of x<\/em> we are getting the same value of y<\/em>.
That means it is many one function.

From the given graph we can see that the range is [2,\u221e) 
and R is the co-domain of the given function.
Hence, Co-domain\u2260Range
Therefore, the given function is into.



 <\/p>\n\n\n\n

Page No 2.72:<\/h4>\n\n\n\n

Question 5:<\/h4>\n\n\n\n

Let the function f<\/em> : R<\/em>-{-b<\/em>}\u2192R<\/em>-{1} be defined by

f<\/em>(x<\/em>)=x<\/em>+a<\/em>x<\/em>+b<\/em>, a<\/em>\u2260b<\/em>. Then,

(a) f<\/em> is one-one but not onto
(b) f<\/em> is onto but not one-one
(c) f<\/em> is both one-one and onto
(d) None of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(c) f<\/em> is both one-one and onto

Injectivity:
Let x<\/em> and y<\/em> be two elements in the domain R- <\/em>{-b<\/em>}, such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>+a<\/em>x<\/em>+b<\/em>=y<\/em>+a<\/em>y<\/em>+b<\/em>\u21d2(x<\/em>+a<\/em>)(y<\/em>+b<\/em>)=(x<\/em>+b<\/em>)(y<\/em>+a<\/em>)\u21d2x<\/em>y<\/em>+b<\/em>x<\/em>+a<\/em>y<\/em>+a<\/em>b<\/em>=x<\/em>y<\/em>+a<\/em>x<\/em>+b<\/em>y<\/em>+a<\/em>b<\/em>\u21d2b<\/em>x<\/em>+a<\/em>y<\/em>=a<\/em>x<\/em>+b<\/em>y<\/em>\u21d2(a<\/em>–b<\/em>)x<\/em>=(a<\/em>–b<\/em>)y<\/em>\u21d2x<\/em>=y<\/em>
So, f<\/em> is one-one.

Surjectivity:
Let y <\/em>be an element in the co-domain of f,<\/em> i.e. R-<\/em>{1}, such that f <\/em>(x<\/em>)=y<\/em>

f<\/em>(x<\/em>)=y<\/em>\u21d2x<\/em>+a<\/em>x<\/em>+b<\/em>=y<\/em>\u21d2x<\/em>+a<\/em>=y<\/em>x<\/em>+y<\/em>b<\/em>\u21d2x<\/em>–y<\/em>x<\/em>=y<\/em>b<\/em>–a<\/em>\u21d2x<\/em>(1-y<\/em>)=y<\/em>b<\/em>–a<\/em>\u21d2x<\/em>=y<\/em>b<\/em>–a<\/em>1-y<\/em>\u2208R<\/em>-{-b<\/em>}
So,  f <\/em>is onto.<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 6:<\/h4>\n\n\n\n

The function f<\/em> : A<\/em>\u2192B<\/em> defined by f<\/em>(x<\/em>)=-x<\/em>2+6x<\/em>-8 is a bijection if
(a) A<\/em>=(-\u221e, 3] and B<\/em>=(-\u221e, 1]
(b) A<\/em>=[-3, \u221e) and B<\/em>=(-\u221e, 1]
(c) A<\/em>=(-\u221e, 3] and B<\/em>=[1, \u221e)
(d) A<\/em>=[3, \u221e) and B<\/em>=[1, \u221e)<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(a) A<\/em>=(-\u221e, 3] and B<\/em>=(-\u221e, 1]


f<\/em>(x<\/em>)=-x<\/em>2+6x<\/em>-8 , is a polynomial functionAnd the domain of polynomial function is real number.\u2234x<\/em>\u2208R<\/em>

f<\/em>(x<\/em>) =-x<\/em>2+6x<\/em>-8       =-(x<\/em>2-6x<\/em>+8)       =-(x<\/em>2-6x<\/em>+9-1)       =-(x<\/em>-3)2+1Maximum value of -(x<\/em>-3)2 woud be 0\u2234Maximum value of -(x<\/em>-3)2+1 woud be 1\u2234 <\/em>f<\/em>(x<\/em>) <\/em>\u2208(-\u221e,1]

\"\"

We can see from the given graph that function is symmetrical about x<\/em>=3& the given function is bijective .So, x<\/em> would be either (-\u221e,3] or [3,\u221e)The correct option which satisfy A and B both is: 
A<\/em>=(-\u221e, 3] and B<\/em>=(-\u221e, 1]<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 7:<\/h4>\n\n\n\n

Let A<\/em>={x<\/em> \u2208 R<\/em> : -1\u2264x<\/em>\u22641}=B<\/em>. Then, the mapping f<\/em> : A<\/em>\u2192B<\/em> given by f<\/em>(x<\/em>)=x<\/em>|x<\/em>| is
(a) injective but not surjective
(b) surjective but not injective
(c) bijective
(d) none of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity:
Let x<\/em> and y<\/em> be any two elements in the domain A.<\/em>

Case-1: Let x<\/em> and y<\/em> be two positive numbers, such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>|x<\/em>|=y<\/em>|y<\/em>|\u21d2x<\/em>(x<\/em>)=y<\/em>(y<\/em>)\u21d2x<\/em>2=y<\/em>2\u21d2x<\/em>=y<\/em>

Case-2: Let x<\/em> and y<\/em> be two negative numbers, such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>|x<\/em>|=y<\/em>|y<\/em>|\u21d2x<\/em>(-x<\/em>)=y<\/em>(-y<\/em>)\u21d2-x<\/em>2=-y<\/em>2\u21d2x<\/em>2=y<\/em>2\u21d2x<\/em>=y<\/em>

Case-3: Let x <\/em>be positive and y<\/em> be negative.
Then, x<\/em>\u2260y<\/em>\u21d2f<\/em>(x<\/em>)=x<\/em>|x<\/em>| is positive and f<\/em>(y<\/em>)=y<\/em>|y<\/em>| is negative\u21d2f<\/em>(x<\/em>)\u2260f<\/em>(y<\/em>)So, x\u2260y<\/em>\u21d2f<\/em>(x<\/em>)\u2260f<\/em>(y<\/em>)
From the 3 cases, we can conclude that  f<\/em> is one-one.

Surjectivity:
Let y<\/em> be an element in the co-domain, such that y<\/em> = f<\/em> (x<\/em>)
Case-1: Let y<\/em>>0. Then, 0<y<\/em>\u22641\u21d2y<\/em>=f<\/em>(x<\/em>)=x<\/em>|x<\/em>|>0\u21d2x<\/em>>0\u21d2|x<\/em>|=x<\/em>f<\/em>(x<\/em>)=y<\/em>\u21d2x<\/em>|x<\/em>|=y<\/em>\u21d2x<\/em>(x<\/em>)=y<\/em>\u21d2x<\/em>2=y<\/em>\u21d2x<\/em>=\u221ay<\/em> \u2208 A<\/em> <\/em>(We do not get \u00b1 because x<\/em>>0)Case-2: Let <\/em>y<\/em><0. Then, -1\u2264y<\/em><0\u21d2y<\/em>=f<\/em>(x<\/em>)=x<\/em>|x<\/em>|<0\u21d2x<\/em><0\u21d2|x<\/em>|=-x<\/em>f<\/em>(x<\/em>)=y<\/em>\u21d2x<\/em>|x<\/em>|=y<\/em>\u21d2x<\/em>(-x<\/em>)=y<\/em>\u21d2-x<\/em>2=y<\/em>\u21d2x<\/em>2=-y<\/em>\u21d2x<\/em>=-\u221a-y<\/em> \u2208 A<\/em> <\/em>(We do not get \u00b1 because x<\/em>>0)
\u21d2f <\/em>is onto.
\u21d2f <\/em>is a bijection.<\/em>
So, the answer is (c).<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 8:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em>\u2192R<\/em> be given by f<\/em>(x<\/em>)=[x<\/em>2]+[x<\/em>+1]-3 where [x<\/em>] denotes the greatest integer less than or equal to x<\/em>. Then, f<\/em>(x<\/em>) is
(a) many-one and onto
(b) many-one and into
(c) one-one and into
(d) one-one and onto<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(b) many-one and into

f<\/em> : R<\/em>\u2192R<\/em>

f<\/em>(x<\/em>)=[x<\/em>2]+[x<\/em>+1]-3

It is many one function because in this case for two different values of x<\/em>
we would get the same value of f<\/em>(x<\/em>) .
For x<\/em>=1.1, 1.2 \u2208R<\/em>f<\/em>(1.1)=[(1.1)2] +[1.1+1]-3           =[1.21]+[2.1]-3           =1+2-3           =0f<\/em>(1.1)=[(1.2)2] +[1.2+1]-3           =[1.44]+[2.2]-3           =1+2-3           =0<\/p>\n\n\n\n


It is into function because for the given domain we would only get the integral values of
f<\/em>(x<\/em>).
but R<\/em> is the codomain of the given function.
That means , Codomain\u2260Range
Hence, the given function is into function.

Therefore, f<\/em>(x<\/em>) is many one and into<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 9:<\/h4>\n\n\n\n

Let M<\/em> be the set of all 2 \u00d7 2 matrices with entries from the set R<\/em> of real numbers. Then, the function f<\/em> : M<\/em> \u2192 R<\/em> defined by f<\/em>(A<\/em>) = |A<\/em>| for every A<\/em> \u2208 M<\/em>, is
(a) one-one and onto
(b) neither one-one nor onto
(c) one-one but-not onto
(d) onto but not one-one<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n


M<\/em>={A<\/em>=[a<\/em>b<\/em>c<\/em>d<\/em>]: a<\/em>, b<\/em>, c<\/em>, d<\/em>\u2208R<\/em> }f<\/em>: M<\/em>\u2192R<\/em> is given by f<\/em>(A<\/em>)=<\/em>|A<\/em>|

Injectivity:
f<\/em>([0000])=|0000|=0and f<\/em>([1000])=|1000|=0\u21d2f<\/em>([0000])=f<\/em>([1000])=0
So, f<\/em> is not one-one.

Surjectivity:
Let y<\/em> be an element of the co-domain, such that
f<\/em>(A<\/em>)=-y<\/em>, A<\/em>=[a<\/em>b<\/em>c<\/em>d<\/em>]\u21d2|a<\/em>b<\/em>c<\/em>d<\/em>|=y<\/em>\u21d2a<\/em>d<\/em>–b<\/em>c<\/em>=y<\/em>\u21d2a<\/em>, b<\/em>, c<\/em>, d<\/em>\u2208R<\/em> \u21d2A<\/em>=[a<\/em>b<\/em>c<\/em>d<\/em>]\u2208M<\/em>
\u21d2f<\/em> is onto.
So, the answer is (d).<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 10:<\/h4>\n\n\n\n

The function f<\/em> : [0, \u221e)\u2192R<\/em> given by f<\/em>(x<\/em>)=x<\/em>x<\/em>+1 is
(a) one-one and onto
(b) one-one but not onto
(c) onto but not one-one
(d) onto but not one-one<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity:
Let x<\/em> and y<\/em> be two elements in the domain, such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>x<\/em>+1=y<\/em>y<\/em>+1\u21d2x<\/em>y<\/em>+x<\/em>=x<\/em>y<\/em>+y<\/em>\u21d2x<\/em>=y<\/em>
So, f<\/em> is one-one.

Surjectivity:
Let y<\/em> be an element in the co domain R<\/em>, such that
y<\/em>=f<\/em>(x<\/em>)\u21d2y<\/em>=x<\/em>x<\/em>+1\u21d2x<\/em>y<\/em>+y<\/em>=x<\/em>\u21d2x<\/em>(y<\/em>-1)=-y<\/em>\u21d2x<\/em>=-y<\/em>y<\/em>-1Range of f<\/em>=R<\/em>-{1}\u2260co domain (R<\/em>)
\u21d2f<\/em> is not onto.
So, the answer is (b).<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 11:<\/h4>\n\n\n\n

The range of the function f<\/em>(x<\/em>)=7-x<\/em>P<\/em>x<\/em>-3 is
(a) {1, 2, 3, 4, 5}
(b) {1, 2, 3, 4, 5, 6}
(c) {1, 2, 3, 4}
(d) {1, 2, 3}<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We know that
7-x<\/em>>0; x<\/em>-3 \u22650 and 7-x<\/em>\u2265x<\/em>-3\u21d2x<\/em><7; x<\/em>\u22653 and 2x<\/em>\u226410\u21d2x<\/em><7; x<\/em>\u22653 and x<\/em>\u22645So, x<\/em>={3, 4, 5}Range of f<\/em>=<\/em>{P<\/em>(3<\/em>–<\/em>3<\/em>)(7<\/em>–<\/em>3<\/em>),<\/em> <\/em>P<\/em>(4<\/em>–<\/em>3<\/em>)(7<\/em>–<\/em>4<\/em> <\/em>),<\/em> <\/em>P<\/em>(7<\/em>–<\/em>5<\/em>)(5<\/em>–<\/em>3<\/em>)}=<\/em>{4<\/em>P<\/em>0<\/em>,<\/em> <\/em>3<\/em>P<\/em>1<\/em>,<\/em> <\/em>2<\/em>P<\/em>2<\/em>}=<\/em>{1<\/em>,<\/em> <\/em>3<\/em>,<\/em> <\/em>2<\/em>}=<\/em>{1<\/em>,<\/em> <\/em>2<\/em>,<\/em> <\/em>3<\/em>}
So, the answer is (d).<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 12:<\/h4>\n\n\n\n

A function f<\/em>  from the set of natural numbers to integers defined by

f<\/em>(n<\/em>)={n<\/em>-12, when n<\/em> is odd-n<\/em>2, when n<\/em> is evenis

(a) neither one-one nor onto
(b) one-one but not onto
(c) onto but not one-one
(d) one-one and onto both<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(d) one-one and onto both

Injectivity:

Let x<\/em> and y<\/em> be any two elements in the domain (N<\/em>).
Case-1: Both x<\/em> and y<\/em> are even.Let f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2-x<\/em>2=-y<\/em>2\u21d2-x<\/em>=-y<\/em>\u21d2x<\/em>=y<\/em>Case-2: Both x<\/em> and y<\/em> are odd.Let f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>-12=y<\/em>-12\u21d2x<\/em>-1=y<\/em>-1\u21d2x<\/em>=y<\/em>Case-3: Let x<\/em> be even and y<\/em> be odd. Then, f<\/em>(x<\/em>)=-x<\/em>2and f<\/em>(y<\/em>)=y<\/em>-12Then, clearly x<\/em>\u2260y<\/em> \u21d2f<\/em>(x<\/em>)\u2260f<\/em>(y<\/em>)From all the cases, f<\/em> is one-one.

Surjectivity:

Co-domain of f<\/em>=Z<\/em>={…,-3, -2, -1, 0, 1, 2, 3, ….}Range of f<\/em>={…, -3-12, -(-2)2,-1-12, 02, 1-12, -22, 3-12, …}\u21d2Range of f<\/em>={…,-2, 1, -1, 0, 0, -1, 1,…}\u21d2Range of f<\/em>={…, -2, -1, 0, 1, 2, ….}\u21d2Co-domain of f<\/em>=Range of f<\/em><\/em><\/strong>
\u21d2 f<\/em> is onto.<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 13:<\/h4>\n\n\n\n

Let f<\/em> be an injective map with domain {x<\/em>, y<\/em>, z<\/em>} and range {1, 2, 3}, such that exactly one of the following statements is correct and the remaining are false.

f<\/em>(x<\/em>)=1, f<\/em>(y<\/em>)\u22601, f<\/em>(z<\/em>)\u22602.

The value of f<\/em>-1 (1) is
(a) x<\/em>
(b) y<\/em>
(c) z<\/em>
(d) none of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Case-1: Let f<\/em>(x<\/em>)=1 be true.Then, f<\/em>(y<\/em>)\u22601 and f<\/em>(z<\/em>)\u22602 are false.So, f<\/em>(y<\/em>)=1 and f<\/em>(z<\/em>)=2\u21d2 f<\/em>(x<\/em>)=1, f<\/em>(y<\/em>)=1\u21d2x<\/em> and y<\/em> have the same images.This contradicts the fact that f<\/em>is one-one.Case-2: Let f<\/em>(y<\/em>)\u22601 be true.Then, f<\/em>(x<\/em>)=1 and f<\/em>(z<\/em>)\u22602 are false.So,  f<\/em>(x<\/em>)\u22601 and f<\/em>(z<\/em>)=2\u21d2 f<\/em>(x<\/em>)\u22601, f<\/em>(y<\/em>)\u22601 and f<\/em>(z<\/em>)=2\u21d2There is no pre-image for 1.This contradicts the fact that range is {1, 2, 3}.Case-3: Let f<\/em>(z<\/em>)\u22602 be true.Then, f<\/em>(x<\/em>)=1 and f<\/em>(y<\/em>)\u22601 are false.So,  f<\/em>(x<\/em>)\u22601 and f<\/em>(y<\/em>)=1\u21d2f<\/em>(x<\/em>)=2, f<\/em>(y<\/em>)=1 and f<\/em>(z<\/em>)=3\u21d2f<\/em>(y<\/em>)=1\u21d2f<\/em>-1(1)=y<\/em>
So, the answer is (b).<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 14:<\/h4>\n\n\n\n

Which of the following functions form Z to itself are bijections?
(a) f<\/em>(x<\/em>)=x<\/em>3
(b) f<\/em>(x<\/em>)=x<\/em>+2
(c) f<\/em>(x<\/em>)=2x<\/em>+1
(d) f<\/em>(x<\/em>)=x<\/em>2+x<\/em><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(a<\/em>) f<\/em> is not onto because for y <\/em>= 3\u2208Co-domain(Z<\/em>), there is no value of x\u2208Domain(Z)x<\/em>3=3\u21d2x<\/em>=3\u221a3\u2209Z<\/em>\u21d2f<\/em> is not onto.So, f<\/em>is not a bijection.

(b) Injectivity:
Let x<\/em> and y<\/em> be two elements of the domain (Z<\/em>), such that
 x<\/em>+2=y<\/em>+2\u21d2x<\/em>=y<\/em>
So, f<\/em> is one-one.

Surjectivity:
Let y<\/em> be an element in the co-domain (Z<\/em>), such that
y<\/em>=f<\/em>(x<\/em>)\u21d2y<\/em>=x<\/em>+2\u21d2x<\/em>=y<\/em>-2\u2208Z<\/em> (Domain)
\u21d2f<\/em> is onto.
So, f<\/em> is a bijection.

(c<\/em>) f<\/em>(x<\/em>)=2x<\/em>+1 is not onto because if we take 4 \u2208 Z<\/em>(co domain), then 4=f<\/em>(x<\/em>)\u21d24=2x<\/em>+1\u21d22x<\/em>=3\u21d2x<\/em>=32\u2209Z<\/em>So, f<\/em>  is not a bijection.(d<\/em>) f<\/em>(0)=02+0=0and f<\/em>(-1)=(-1)2+(-1)=1-1=0\u21d20 and -1 have the same image.\u21d2f<\/em> is not one-one.So, f<\/em> is not a bijection.

So, the answer is (b).<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 15:<\/h4>\n\n\n\n

Which of the following functions from A<\/em>={x<\/em> : -1\u2264x<\/em>\u22641} to itself are bijections?
(a) f<\/em>(x<\/em>)=x<\/em>2
(b) g<\/em>(x<\/em>)=sin(\u03c0<\/em> x<\/em>2)
(c) h<\/em>(x<\/em>)=|x<\/em>|
(d) k<\/em>(x<\/em>)=x<\/em>2<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(a<\/em>) Range of f<\/em> =[-12, 12] \u2260 A<\/em>So, f<\/em> is not a bijection.(b<\/em>) Range =[sin(-\u03c02), sin(\u03c02)]=[-1,1]=A<\/em>So, g<\/em> is a bijection.(c<\/em>) h<\/em>(-1)=|-1|=1and h<\/em>(1)=|1|=1\u21d2-1 and 1 have the same imagesSo, h<\/em> is not a bijection.(d<\/em>) k<\/em>(-1)=(-1)2=1and k<\/em>(1)=(1)2=1\u21d2-1 and 1 have the same imagesSo, k<\/em> is not a bijection.
So, the answer is (b).<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 16:<\/h4>\n\n\n\n

Let A<\/em>={x<\/em> : -1\u2264x<\/em>\u22641} a<\/em>n<\/em>d<\/em> f<\/em> : A<\/em>\u2192A<\/em> such that f<\/em>(x<\/em>)=x<\/em>|x<\/em>|, then f<\/em> is
(a) a bijection
(b) injective but not surjective
(c) surjective but not injective
(d) neither injective nor surjective<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity:
Let x<\/em> and y<\/em> be any two elements in the domain A<\/em>.

Case-1: Let x<\/em> and y<\/em> be two positive numbers, such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>|x<\/em>|=y<\/em>|y<\/em>|\u21d2x<\/em>(x<\/em>)=y<\/em>(y<\/em>)\u21d2x<\/em>2=y<\/em>2\u21d2x<\/em>=y<\/em>

Case-2: Let x<\/em> and y<\/em> be two negative numbers, such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>|x<\/em>|=y<\/em>|y<\/em>|\u21d2x<\/em>(-x<\/em>)=y<\/em>(-y<\/em>)\u21d2-x<\/em>2=-y<\/em>2\u21d2x<\/em>2=y<\/em>2\u21d2x<\/em>=y<\/em>

Case-3: Let x <\/em>be positive and y<\/em> be negative.
Then, x<\/em>\u2260y<\/em>\u21d2f<\/em>(x<\/em>)=x<\/em>|x<\/em>| is positive and f<\/em>(y<\/em>)=y<\/em>|y<\/em>| is negative\u21d2f<\/em>(x<\/em>)\u2260f<\/em>(y<\/em>)So, x\u2260y<\/em>\u21d2f<\/em>(x<\/em>)\u2260f<\/em>(y<\/em>)
So, f<\/em> is one-one.

Surjectivity:
Let y<\/em> be an element in the co-domain, such that y<\/em> = f<\/em> (x<\/em>)
Case-1: Let y<\/em>>0. Then, 0<y<\/em>\u22641y<\/em>=f<\/em>(x<\/em>)=x<\/em>|x<\/em>|>0\u21d2x<\/em>>0\u21d2|x<\/em>|=x<\/em>\u21d2f<\/em>(x<\/em>)=y<\/em>\u21d2x<\/em>|x<\/em>|=y<\/em>\u21d2x<\/em>(x<\/em>)=y<\/em>\u21d2x<\/em>2=y<\/em>\u21d2x<\/em>=\u221ay<\/em> \u2208 A<\/em> <\/em>(We do not get <\/em>\u00b1, <\/em>as x<\/em>>0)Case-2: Let <\/em>y<\/em><0. <\/em>Then, <\/em>-1\u2264y<\/em><0y<\/em>=f<\/em>(x<\/em>)=x<\/em>|x<\/em>|<0\u21d2x<\/em><0\u21d2|x<\/em>|=-x<\/em>\u21d2f<\/em>(x<\/em>)=y<\/em>\u21d2x<\/em>|x<\/em>|=y<\/em>\u21d2x<\/em>(-x<\/em>)=y<\/em>\u21d2-x<\/em>2=y<\/em>\u21d2x<\/em>2=-y<\/em>\u21d2x<\/em>=-\u221a-y<\/em> \u2208 A<\/em> <\/em>(We do not get <\/em>\u00b1, as x>0<\/em>)
\u21d2f <\/em>is onto
\u21d2f <\/em>is a bijection.<\/em>
So, the answer is (a).<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 17:<\/h4>\n\n\n\n

If the function f : R\u2192A given by f(x)=x2x2+1 <\/em>is a surjection, then A =<\/em>
(a) R<\/em>
(b) [0, 1]
(c) [0, 1)
(d) [0, 1)<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n


As f<\/em> is surjective, range of f<\/em>=co-domain of f<\/em>\u21d2A<\/em>= range of f<\/em> \u2235f<\/em>(x<\/em>)= x<\/em>2x<\/em>2+1,  y<\/em>=x<\/em>2x<\/em>2+1\u21d2y<\/em>(x<\/em>2+1)= x<\/em>2\u21d2(y<\/em>-1)x<\/em>2+y<\/em>= 0\u21d2x<\/em>2= –y<\/em>(y<\/em>-1)\u21d2x<\/em>=\u221ay<\/em>(1-y<\/em>)\u21d2y<\/em>(1-y<\/em>)\u22650\u21d2y<\/em>\u2208[0, 1)\u21d2Range of f<\/em>= [0, 1)\u21d2A<\/em>= [0, 1)
So, the answer is (d).<\/p>\n\n\n\n

Page No 2.73:<\/h4>\n\n\n\n

Question 18:<\/h4>\n\n\n\n

If a function f<\/em> : [2, \u221e) \u2192 B<\/em> defined by f<\/em>(x<\/em>)=x<\/em>2-4x<\/em>+5 is a bijection, then B<\/em> =
(a) R<\/em>
(b) [1, \u221e)
(c) [4, \u221e)
(d) [5, \u221e)<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Since f<\/em> is a bijection, co-domain of f = <\/em>range of f<\/em>
\u21d2B<\/em> = range of f<\/em>
Given: f<\/em>(x<\/em>)=x<\/em>2-4x<\/em>+5Let f<\/em>(x<\/em>)=y<\/em>\u21d2y<\/em>=x<\/em>2-4x<\/em>+5\u21d2x<\/em>2-4x<\/em>+(5-y<\/em>)=0\u2235Discrimant, D<\/em>=b<\/em>2-4a<\/em>c<\/em>\u22650,(-4)2-4\u00d71\u00d7(5-y<\/em>)\u22650\u21d216-20+4y<\/em>\u22650\u21d24y<\/em>\u22654\u21d2y<\/em>\u22651\u21d2y<\/em>\u2208[1, \u221e)\u21d2Range of f<\/em>=[1, \u221e)\u21d2B<\/em>=[1, \u221e)
So, the answer is (b).<\/p>\n\n\n\n

Page No 2.74:<\/h4>\n\n\n\n

Question 19:<\/h4>\n\n\n\n

The function f<\/em> : R<\/em>\u2192R<\/em> defined by
f<\/em>(x<\/em>)=(x<\/em>-1) (x<\/em>-2) (x<\/em>-3) is
(a) one-one but not onto
(b) onto but not one-one
(c) both one and onto
(d) neither one-one nor onto<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em>(x<\/em>)=(x<\/em>-1)(x<\/em>-2)(x<\/em>-3)

Injectivity:
f<\/em>(1)=(1-1)(1-2)(1-3)=0f<\/em>(2)=(2-1)(2-2)(2-3)=0f<\/em>(3)=(3-1)(3-2)(3-3)=0\u21d2 f<\/em>(1)=f<\/em>(2)=f<\/em>(3)=0
So, f <\/em>is not one-one.

Surjectivity:
Let y<\/em> be an element in the co domain R,<\/em> such that
y<\/em>=f<\/em>(x<\/em>)\u21d2y<\/em>=(x<\/em>-1)(x<\/em>-2)(x<\/em>-3)Since y<\/em>\u2208R<\/em> and x<\/em>\u2208R<\/em>, f<\/em> is onto.

So, the answer is (b).<\/p>\n\n\n\n

Page No 2.74:<\/h4>\n\n\n\n

Question 20:<\/h4>\n\n\n\n

The function f<\/em> : [-1\/2, 1\/2, 1\/2]\u2192[-\u03c0<\/em>\/2, \u03c0<\/em>\/2] , defined by f<\/em>(x<\/em>)=sin-1 (3x<\/em>-4x<\/em>3), is
(a) bijection
(b) injection but not a surjection
(c) surjection but not an injection
(d) neither an injection nor a surjection<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em>(x<\/em>)=sin-1(3x<\/em>-4x<\/em>3)\u21d2f<\/em>(x<\/em>)=3sin-1x<\/em>

Injectivity:
Let x <\/em>and y <\/em>be two elements in the domain [-12, 12] , such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d23sin-1x<\/em>=3sin-1y<\/em>\u21d2sin-1x<\/em>=sin-1y<\/em>\u21d2x<\/em>=y<\/em>
So, f<\/em> is one-one.

Surjectivity:
Let y<\/em> be any element in the co-domain, such that
f<\/em>(x<\/em>)=y<\/em>\u21d23sin-1(x<\/em>)=y<\/em>\u21d2sin-1(x<\/em>)=y<\/em>3\u21d2x<\/em>=siny<\/em>3\u2208[-12, 12]
\u21d2f <\/em>is onto.
\u21d2f<\/em> is a bijection.
So, the answer is (a).<\/p>\n\n\n\n

Page No 2.74:<\/h4>\n\n\n\n

Question 21:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em>\u2192R<\/em> be a function defined by f<\/em>(x<\/em>)=e<\/em>|x<\/em>|-e<\/em>–x<\/em>e<\/em>x<\/em>+e<\/em>–x<\/em>. Then,
(a) f<\/em> is a bijection
(b) f<\/em> is an injection only
(c) f<\/em> is surjection on only
(d) f<\/em> is neither an injection nor a surjection<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(d) f<\/em> is neither an injection nor a surjection

f<\/em> : R<\/em>\u2192R<\/em>

f<\/em>(x<\/em>)=e<\/em>|x<\/em>|-e<\/em>–x<\/em>e<\/em>x<\/em>+e<\/em>–x<\/em>For x<\/em>=-2 and -3\u2208 R<\/em> f<\/em>(-2) =e<\/em>|-2|-e<\/em>2e<\/em>-2+e<\/em>2           =e<\/em>2-e<\/em>2e<\/em>-2+e<\/em>2           = 0& f<\/em>(-3) =e<\/em>|-3|-e<\/em>3e<\/em>-3+e<\/em>3              =e<\/em>3-e<\/em>3e<\/em>-3+e<\/em>3              = 0Hence, for different values of x<\/em> we are getting same values of f<\/em>(x<\/em>)That means , the given function is many one .
Therefore, this function is not injective.

For x<\/em><0f<\/em>(x<\/em>) =0For x<\/em>>0f<\/em>(x<\/em>) =e<\/em>x<\/em>–e<\/em>–x<\/em>e<\/em>x<\/em>+e<\/em>–x<\/em>        =e<\/em>x<\/em>+e<\/em>–x<\/em>e<\/em>x<\/em>+e<\/em>–x<\/em>-2e<\/em>–x<\/em>e<\/em>x<\/em>+e<\/em>–x<\/em>        =1-2e<\/em>–x<\/em>e<\/em>x<\/em>+e<\/em>–x<\/em>The value of 2e<\/em>–x<\/em>e<\/em>x<\/em>+e<\/em>–x<\/em> is always  positive.Therefore, the value of f<\/em>(x<\/em>) is always less than 1Numbers more than 1 are not included in the range but they are included in codomain.As the codomain is R.\u2234 Codomain\u2260RangeHence, the given function is not onto .
Therefore, this function is not surjective .<\/p>\n\n\n\n

Page No 2.74:<\/h4>\n\n\n\n

Question 22:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em>-{n<\/em>}\u2192R<\/em> be a function defined by

f<\/em>(x<\/em>)=x<\/em>–m<\/em>x<\/em>–n<\/em>, where m<\/em>\u2260n<\/em>. Then,

(a) f<\/em> is one-one onto
(b) f<\/em> is one-one into
(c) f<\/em> is many one onto
(d) f<\/em> is many one into<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity:
Let x<\/em> and y <\/em>be two elements in the domain R<\/em>-{n<\/em>}, such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>–m<\/em>x<\/em>–n<\/em>=y<\/em>–m<\/em>y<\/em>–n<\/em>\u21d2(x<\/em>–m<\/em>)(y<\/em>–n<\/em>)=(x<\/em>–n<\/em>)(y<\/em>–m<\/em>)\u21d2x<\/em>y<\/em>–n<\/em>x<\/em>–m<\/em>y<\/em>+m<\/em>n<\/em>=x<\/em>y<\/em>–m<\/em>x<\/em>–n<\/em>y<\/em>+m<\/em>n<\/em>\u21d2(m<\/em>–n<\/em>)x<\/em>=(m<\/em>–n<\/em>)y<\/em>\u21d2x<\/em>=y<\/em>
So, f<\/em> is one-one.

Surjectivity:
Let y<\/em> be an element in the co domain R,<\/em> such that
f<\/em>(x<\/em>)=y<\/em>\u21d2x<\/em>–m<\/em>x<\/em>–n<\/em>=y<\/em>\u21d2x<\/em>–m<\/em>=x<\/em>y<\/em>–n<\/em>y<\/em>\u21d2n<\/em>y<\/em>–m<\/em>=x<\/em>y<\/em>–x<\/em>\u21d2n<\/em>y<\/em>–m<\/em>=x<\/em>(y<\/em>-1)\u21d2x<\/em>=n<\/em>y<\/em>–m<\/em>y<\/em>-1, which is not defined for y<\/em> =1So, 1 \u2208 R<\/em> (co domain) has no pre image in R<\/em>-{n<\/em>}
\u21d2f<\/em> is not onto.
Thus, the answer is (b).<\/p>\n\n\n\n

Page No 2.74:<\/h4>\n\n\n\n

Question 23:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em>\u2192R<\/em> be a function defined by f<\/em>(x<\/em>)=x<\/em>2-8x<\/em>2+2. Then,  f<\/em> is
(a) one-one but not onto
(b) one-one and onto
(c) onto but not one-one
(d) neither one-one nor onto<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity:
Let x<\/em> and y<\/em> be two elements in the domain (R<\/em>), such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>2-8x<\/em>2+2=y<\/em>2-8y<\/em>2+2\u21d2(x<\/em>2-8)(y<\/em>2+2)=(x<\/em>2+2)(y<\/em>2-8)\u21d2x<\/em>2y<\/em>2+2x<\/em>2-8y<\/em>2-16=x<\/em>2y<\/em>2-8x<\/em>2+2y<\/em>2-16\u21d210x<\/em>2=10y<\/em>2\u21d2x<\/em>2=y<\/em>2\u21d2x<\/em>=\u00b1y<\/em>
So, f <\/em>is not one-one.

Surjectivity:
f<\/em>(-1)=(-1)2-8(-1)2+2=1-81+2=-73 and f<\/em>(1)=(1)2-8(1)2+2=1-81+2=-73\u21d2f<\/em>(-1)=f<\/em>(1)=-73
\u21d2f<\/em> is not onto.
The correct answer is (d).<\/p>\n\n\n\n

Page No 2.74:<\/h4>\n\n\n\n

Question 24:<\/h4>\n\n\n\n

f<\/em> : R<\/em>\u2192R<\/em> is defined by f<\/em>(x<\/em>)=e<\/em>x<\/em>2-e<\/em>–x<\/em>2e<\/em>x<\/em>2+e<\/em>–x<\/em>2 is
(a) one-one but not onto
(b) many-one but onto
(c) one-one and onto
(d) neither one-one nor onto<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(d) neither one-one nor onto

We have,f<\/em>(x<\/em>)=e<\/em>x<\/em>2-e<\/em>–x<\/em>2e<\/em>x<\/em>2+e<\/em>–x<\/em>2Here, -2, 2\u2208R<\/em>Now, 2\u2260-2But, f<\/em>(2)=f<\/em>(-2)Therefore, function is not one-one.And,The minimum value of the function is 0 and maximum value is 1That is range of the function is [0, 1] but the co-domain of the function is given R.Therefore, function is not onto.\u2234function is neither one-one nor onto.<\/p>\n\n\n\n

Page No 2.74:<\/h4>\n\n\n\n

Question 25:<\/h4>\n\n\n\n

The function f<\/em> : R<\/em>\u2192R<\/em>, f<\/em>(x<\/em>)=x<\/em>2 is
(a) injective but not surjective
(b) surjective but not injective
(c) injective as well as surjective
(d) neither injective nor surjective<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity:
Let x<\/em> and y<\/em> be any two elements in the domain (R<\/em>), such that f(x) = f(y)<\/em>. Then,

x2=y2\u21d2x=\u00b1y<\/em>

So, f<\/em> is not one-one.

Surjectivity:
As f<\/em>(-1)=(-1)2=1and f<\/em>(1)=12=1, f<\/em>(-1)=f<\/em>(1)
So, both -1 and 1 have the same images.
\u21d2f<\/em> is not onto.
So, the answer is (d).

x<\/em>2<\/em>+x+1=y2+y+1(x2\u2212y2)+(x\u2212y)=0(x+y)(x\u2212y)+(x\u2212y)=0(x\u2212y)(x+y+1)=0x\u2212y=0 (x+y+1) cannot be zero because x and y are natural numbers)x=y<\/em><\/p>\n\n\n\n

Page No 2.74:<\/h4>\n\n\n\n

Question 26:<\/h4>\n\n\n\n

A function f<\/em> from the set of natural numbers to the set of integers defined by

f<\/em>(n<\/em>){n<\/em>-12,when n<\/em> is odd-n<\/em>2,when n<\/em> is evenis

(a) neither one-one nor onto
(b) one-one but not onto
(c) onto but not one-one
(d) one-one and onto<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity:
Let x<\/em> and y<\/em> be any two elements in the domain (N<\/em>).
Case-1: Both x<\/em> and y<\/em> are even.Let f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2-x<\/em>2=-y<\/em>2\u21d2-x<\/em>=-y<\/em>\u21d2x<\/em>=y<\/em>Case-2: Both x<\/em> and y<\/em> are odd.Let f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>-12=y<\/em>-12\u21d2x<\/em>-1=y<\/em>-1\u21d2x<\/em>=y<\/em>Case-3: Let x<\/em> be even and y<\/em> be odd. Then, f<\/em>(x<\/em>)=-x<\/em>2and f<\/em>(y<\/em>)=y<\/em>-12Then, clearly x<\/em>\u2260y<\/em> \u21d2f<\/em>(x<\/em>)\u2260f<\/em>(y<\/em>)From all the cases, f<\/em> is one-one.

Surjectivity:
Co-domain of f<\/em>=Z<\/em>={…,-3, -2, -1, 0, 1, 2, 3, ….}Range of f<\/em>={…, -3-12, -(-2)2,-1-12, 02, 1-12, -22, 3-12, …}\u21d2Range of f<\/em>={…,-2, 1, -1, 0, 0, -1, 1,…}\u21d2Range of f<\/em>={…, -2, -1, 0, 1, 2, ….}\u21d2Co-domain of f<\/em>=Range of f<\/em><\/em><\/strong>
\u21d2 f<\/em> is onto.
So, the answer is (d).<\/p>\n\n\n\n

Page No 2.74:<\/h4>\n\n\n\n

Question 27:<\/h4>\n\n\n\n

Which of the following functions from A<\/em>={x<\/em> \u2208 R<\/em> : -1\u2264x<\/em>\u22641} to itself are bijections?

(a) f<\/em>(x<\/em>)=|x<\/em>|
(b) f<\/em>(x<\/em>)=sin\u03c0<\/em> x<\/em>2
(c) f<\/em>(x<\/em>)=sin\u03c0<\/em> x<\/em>4
(d) None of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(b) f<\/em>(x<\/em>)=sin\u03c0<\/em> x<\/em>2

It is clear that  f<\/em>(x<\/em>) is one-one.

Range of f<\/em>=[sin\u03c0(-1)2, sin\u03c0(1)2]=[sin -\u03c02, sin\u03c02]=[-1,1]= A<\/em>=Co domain of f<\/em>
\u21d2 f<\/em> is onto.
So, f<\/em> is a bijection.<\/p>\n\n\n\n

Page No 2.74:<\/h4>\n\n\n\n

Question 28:<\/h4>\n\n\n\n

Let f<\/em> : Z<\/em>\u2192Z<\/em> be given by f<\/em>(x<\/em>)={x<\/em>2, if x<\/em> is even0, if x<\/em> is odd
Then,  f<\/em> is
(a) onto but not one-one
(b) one-one but not onto
(c) one-one and onto
(d) neither one-one nor onto<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Injectivity:
Let x<\/em> and y<\/em> be two elements in the domain (Z<\/em>), such that
f<\/em>(x<\/em>)=f<\/em>(y<\/em>)Case-1: Let both x<\/em> and y<\/em> be even.Then,f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d2x<\/em>2=y<\/em>2\u21d2x<\/em>=y<\/em>Case-2: Let both x<\/em> and y<\/em> be odd.Then,f<\/em>(x<\/em>)=f<\/em>(y<\/em>)\u21d20=0Here, we cannot determine whether x<\/em>=y<\/em>.
So, f<\/em> is not one-one.

Surjectivity:
Let y<\/em> be an element in the co-domain (Z<\/em>), such that
Co-domain of f<\/em> =Z<\/em>={0, \u00b11, \u00b12, \u00b13, \u00b14, …}Range of f<\/em>={0, 0, \u00b122, 0, \u00b142 ,…}={0, \u00b11, \u00b12, …}\u21d2Co-domain of f<\/em>=Range of f<\/em>
\u21d2 f<\/em> is onto.
So, the answer is (a).<\/p>\n\n\n\n

Page No 2.74:<\/h4>\n\n\n\n

Question 29:<\/h4>\n\n\n\n

The function f<\/em> : R<\/em>\u2192R<\/em> defined by f<\/em>(x<\/em>)=6x<\/em>+6|x<\/em>| is
(a) one-one and onto
(b) many one and onto
(c) one-one and into
(d) many one and into<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(d) many one and into

Graph of the given function is as follows :

\"\"

A line parallel to X axis is cutting the graph at two different values.

Therefore, for two different values of x we are getting the same value of y<\/em> .
That means it is many one function .

From the given graph we can see that the range is [2,\u221e) 
and R is the codomain of the given function .
Hence, Codomain\u2260Range
Therefore, the given function is into .
 <\/p>\n\n\n\n

Page No 2.75:<\/h4>\n\n\n\n

Question 30:<\/h4>\n\n\n\n

Let f<\/em>(x<\/em>)=x<\/em>2 and g<\/em>(x<\/em>)=2x<\/em>. Then, the solution set of the equation f<\/em>o<\/em>g<\/em> (x<\/em>)=g<\/em>o<\/em>f<\/em> (x<\/em>) is
(a) R<\/em>
(b) {0}
(c) {0, 2}
(d) none of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n


Since (f<\/em>o<\/em>g<\/em>)(x<\/em>)=(g<\/em>o<\/em>f<\/em>)(x<\/em>), f<\/em>(g<\/em>(x<\/em>))=g<\/em>(f<\/em>(x<\/em>))\u21d2f<\/em>(2x<\/em>)=g<\/em>(x<\/em>2)\u21d2(2x<\/em>)2=2x<\/em>2\u21d222x<\/em>=2x<\/em>2\u21d2x<\/em>2=2x<\/em>\u21d2x<\/em>2-2x<\/em>=0\u21d2x<\/em>(x<\/em>-2)=0\u21d2x<\/em>=0, 2\u21d2x<\/em>\u2208{0, 2}
So, the answer is (c).<\/p>\n\n\n\n

Page No 2.75:<\/h4>\n\n\n\n

Question 31:<\/h4>\n\n\n\n

If f<\/em> : R<\/em>\u2192R<\/em> is given by f<\/em>(x<\/em>)=3x<\/em>-5, then f<\/em>-1(x<\/em>)
(a) is given by 13x<\/em>-5
(b) is given by x<\/em>+53
(c) does not exist because f<\/em> is not one-one
(d) does not exist because f<\/em> is not onto<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Clearly, f<\/em> is a bijection.
So, f<\/em> -1<\/sup> exists.
Let f<\/em>-1(x<\/em>)=y<\/em>                   …(1)\u21d2f<\/em>(y<\/em>)=x<\/em>\u21d23y<\/em>-5=x<\/em>\u21d23y<\/em>=x<\/em>+5\u21d2y<\/em>=x<\/em>+53\u21d2f<\/em>-1(x<\/em>)=x<\/em>+53          [from (1)]
So, the answer is (b).<\/p>\n\n\n\n

Page No 2.75:<\/h4>\n\n\n\n

Question 32:<\/h4>\n\n\n\n

If g<\/em> (f<\/em> (x<\/em>))=|sin x<\/em>| and f<\/em> (g<\/em> (x<\/em>))=(sin \u221ax<\/em>)2, then
(a) f<\/em>(x<\/em>)=sin2 x<\/em>, g<\/em>(x<\/em>)=\u221ax<\/em>
(b) f<\/em>(x<\/em>)=sin x<\/em>, g<\/em>(x<\/em>)=|x<\/em>|
(c) f<\/em>(x<\/em>)=x<\/em>2, g<\/em>(x<\/em>)=sin \u221ax<\/em>
(d) f<\/em> and g<\/em> cannot be determined.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

If we solve it  by the trial-and-error method, we can see that (a) satisfies the given condition.
From (a):
f<\/em>(x<\/em>)=sin2x<\/em> and g<\/em>(x<\/em>)=\u221ax<\/em>\u21d2f<\/em>(g<\/em>(x<\/em>))=f<\/em>(\u221ax<\/em>)=sin2 \u221ax<\/em>=(sin \u221ax<\/em>)2
So, the answer is (a).<\/p>\n\n\n\n

Page No 2.75:<\/h4>\n\n\n\n

Question 33:<\/h4>\n\n\n\n

The inverse of the function f<\/em> : R<\/em>\u2192{x<\/em> \u2208 R<\/em> : x<\/em> < 1} given by f<\/em>(x<\/em>)=e<\/em>x<\/em>–e<\/em>–x<\/em>e<\/em>x<\/em>+e<\/em>–x<\/em> is
(a) 12 log 1+x<\/em>1-x<\/em>

(b) 12 log 2+x<\/em>2-x<\/em>

(c) 12 log 1-x<\/em>1+x<\/em>

(d) none of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(x<\/em>)=y<\/em>         …(1)\u21d2f<\/em>(y<\/em>)=x<\/em>\u21d2e<\/em>y<\/em>–e<\/em>–y<\/em>e<\/em>y<\/em>+e<\/em>–y<\/em>=x<\/em>\u21d2e<\/em>–y<\/em>(e<\/em>2y<\/em>-1)e<\/em>–y<\/em>(e<\/em>2y<\/em>+1)=x<\/em>\u21d2(e<\/em>2y<\/em>-1)=x<\/em>(e<\/em>2y<\/em>+1)\u21d2e<\/em>2y<\/em>-1=x<\/em>e<\/em>2y<\/em>+x<\/em>\u21d2e<\/em>2y<\/em>(1-x<\/em>)=x<\/em>+1\u21d2e<\/em>2y<\/em>=1+x<\/em>1-x<\/em>\u21d22y<\/em>=loge<\/em> (1+x<\/em>1-x<\/em>)\u21d2y<\/em>=12l<\/em>oge<\/em> (1+x<\/em>1-x<\/em>)\u21d2f<\/em>-1(x<\/em>)=12l<\/em>oge<\/em> (1+x<\/em>1-x<\/em>)                [from (1)]
So, the answer is (a).<\/p>\n\n\n\n

Page No 2.75:<\/h4>\n\n\n\n

Question 34:<\/h4>\n\n\n\n

Let A<\/em>={x<\/em> \u2208 R<\/em> : x<\/em> \u2265 1}. The inverse of the function, f<\/em> : A<\/em>\u2192A<\/em> given by f<\/em>(x<\/em>)=2x<\/em> (x<\/em>-1), is

(a) (12)x<\/em> (x<\/em>-1)

(b) 12 {1+\u221a1+4 log2 x<\/em>}

(c) 12 {1-\u221a1+4 log2 x<\/em>}

(d) not defined<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(x<\/em>)=y<\/em>…(1)\u21d2f<\/em>(y<\/em>)=x<\/em>\u21d22y<\/em>(y<\/em>-1)=x<\/em>\u21d22y<\/em>2-y<\/em>=x<\/em>\u21d2y<\/em>2-y<\/em>=log2 x<\/em>\u21d2y<\/em>2-y<\/em>+14=log2 x<\/em>+14\u21d2(y<\/em>-12)2=4log2 x<\/em>+14\u21d2y<\/em>-12=\u00b1\u221a4log2 x<\/em>+12\u21d2y<\/em>=12\u00b1\u221a4log2 x<\/em>+12\u21d2y<\/em>=12+\u221a4log2 x<\/em>+12                      (\u2235 y \u22651)So, f<\/em>-1(x<\/em>)=12(1+\u221a1+4log2 x<\/em> )       [from (1)]
So, the answer is (b).<\/p>\n\n\n\n

Page No 2.75:<\/h4>\n\n\n\n

Question 35:<\/h4>\n\n\n\n

Let A<\/em>={x<\/em> \u2208 R<\/em> : x<\/em> \u22641} and f<\/em> : A<\/em>\u2192A<\/em>  be defined as f<\/em>(x<\/em>)=x<\/em> (2-x<\/em>). Then, f<\/em>-1 (x<\/em>) is
(a) 1+\u221a1-x<\/em>
(b) 1-\u221a1-x<\/em>
(c) \u221a1-x<\/em>
(d) 1\u00b1\u221a1-x<\/em><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let y<\/em> be the element in the codomain R<\/em> such that        f<\/em>-1(x<\/em>)=y<\/em> …(1)\u21d2f<\/em>(y<\/em>)=x<\/em> and y \u22641\u21d2y<\/em>(2-y<\/em>)=x<\/em>\u21d22y<\/em>–y<\/em>2=x<\/em>\u21d2y<\/em>2-2y<\/em>+x<\/em>=0\u21d2y<\/em>2-2y<\/em>=-x<\/em>\u21d2y<\/em>2-2y<\/em>+1=1-x<\/em>\u21d2(y<\/em>-1)2=1-x<\/em>\u21d2y<\/em>-1=\u00b1\u221a1-x<\/em>\u21d2y<\/em>=1\u00b1\u221a1-x<\/em>\u21d2y<\/em>=1-\u221a1-x<\/em>        (\u2235 y \u22641)
The correct answer is (b).<\/p>\n\n\n\n

Page No 2.75:<\/h4>\n\n\n\n

Question 36:<\/h4>\n\n\n\n

Let f<\/em>(x<\/em>)=11-x<\/em>. Then, {f<\/em> o<\/em> (f<\/em>o<\/em>f<\/em>)} (x<\/em>)
(a) x<\/em> for all x<\/em> \u2208 R<\/em>
(b) x<\/em> for all x<\/em> \u2208 R<\/em>-{1}
(c) x<\/em> for all x<\/em> \u2208 R<\/em>-{0, 1}
(d) none of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Domain of f<\/em>:1-x<\/em>\u22600\u21d2x<\/em>\u22601Domain of f<\/em>=R<\/em>-{1}Range of f<\/em>:y<\/em>=11-x<\/em>\u21d21-x<\/em>=1y<\/em>\u21d2x<\/em>=1-1y<\/em>\u21d2x<\/em>=y<\/em>-1y<\/em>\u21d2y<\/em>\u22600Range of f<\/em>=R<\/em>-{0}So, f<\/em>:R<\/em>-{1}\u2192R<\/em>-{0} and f<\/em>:R<\/em>-{1}\u2192R<\/em>-{0} Range of f<\/em> is not a subset of the domain of f<\/em>.Domain (f<\/em>o<\/em>f<\/em>)={x<\/em>: x<\/em>\u2208domain of f<\/em> and f<\/em>(x<\/em>)\u2208domain of f<\/em>}Domain (f<\/em>o<\/em>f<\/em>)={x<\/em>: x<\/em>\u2208R<\/em>-{1} and 11-x<\/em>\u2208R<\/em>-{1}} Domain (f<\/em>o<\/em>f<\/em>)={x<\/em>: x<\/em> \u22601 and 11-x<\/em>\u22601}Domain (f<\/em>o<\/em>f<\/em>)={x<\/em>: x<\/em> \u22601 and 1-x<\/em>\u22601}Domain (f<\/em>o<\/em>f<\/em>)={x<\/em>: x<\/em> \u22601 and x<\/em>\u22600}Domain (f<\/em>o<\/em>f<\/em>)=R<\/em>-{0, 1}(f<\/em>o<\/em>f<\/em>)(x<\/em>)=f<\/em>(f<\/em>(x<\/em>))=f<\/em>(11-x<\/em>)=11-11-x<\/em>=1-x<\/em>1-x<\/em>-1=1-x<\/em>–x<\/em>=x<\/em>-1x<\/em>For range of f<\/em>o<\/em>f<\/em>, x<\/em>\u22600Now, f<\/em>o<\/em>f<\/em>:R<\/em>-{0, 1}\u2192R<\/em> -{0} and f<\/em>:R<\/em>-{1}\u2192R<\/em>-{0}Range of f<\/em>o<\/em>f<\/em> is not a subset of domain of f<\/em>.Domain(f o <\/em>(f<\/em>o<\/em>f<\/em>))={x<\/em>: x<\/em>\u2208domain of f<\/em>o<\/em>f<\/em> and (f<\/em>o<\/em>f<\/em>)(x<\/em>)\u2208domain of f<\/em>}Domain (f o <\/em>(f<\/em>o<\/em>f<\/em>))={x<\/em>: x<\/em>\u2208R<\/em>-{0, 1} and x<\/em>-1x<\/em>\u2208R<\/em>-{1}} Domain(f o <\/em>(f<\/em>o<\/em>f<\/em>))={x<\/em>: x<\/em> \u22600, 1 and x<\/em>-1x<\/em>\u22601}Domain (f o <\/em>(f<\/em>o<\/em>f<\/em>))={x<\/em>: x<\/em> \u22600, 1 and x<\/em>-1\u2260x<\/em>}Domain (f o <\/em>(f<\/em>o<\/em>f<\/em>))={x<\/em>: x<\/em> \u22600, 1 and x<\/em>\u2208R<\/em>}Domain (f o <\/em>(f<\/em>o<\/em>f<\/em>))=R<\/em>-{0, 1}(f<\/em>o<\/em>(f<\/em>o<\/em>f<\/em>))(x<\/em>)=f<\/em>((f<\/em>o<\/em>f<\/em>)(x<\/em>))=f<\/em>(x<\/em>-1x<\/em>)=11-x<\/em>-1x<\/em>=x<\/em>x<\/em>–x<\/em>+1=x<\/em>So, (f<\/em>o<\/em>(f<\/em>o<\/em>f<\/em>))(x<\/em>)=x<\/em>, where x<\/em>\u22600,1
So, the answer is (c).<\/p>\n\n\n\n

Page No 2.75:<\/h4>\n\n\n\n

Question 37:<\/h4>\n\n\n\n

If the function f<\/em> : R<\/em>\u2192R<\/em> be such that f<\/em>(x<\/em>)=x<\/em>-[x<\/em>], where [x<\/em>] denotes the greatest integer less than or equal to x<\/em>, then f<\/em>-1 (x<\/em>) is

(a) 1x<\/em>-[x<\/em>]

(b) [x<\/em>] \u2212 x<\/em>

(c) not defined

(d) none of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em>(x<\/em>) = x<\/em> – [x<\/em>]
We know that the range of f<\/em> is [0, 1).
Co-domain of f<\/em> = R<\/em>
As range of f \u2260<\/em>Co-domain of f<\/em>,  f<\/em> is not onto.
\u21d2 f<\/em> is not a bijective function.
So,  f<\/em> -1<\/sup> does not exist.
Thus, the answer is (c).<\/p>\n\n\n\n

Page No 2.75:<\/h4>\n\n\n\n

Question 38:<\/h4>\n\n\n\n

If F<\/em> : [1, \u221e)\u2192[2, \u221e) is given by f<\/em>(x<\/em>)=x<\/em>+1x<\/em>, then f<\/em>-1 (x<\/em>) equals

(a) x<\/em>+\u221ax<\/em>2-42

(b) x<\/em>1+x<\/em>2

(c) x<\/em>-\u221ax<\/em>2-42

(d) 1+\u221ax<\/em>2-4<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(x<\/em>) = y<\/em>\u21d2f<\/em>(y<\/em>)=x<\/em>\u21d2y<\/em>+1y<\/em>=x<\/em>\u21d2y<\/em>2+1=x<\/em>y<\/em>\u21d2y<\/em>2-x<\/em>y<\/em>+1=0\u21d2y<\/em>2-2\u00d7y<\/em>\u00d7x<\/em>2+(x<\/em>2)2-(x<\/em>2)2+1=0\u21d2y<\/em>2-2\u00d7y<\/em>\u00d7x<\/em>2+(x<\/em>2)2=x<\/em>2-14\u21d2(y<\/em>–x<\/em>2)2=x<\/em>2-14\u21d2y<\/em>–x<\/em>2=\u221ax<\/em>2-42\u21d2y<\/em>=x<\/em>2+\u221ax<\/em>2-42\u21d2y<\/em>=x<\/em>+\u221ax<\/em>2-42\u21d2f<\/em>-1(x<\/em>)=x<\/em>+\u221ax<\/em>2-42

So, the answer is (a).<\/p>\n\n\n\n

Page No 2.75:<\/h4>\n\n\n\n

Question 39:<\/h4>\n\n\n\n

Let g<\/em>(x<\/em>)=1+x<\/em>-[x<\/em>] and f<\/em>(x<\/em>)={-1,x<\/em><00,x<\/em>=0, 1,x<\/em>>0, where [x<\/em>] denotes the greatest integer less than or equal to x<\/em>. Then for all x<\/em>, f<\/em> (g<\/em> (x<\/em>)) is equal to
(a) x<\/em>
(b) 1
(c) f<\/em>(x<\/em>)
(d) g<\/em>(x<\/em>)<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(b) 1

When, -1<x<\/em><0Then, g<\/em>(x<\/em>)=1+x<\/em>-[x<\/em>]               =1+x<\/em>-(-1)=2+x<\/em>\u2234f<\/em>(g<\/em>(x<\/em>))=1 When, x<\/em>=0Then, g<\/em>(x<\/em>)=1+x<\/em>-[x<\/em>]               =1+x<\/em>-0=1+x<\/em>\u2234f<\/em>(g<\/em>(x<\/em>))=1When, x<\/em>>1Then, g<\/em>(x<\/em>)=1+x<\/em>-[x<\/em>]               =1+x<\/em>-1=x<\/em>\u2234f<\/em>(g<\/em>(x<\/em>))=1

Therefore, for each interval f<\/em>(g<\/em>(x<\/em>))=1<\/p>\n\n\n\n

Page No 2.75:<\/h4>\n\n\n\n

Question 40:<\/h4>\n\n\n\n

Let f<\/em>(x<\/em>)=\u03b1<\/em> x<\/em>x<\/em>+1, x<\/em>\u2260-1. Then, for what value of \u03b1 is f<\/em> (f<\/em>(x<\/em>))=x<\/em>?
(a) \u221a2
(b) -\u221a2
(c) 1
(d) \u22121<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(d) \u22121

  f<\/em>(f<\/em>(x<\/em>))=x<\/em>\u21d2 f<\/em>(\u03b1<\/em>x<\/em>x<\/em>+1)=x<\/em>\u21d2\u03b1<\/em>(\u03b1<\/em>x<\/em>x<\/em>+1)(\u03b1<\/em>x<\/em>x<\/em>+1)+1=x<\/em>\u21d2\u03b1<\/em>2x<\/em>\u03b1<\/em>x<\/em>+x<\/em>+1=x<\/em>\u21d2\u03b1<\/em>2x<\/em>=\u03b1<\/em>x<\/em>2+x<\/em>2+x<\/em>\u21d2\u03b1<\/em>2x<\/em>–\u03b1<\/em>x<\/em>2-x<\/em>2-x<\/em>=0\u21d2\u03b1<\/em>2x<\/em>–\u03b1<\/em>x<\/em>2-(x<\/em>2+x<\/em>)=0Solving for the \u03b1 we get,\u03b1<\/em>=-(-x<\/em>2)\u00b1\u221a(-x<\/em>2)2-4\u00d7x<\/em>\u00d7[-(x<\/em>2+x<\/em>)]2x<\/em>  =x<\/em>2\u00b1\u221ax<\/em>4+4x<\/em>3+4x<\/em>22x<\/em>  =x<\/em>+1, -1Here, -1 is independent of x<\/em>,\u2234for, \u03b1<\/em>=-1, f<\/em>(f<\/em>(x<\/em>))=x<\/em>   <\/p>\n\n\n\n

Page No 2.76:<\/h4>\n\n\n\n

Question 41:<\/h4>\n\n\n\n

The distinct linear functions that map [\u22121, 1] onto [0, 2] are
(a) f<\/em>(x<\/em>)=x<\/em>+1, g<\/em>(x<\/em>)=-x<\/em>+1
(b) f<\/em>(x<\/em>)=x<\/em>-1, g<\/em>(x<\/em>)=x<\/em>+1
(c) f<\/em>(x<\/em>)=-x<\/em>-1, g<\/em>(x<\/em>)=x<\/em>-1
(d) None of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let us substitute the end-points of the intervals in the given functions. Here, domain = [-1, 1] and range =[0, 2]
By substituting -1 or 1 in each option, we get:

Option (a):
f<\/em>(-1)=-1+1=0 and f<\/em>(1)=1+1=2g<\/em>(-1)=1+1=2 and g<\/em>(1)=-1+1=0
So, option (a) is correct.

Option (b):
f<\/em>(-1)=-1-1=-2 and f<\/em>(1)=1-1=0g<\/em>(-1)=-1+1=0 and g<\/em>(1)=1+1=2
Here, f(-1) gives -2\u2209[0, 2]
So, (b) is not correct.

Similarly, we can see that (c) is also not correct.<\/p>\n\n\n\n

Page No 2.76:<\/h4>\n\n\n\n

Question 42:<\/h4>\n\n\n\n

Let f<\/em> : [2, \u221e)\u2192X<\/em> be defined by f<\/em>(x<\/em>)=4x<\/em>–x<\/em>2. Then, f<\/em> is invertible if X<\/em> =
(a) [2, \u221e)
(b) (-\u221e, 2]
(c) (-\u221e, 4]
(d) [4, \u221e)<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Since f<\/em> is invertible, range of f<\/em> = co domain of f<\/em> = X<\/em>
So, we need to find the range of f<\/em> to find X<\/em>.
For finding the range, let
f<\/em>(x<\/em>)=y<\/em>\u21d24x<\/em>–x<\/em>2=y<\/em>\u21d2x<\/em>2-4x<\/em>=-y<\/em>\u21d2x<\/em>2-4x<\/em>+4=4-y<\/em>\u21d2(x<\/em>-2)2=4-y<\/em>\u21d2x<\/em>-2=\u00b1\u221a4-y<\/em>\u21d2x<\/em>=2\u00b1\u221a4-y<\/em>This is defined only when 4-y<\/em>\u22650\u21d2y<\/em>\u22644X<\/em>=Range of f<\/em>=(-\u221e,4]
So, the answer is (c).<\/p>\n\n\n\n

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Question 43:<\/h4>\n\n\n\n

If f<\/em> : R<\/em>\u2192 (-1, 1) is defined by f<\/em>(x<\/em>)=-x<\/em>|x<\/em>|1+x<\/em>2, then f<\/em>-1 (x<\/em>) equals
(a) \u221a|x<\/em>|1-|x<\/em>|

(b) S<\/em>g<\/em>n<\/em> (x<\/em>) \u221a|x<\/em>|1-|x<\/em>|

(c) -\u221ax<\/em>1-x<\/em>

(d) None of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(b) -Sgn (x<\/em>) \u221a|x<\/em>|1-|x<\/em>|

We have, f<\/em>(x<\/em>)=-x<\/em>|x<\/em>|1+x<\/em>2    x<\/em>\u2208(-1, 1)Case-(I)When, x<\/em><0,Then, |x<\/em>|=-x<\/em>And f<\/em>(x<\/em>)>0Now,f<\/em>(x<\/em>)=-x<\/em>(-x<\/em>)1+x<\/em>2\u21d2y<\/em>=x<\/em>21+x<\/em>2\u21d2y<\/em>1=x<\/em>21+x<\/em>2\u21d2y<\/em>+1y<\/em>-1=x<\/em>2+1+x<\/em>2x<\/em>2-1-x<\/em>2           [Using Componendo and dividendo]\u21d2y<\/em>+1y<\/em>-1=2x<\/em>2+1-1\u21d2-y<\/em>+1y<\/em>-1=2x<\/em>2+1\u21d22y<\/em>1-y<\/em>=2x<\/em>2\u21d2y<\/em>1-y<\/em>=x<\/em>2\u21d2x<\/em>=-\u221ay<\/em>1-y<\/em>                         (As x<\/em><0)\u21d2x<\/em>=-\u221a|y<\/em>|1-|y<\/em>|                        [As y<\/em>>0]To find the inverse interchanging x<\/em> and y<\/em> we get,f<\/em>-1(x<\/em>)=-\u221a|x<\/em>|1-|x<\/em>|            …(i)Case-(II)When, x<\/em>>0,Then, |x<\/em>|=x<\/em>And f<\/em>(x<\/em>)<0Now,f<\/em>(x<\/em>)=-x<\/em>(x<\/em>)1+x<\/em>2\u21d2y<\/em>=-x<\/em>21+x<\/em>2\u21d2y<\/em>1=-x<\/em>21+x<\/em>2\u21d2y<\/em>+1y<\/em>-1=-x<\/em>2+1+x<\/em>2-x<\/em>2-1-x<\/em>2           [Using Componendo and dividendo]\u21d2y<\/em>+1y<\/em>-1=1-2x<\/em>2-1\u21d21+y<\/em>1-y<\/em>=12x<\/em>2+1\u21d21-y<\/em>1+y<\/em>=2x<\/em>2+1\u21d2-2y<\/em>1+y<\/em>=2x<\/em>2\u21d2x<\/em>2=-y<\/em>1+y<\/em>\u21d2x<\/em>=\u221a-y<\/em>1+y<\/em>                         (As x<\/em>>0)\u21d2x<\/em>=\u221a|y<\/em>|1-|y<\/em>|                        [As y<\/em><0]To find the inverse interchanging x<\/em> and y<\/em> we get,f<\/em>-1(x<\/em>)=\u221a|x<\/em>|1-|x<\/em>|            …(ii)Case-(III)When, x<\/em>=0,Then, f<\/em>(x<\/em>)=0Hence, f<\/em>-1(x<\/em>)=0     …(iii)Combinig equation (i) , (ii) and (iii) we get,f<\/em>-1(x<\/em>)=-Sgn(x<\/em>)\u221a|x<\/em>|1-|x<\/em>|  <\/p>\n\n\n\n

Page No 2.76:<\/h4>\n\n\n\n

Question 44:<\/h4>\n\n\n\n

Let [x<\/em>] denote the greatest integer less than or equal to x<\/em>. If f<\/em>(x<\/em>)=sin-1x<\/em>, g<\/em>(x<\/em>)=[x<\/em>2] and h<\/em>(x<\/em>)=2x<\/em>,12\u2264x<\/em>\u22641\u221a2, then
(a) f<\/em>o<\/em>g<\/em>o<\/em>h<\/em>(x<\/em>)=\u03c0<\/em>2
(b) f<\/em>o<\/em>g<\/em>o<\/em>h<\/em>(x<\/em>)=\u03c0<\/em>
(c) h<\/em>o<\/em>f<\/em>o<\/em>g<\/em>=h<\/em>o<\/em>g<\/em>o<\/em>f<\/em>
(d) h<\/em>o<\/em>f<\/em>o<\/em>g<\/em>\u2260h<\/em>o<\/em>g<\/em>o<\/em>f<\/em><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(c) h<\/em>o<\/em>f<\/em>o<\/em>g<\/em>=h<\/em>o<\/em>g<\/em>o<\/em>f<\/em>

We have,g<\/em>(x<\/em>)=[x<\/em>2]       =0          ( A<\/em>s<\/em> 12\u2264x<\/em>\u2264 1\u221a2\u2234 14\u2264x<\/em>2\u2264 12)f<\/em>o<\/em>g<\/em>(x<\/em>)=f<\/em>(g<\/em>(x<\/em>))=sin-1(0)                       =0h<\/em>o<\/em>f<\/em>o<\/em>g<\/em>(x<\/em>)=h<\/em>(f<\/em>(g<\/em>(x<\/em>)))=2\u00d70=0

Andf<\/em>(x<\/em>)=sin-1x<\/em>Now,f<\/em>o<\/em>r<\/em>, x<\/em>\u2208[12, 1\u221a2]f<\/em>(x<\/em>)\u2208[\u03c06, \u03c04]f<\/em>(x<\/em>)\u2208[0.52, 0.78]g<\/em>o<\/em>f<\/em>(x<\/em>)=0    (As, f<\/em>(x<\/em>)\u2208[0.52, 0.78])                       =0h<\/em>o<\/em>g<\/em>o<\/em>f<\/em>(x<\/em>)=h<\/em>(g<\/em>(f<\/em>(x<\/em>)))=2\u00d70=0

 \u2234 h<\/em>o<\/em>f<\/em>o<\/em>g<\/em>=h<\/em>o<\/em>g<\/em>o<\/em>f<\/em>=0
 <\/p>\n\n\n\n

Page No 2.76:<\/h4>\n\n\n\n

Question 45:<\/h4>\n\n\n\n

If g<\/em>(x<\/em>)=x<\/em>2+x<\/em>-2 and 12 g<\/em>o<\/em>f<\/em>(x<\/em>)=2x<\/em>2-5x<\/em>+2, then f<\/em>(x<\/em>) is equal to
(a) 2 x<\/em>-3
(b) 2 x<\/em>+3
(c) 2 x<\/em>2+3x<\/em>+1
(d) 2 x<\/em>2-3x<\/em>-1<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We will solve this problem by the trial-and-error method.
Let us check option (a) first.
If f<\/em>(x<\/em>) = 2x<\/em>-312(g<\/em>o<\/em>f<\/em>)(x<\/em>) = g<\/em>(f<\/em>(x<\/em>))= 12g<\/em>(2x<\/em>-3)= 12[(2x<\/em>-3)2+(2x<\/em>-3)-2]= 12[4x<\/em>2+9-12x<\/em>+2x<\/em>-3-2]= 12[4x<\/em>2-10x<\/em>+4]= 2x<\/em>2-5x<\/em>+2
The given condition is satisfied by (a).
So, the answer is (a).<\/p>\n\n\n\n

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Question 46:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>)=sin2 x<\/em> and the composite function g<\/em>(f<\/em>(x<\/em>))=|sin x<\/em>|, then g<\/em>(x<\/em>) is equal to
(a) \u221ax<\/em>-1
(b) \u221ax<\/em>
(c) \u221ax<\/em>+1
(d) -\u221ax<\/em><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(b)

 If we take g<\/em>(x<\/em>) = \u221ax<\/em>, theng<\/em>(f<\/em>(x<\/em>)) = g<\/em>(sin2x<\/em>) = \u221asin2x<\/em> = \u00b1sin x<\/em> = |sin x<\/em>|

So, the answer is (b).<\/p>\n\n\n\n

Page No 2.76:<\/h4>\n\n\n\n

Question 47:<\/h4>\n\n\n\n

If f<\/em> : R<\/em>\u2192R<\/em> is given by f<\/em>(x<\/em>)=x<\/em>3+3, then f<\/em>-1(x<\/em>) is equal to
(a) x<\/em>1\/3-3
(b) x<\/em>1\/3+3
(c) (x<\/em>-3)1\/3
(d) x<\/em>+31\/3<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(c)

Let f<\/em>-1(x<\/em>) = y<\/em>f<\/em>(y<\/em>) = x<\/em>\u21d2y<\/em>3+3 = x<\/em>\u21d2y<\/em>3 = x<\/em>-3\u21d2y<\/em> = 3\u221ax<\/em>-3 \u21d2y<\/em> = (x<\/em>-3)13
So, the answer is (c).<\/p>\n\n\n\n

Page No 2.76:<\/h4>\n\n\n\n

Question 48:<\/h4>\n\n\n\n

Let f<\/em>(x<\/em>)=x<\/em>3 be a function with domain {0, 1, 2, 3}. Then domain of f<\/em>-1 is
(a) {3, 2, 1, 0}
(b) {0, \u22121, \u22122, \u22123}
(c) {0, 1, 8, 27}
(d) {0, \u22121, \u22128, \u221227}<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(c) {0, 1, 8, 27}<\/p>\n\n\n\n

f<\/em>(x<\/em>)=x<\/em>3Domain = {0, 1, 2, 3}Range = {03, 13, 23, 33} = {0, 1, 8, 27}So, f<\/em> = {(0, 0), (1, 1), (2, 8), (3, 27)}f-1 = {(0, 0), (1, 1), (8, 2), (27, 3)}Domain of f<\/em>-1 = {0, 1, 8, 27}<\/p>\n\n\n\n

Page No 2.76:<\/h4>\n\n\n\n

Question 49:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em>\u2192R<\/em> be given by f<\/em>(x<\/em>)=x<\/em>2-3. Then, f<\/em>-1 is given by
(a) \u221ax<\/em>+3
(b) \u221ax<\/em>+3
(c) x<\/em>+\u221a3
(d) None of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(d)

Let f<\/em>-1(x<\/em>)=y<\/em>f<\/em>(y<\/em>)=x<\/em>y<\/em>2-3=x<\/em>y<\/em>2=x<\/em>+3y<\/em>=\u00b1\u221ax<\/em>+3
So, the answer is (d).<\/p>\n\n\n\n

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Question 50:<\/h4>\n\n\n\n

Mark the correct alternative in the following question:

Let f<\/em> : R<\/strong> \u2192 R<\/strong> be given by f<\/em>(x<\/em>) = tanx<\/em>. Then, f<\/em> -1<\/sup>(1) is

(a) \u03c04                              (b) {n<\/em>\u03c0+\u03c04:n<\/em>\u2208Z<\/strong>}                              (c) does not exist                              (d) none of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,f<\/em>:R<\/strong>\u2192R<\/strong> is given byf<\/em>(x<\/em>)=tanx<\/em>\u21d2f<\/em>-1(x<\/em>)=tan-1x<\/em>\u2234 f<\/em>-1(1)=tan-11={n<\/em>\u03c0+\u03c04:n<\/em>\u2208Z<\/strong>}

Hence, the correct alternative is option (b).<\/p>\n\n\n\n

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Question 51:<\/h4>\n\n\n\n

Mark the correct alternative in the following question:

Let f<\/em> : R<\/strong> \u2192 R<\/strong> be defined as f<\/em>(x<\/em>) = {2x<\/em>, if x<\/em>>3     x<\/em>2, if 1<x<\/em>\u226433x<\/em>, if x<\/em>\u22641      

Then, find f<\/em>(-1) + f<\/em>(2) + f<\/em>(4)

(a) 9                              (b) 14                              (c) 5                              (d) none of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,f<\/em>(x<\/em>)={2x<\/em>, if x<\/em>>3     x<\/em>2, if 1<x<\/em>\u226433x<\/em>, if x<\/em>\u22641      Now,f<\/em>(-1)+f<\/em>(2)+f<\/em>(4)=3(-1)+22+2(4)=-3+4+8=9

Hence, the correct alternative is option (a).<\/p>\n\n\n\n

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Question 52:<\/h4>\n\n\n\n

Mark the correct alternative in the following question:

Let A<\/em> = {1, 2, … , n<\/em>} and B<\/em> = {a<\/em>, b<\/em>}. Then the number of subjections from A<\/em> into B<\/em> is

(a) n<\/em><\/sup>P2<\/sub>                               (b) 2n<\/em><\/sup> – 2                               (c) 2n<\/em><\/sup> – 1                               (d) n<\/em><\/sup>C2<\/sub><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

As, the number of surjections from A<\/em> to B<\/em> is equal to the number of functions from A<\/em> to B<\/em> minus the number of functions from A<\/em> to B<\/em> whose images are proper subsets of B<\/em>.

And, the number of functions from a set with n<\/em> number of elements into a set with m<\/em> number of elements = mn<\/sup><\/em>

So, the number of subjections from A<\/em> into B<\/em> where A<\/em> = {1, 2, … , n<\/em>} and B<\/em> = {a<\/em>, b<\/em>} is 2n<\/em><\/sup> – 2.           (As, two functions can be many-one into functions)

Hence, the correct alternative is option (b).<\/p>\n\n\n\n

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Question 53:<\/h4>\n\n\n\n

Mark the correct alternative in the following question:

If the set A<\/em> contains 5 elements and the set B<\/em> contains 6 elements, then the number of one-one and onto mappings from A<\/em> to B<\/em> is

(a) 720                                                 (b) 120                                                 (c) 0                                                 (d) none of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

As, the number of bijection from A<\/em> into B<\/em> can only be possible when provided n<\/em>(A<\/em>)\u2265n<\/em>(B<\/em>)But here n<\/em>(A<\/em>)<n<\/em>(B<\/em>)So, the number of bijection i.e. one-one and onto mappings from A<\/em> to B<\/em>=0

Hence, the correct alternative is option (c).<\/p>\n\n\n\n

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Question 54:<\/h4>\n\n\n\n

Mark the correct alternative in the following question:

If the set A<\/em> contains 7 elements and the set B<\/em> contains 10 elements, then the number one-one functions from A<\/em> to B<\/em> is

(a) 10<\/sup>C7<\/sub>                                            (b) 10<\/sup>C7<\/sub> \u00d7 7!                                              (c) 710<\/sup>                                              (d)107<\/sup><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

As, the number of one-one functions from A<\/em> to B<\/em> with m<\/em> and n<\/em> elements, respectively = n<\/em><\/sup>Pm<\/em><\/sub> = n<\/em><\/sup>Cm<\/em><\/sub> \u00d7 m<\/em>!

So, the number of one-one functions from A<\/em> to B<\/em> with 7 and 10 elements, respectively = 10<\/sup>P7<\/sub> = 10<\/sup>C7<\/sub> \u00d7 7!

Hence, the correct alternative is option (b).<\/p>\n\n\n\n

Page No 2.76:<\/h4>\n\n\n\n

Question 55:<\/h4>\n\n\n\n

Mark the correct alternative in the following question:

Let f<\/em> : R<\/strong> – {35} \u2192 R<\/strong> be defined by f<\/em>(x<\/em>) = 3x<\/em>+25x<\/em>-3. Then,

(a) f<\/em> -l<\/sup>(x<\/em>) = f<\/em>(x<\/em>)                         (b) f<\/em> -1<\/sup>(x<\/em>) = –f<\/em>(x<\/em>)                         (c) f<\/em>of<\/em>(x<\/em>) = –x<\/em>                         (d) f <\/em>-1<\/sup>(x<\/em>) = 119f<\/em>(x<\/em>)<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,

f<\/em> : R<\/strong> – {35} \u2192 R<\/strong> is defined by f<\/em>(x<\/em>) = 3x<\/em>+25x<\/em>-3

f<\/em>of<\/em>(x<\/em>)=f<\/em>(f<\/em>(x<\/em>))=f<\/em>(3x<\/em>+25x<\/em>-3)=3(3x<\/em>+25x<\/em>-3)+25(3x<\/em>+25x<\/em>-3)-3=(9x<\/em>+65x<\/em>-3)+2(15x<\/em>+105x<\/em>-3)-3=(9x<\/em>+6+10x<\/em>-65x<\/em>-3)(15x<\/em>+10-15x<\/em>+95x<\/em>-3)=19x<\/em>19=x<\/em>Let y<\/em>=3x<\/em>+25x<\/em>-3\u21d25x<\/em>y<\/em>-3y<\/em>=3x<\/em>+2\u21d25x<\/em>y<\/em>-3x<\/em>=3y<\/em>+2\u21d2x<\/em>(5y<\/em>-3)=3y<\/em>+2\u21d2x<\/em>=3y<\/em>+25y<\/em>-3\u21d2f<\/em>-1(y<\/em>)=3y<\/em>+25y<\/em>-3So, f<\/em>-1(x<\/em>)=3x<\/em>+25x<\/em>-3=f<\/em>(x<\/em>)

Hence, the correct alternative is option (a).<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 56:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em> \u2192 R<\/em> be defined by f<\/em>(x<\/em>)=1x<\/em>. Then,  f<\/em> is
(a) one-one
(b) onto
(e) bijective
(d) not defined<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: The function f<\/em> : R<\/em> \u2192 R<\/em> be defined by f<\/em>(x<\/em>)=1x<\/em>.

To check f<\/em> is one-one:

Let f<\/em>(x<\/em>1)=f<\/em>(x<\/em>2)\u21d21x<\/em>1=1x<\/em>2\u21d2x<\/em>1=x<\/em>2Hence, f<\/em> is one-one.To check <\/em>f<\/em> is onto: Since, y<\/em>=1x<\/em>\u21d2x<\/em>=1y<\/em>\u21d2y<\/em>\u2208R<\/em>-{0}\u2260R<\/em>There is no pre-image of y<\/em>=0.Hence, f<\/em> is not onto.

Hence, the correct option is (a).

\u200b<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 57:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em> \u2192 R<\/em> be defined by f<\/em>(x<\/em>) = 3x<\/em>2<\/sup> \u2013 5 and g<\/em> : R<\/em> \u2192 R<\/em> by g<\/em>(x<\/em>)=x<\/em>x<\/em>2+1. Then (gof<\/em>) (x<\/em>) is
(a) 3x<\/em>2\u221259x<\/em>4\u221230x<\/em>2+26

(b) 3x<\/em>2\u221259x<\/em>4\u22126x<\/em>2+26

(c) 3x<\/em>2x<\/em>4+2x<\/em>2\u22124

(d) 3x<\/em>29x<\/em>4+30x<\/em>2\u22122<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = 3x<\/em>2<\/sup> \u2013 5 and g<\/em>(x<\/em>)=x<\/em>x<\/em>2+1

(g<\/em>o<\/em>f<\/em>)(x<\/em>)=g<\/em>(f<\/em>(x<\/em>))            =g<\/em>(3x<\/em>2-5)            =3x<\/em>2-5(3x<\/em>2-5)2+1            =3x<\/em>2-5(3x<\/em>2)2+52-2(3x<\/em>2)(5)+1            =3x<\/em>2-59x<\/em>4+25-30x<\/em>2+1            =3x<\/em>2-59x<\/em>4-30x<\/em>2+26

\u200bHence, the correct option is (a).

\u200b<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 58:<\/h4>\n\n\n\n

Which of the following functions from Z<\/em> to Z<\/em> are bijections?
(a) f<\/em>(x<\/em>) = x<\/em>3<\/sup>
(b) f<\/em>(x<\/em>) = x<\/em> + 2
(c) f<\/em>(x<\/em>) = 2x<\/em> + 1
(d) f<\/em>(x<\/em>) = x<\/em>2<\/sup> + 1<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em> : Z<\/em> \u2192 Z<\/em>

(a) f<\/em>(x<\/em>) = x<\/em>3<\/sup>

It is one-one but not onto.

Thus, it is not bijective.

(b) f<\/em>(x<\/em>) = x<\/em> + 2

It is one-one and onto.

Thus, it is bijective.

(c) f<\/em>(x<\/em>) = 2x<\/em> + 1

It is one-one but not onto.

Thus, it is not bijective.

(d) f<\/em>(x<\/em>) = x<\/em>2<\/sup> + 1

It is neither one-one nor onto.

Thus, it is not bijective.

Hence, the correct option is (b).<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 59:<\/h4>\n\n\n\n

Let f<\/em> : A<\/em> \u2192 B<\/em> and g<\/em> : B<\/em> \u2192 C<\/em> be the bijective functions. Then, (gof<\/em>)\u20131<\/sup> =
(a) f<\/em>\u20131<\/sup>o g<\/em>\u20131<\/sup>
(b) fog<\/em>
(c) g<\/em>\u20131<\/sup>of<\/em>\u20131<\/sup>
(d) gof<\/em><\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em> : A<\/em> \u2192 B<\/em> and g<\/em> : B<\/em> \u2192 C<\/em> be the bijective functions

Since, f<\/em>:A<\/em>\u2192B<\/em>Thus, f<\/em>-1:B<\/em>\u2192A<\/em>       …(1)Since, g<\/em>:B<\/em>\u2192C<\/em>Thus, g<\/em>-1:C<\/em>\u2192B<\/em>       …(2)From (1) and (2), we getf<\/em>-1o<\/em>g<\/em>-1:C<\/em>\u2192A<\/em>          …(3)Also, g<\/em>o<\/em>f<\/em>:A<\/em>\u2192C<\/em>\u21d2(g<\/em>o<\/em>f<\/em>)-1:C<\/em>\u2192A<\/em>       …(4)Therefore, (g<\/em>o<\/em>f<\/em>)-1=f<\/em>-1o<\/em>g<\/em>-1

\u200bHence, the correct option is (a).

\u200b<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 60:<\/h4>\n\n\n\n

Let f<\/em> : N<\/em> \u2192 R<\/em> be the function defined by f<\/em>(x<\/em>)=2x<\/em>\u221212 and g<\/em> : Q<\/em> \u2192 R<\/em> be another function defined by g<\/em>(x<\/em>) = x<\/em> + 2. Then, (gof<\/em>) (3\/2) is
(a) 1

(b) 2

 (c) 72

(d) none of these<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>)=2x<\/em>\u221212 and g<\/em>(x<\/em>) = x<\/em> + 2

(g<\/em>o<\/em>f<\/em>)(x<\/em>)=g<\/em>(f<\/em>(x<\/em>))            =g<\/em>(2x<\/em>-12)            =2x<\/em>-12+2            =2x<\/em>-1+42            =2x<\/em>+32(g<\/em>o<\/em>f<\/em>)(32)=2(32)+32              =3+32              =62              =3

\u200bHence, the correct option is (d).

\u200b<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 1:<\/h4>\n\n\n\n

The total number of functions from the set A<\/em> = (1, 2, 3, 4 to the set B<\/em> = a<\/em>, b<\/em>, c<\/em>) is _________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>:A<\/em>\u2192B<\/em> where A<\/em>={1, 2, 3, 4} and B<\/em>={a<\/em>, b<\/em>, c<\/em>}

Number of elements in A <\/em>= 4
Number of elements in B <\/em>= 3
Each element of A<\/em> have 3 options to form an image.

Thus, Number of functions that can be formed = 3 \u00d7 3 \u00d7 3 \u00d7 3 = 81

\u200bHence, the total number of functions from the set A<\/em> = {1, 2, 3, 4} to the set B<\/em> = {a<\/em>, b<\/em>, c<\/em>} is 81.

\u200b<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 2:<\/h4>\n\n\n\n

The total number of one-one functions from the set A<\/em> = {a, b<\/em>, c<\/em>} to the set B<\/em> = {x<\/em>, y<\/em>, z<\/em>, t<\/em>} is _________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>:A<\/em>\u2192B<\/em> where A<\/em>={a<\/em>, b<\/em>, c<\/em>} and B<\/em>={x<\/em>, y<\/em>, z<\/em>, t<\/em>}

Number of elements in A <\/em>= 3
Number of elements in B <\/em>= 4

To form a one-one function,

Element a<\/em> \u2208 A<\/em> have 4 options to form an image.
Element b<\/em> \u2208 A<\/em> have 3 options to form an image.
Element c<\/em> \u2208 A<\/em> have 2 options to form an image.

Thus, Number of one-one functions that can be formed = 4 \u00d7 3 \u00d7 2 = 24

\u200bHence, the total number of one-one functions from the set A<\/em> = {a, b<\/em>, c<\/em>} to the set B<\/em> = {x<\/em>, y<\/em>, z<\/em>, t<\/em>} is 24.

\u200b<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 3:<\/h4>\n\n\n\n

The total number of onto functions from the set A<\/em> = (1, 2, 3, 4, 5) to the set B<\/em> = {x<\/em>, y<\/em>} is _________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>:A<\/em>\u2192B<\/em> where A<\/em>={1, 2, 3, 4, 5} and B<\/em>={x<\/em>, y<\/em>}

Number of elements in A <\/em>= 5
Number of elements in B <\/em>= 2


Each Element of A<\/em> have 2 options to form an image.

Thus, Total number of functions that can be formed = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 = 32

Number of functions having only one image i.e., {x<\/em>} = 1
Number of functions having only one image i.e., {y<\/em>} = 1

Thus, Number of onto functions that can be formed = 32 \u2212 1 \u2212 1 = 30

\u200bHence, the total number of onto functions from the set A<\/em> = {1, 2, 3, 4, 5} to the set B<\/em> = {x<\/em>, y<\/em>} is 30.<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 4:<\/h4>\n\n\n\n

The domain of the real function f<\/em>(x<\/em>)=\u221a16\u2212x<\/em>2 is _________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>)=\u221a16\u2212x<\/em>2

To find the domain, we find the real values of x<\/em> for which the function is defined.

16-x<\/em>2\u22650\u21d216\u2265x<\/em>2\u21d2x<\/em>2\u226416\u21d2x<\/em>\u22644 and x<\/em>\u2265-4\u21d2-4\u2264x<\/em>\u22644\u21d2x<\/em>\u2208[-4, 4]

Hence, the domain of the real function f<\/em>(x<\/em>)=\u221a16\u2212x<\/em>2 is [\u22124, 4].<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 5:<\/h4>\n\n\n\n

The domain of the real function f<\/em>(x<\/em>)=x<\/em>\u221a9\u2212x<\/em>2 is ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>)=x<\/em>\u221a9\u2212x<\/em>2 

To find the domain, we find the real values of x<\/em> for which the function is defined.

x<\/em>\u2208R<\/em> and 9-x<\/em>2>0\u21d2x<\/em>\u2208R<\/em> and 9>x<\/em>2\u21d2x<\/em>\u2208R<\/em> and x<\/em>2<9\u21d2x<\/em>\u2208R<\/em> and -3<x<\/em><3\u21d2-3<x<\/em><3\u21d2x<\/em>\u2208(-3, 3)

Hence, the domain of the real function f<\/em>(x<\/em>)=x<\/em>\u221a9\u2212x<\/em>2 is  (\u22123, 3).<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 6:<\/h4>\n\n\n\n

The range of the function f<\/em> : R<\/em> \u2192 R<\/em> given by f<\/em>(x<\/em>)=x<\/em>+\u221ax<\/em>2 is  _________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>)=x<\/em>+\u221ax<\/em>2 

f<\/em>(x<\/em>)=x<\/em>+\u221ax<\/em>2      =x<\/em>+|x<\/em>|      ={x<\/em>+x<\/em>         ,x<\/em>\u22650x<\/em>–x<\/em>         ,x<\/em><0      ={2x<\/em>         ,x<\/em>\u226500           ,x<\/em><0

To find the range, we find the real values of y<\/em> obtained.

y<\/em>=2x<\/em> when x<\/em>\u22650\u21d2x<\/em>=y<\/em>2\u22650\u21d2y<\/em>\u22650\u21d2y<\/em>\u2208[0, \u221e)         …(1)y<\/em>=0 when x<\/em><0    …(2)Thus, from (1) and (2),y<\/em>\u2208[0, \u221e)

Hence, the range of the function f<\/em> : R<\/em> \u2192 R<\/em> given by f<\/em>(x<\/em>)=x<\/em>+\u221ax<\/em>2 is [0, \u221e).<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 7:<\/h4>\n\n\n\n

The range of the function f<\/em> : R<\/em> \u2013{\u20132) \u2192 R<\/em> given by f<\/em>(x<\/em>)=x<\/em>+2|x<\/em>+2| is _________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>)=x<\/em>+2|x<\/em>+2|

f<\/em>(x<\/em>)=x<\/em>+2|x<\/em>+2|      ={x<\/em>+2x<\/em>+2                , x<\/em>+2\u22650x<\/em>+2-(x<\/em>+2)          , x<\/em>+2<0      ={1                , x<\/em>+2\u22650-1             , x<\/em>+2<0

To find the range, we find the real values of y<\/em> obtained.

Hence, the range of the function f<\/em> : R<\/em> \u2013{\u20132) \u2192 R<\/em> given by f<\/em>(x<\/em>)=x<\/em>+2|x<\/em>+2| is {\u20131, 1}.<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 8:<\/h4>\n\n\n\n

If f<\/em> : C<\/em> \u2192 C<\/em> is defined by f<\/em>(x<\/em>) = 8x<\/em>3<\/sup>, then f<\/em>\u20131<\/sup>(8) = . _________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = 8x<\/em>3<\/sup>

f<\/em>(x<\/em>)=8x<\/em>3\u21d2y<\/em>=8x<\/em>3\u21d2x<\/em>3=y<\/em>8\u21d2x<\/em>=(y<\/em>8)13Thus, f<\/em>-1(x<\/em>)=(x<\/em>8)13f<\/em>-1(8)=(88)13         =113         =1, \u03c9<\/em>, \u03c9<\/em>2       (\u2235 f<\/em>:C<\/em>\u2192C<\/em>)where, \u03c9<\/em> is the cube root of unity.

Hence, if f<\/em> : C<\/em> \u2192 C<\/em> is defined by f<\/em>(x<\/em>) = 8x<\/em>3<\/sup>, then f\u2212<\/sup><\/em>1<\/sup>(8) = 1, \u03c9<\/em>, \u03c9<\/em>2.<\/p>\n\n\n\n

Page No 2.77:<\/h4>\n\n\n\n

Question 9:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> is defined by f<\/em>(x<\/em>) = 8x<\/em>3<\/sup> then, f<\/em>\u20131<\/sup>(8) = _________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = 8x<\/em>3<\/sup>

f<\/em>(x<\/em>)=8x<\/em>3\u21d2y<\/em>=8x<\/em>3\u21d2x<\/em>3=y<\/em>8\u21d2x<\/em>=(y<\/em>8)13Thus, f<\/em>-1(x<\/em>)=(x<\/em>8)13f<\/em>-1(8)=(88)13         =113         =1                 (\u2235 f<\/em>:R<\/em>\u2192R<\/em>)

Hence, if f<\/em> : R<\/em> \u2192 R<\/em> is defined by f<\/em>(x<\/em>) = 8x<\/em>3<\/sup> then f\u2212<\/sup><\/em>1<\/sup>(8) = 1.<\/p>\n\n\n\n

Page No 2.78:<\/h4>\n\n\n\n

Question 10:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2013 {0} \u2192 R<\/em> \u2013 {0} is defined as f<\/em>(x<\/em>)=23x<\/em>, then f<\/em>\u20131<\/sup>(x<\/em>) = ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: A function f<\/em> : R<\/em> \u2013 {0} \u2192 R<\/em> \u2013 {0} is defined as f<\/em>(x<\/em>)=23x<\/em>

f<\/em>(x<\/em>)=23x<\/em>\u21d2y<\/em>=23x<\/em>\u21d23x<\/em>=2y<\/em>\u21d2x<\/em>=23y<\/em>Thus, f<\/em>-1(x<\/em>)=23x<\/em>

Hence, if f<\/em> : R<\/em> \u2013 {0} \u2192 R<\/em> \u2013 {0} is defined as f<\/em>(x<\/em>)=23x<\/em>, then f\u2212<\/sup><\/em>1<\/sup>(x<\/em>) = 23x<\/em>.<\/p>\n\n\n\n

Page No 2.78:<\/h4>\n\n\n\n

Question 11:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> is defined by f<\/em>(x<\/em>) = 6 \u2013 (x<\/em> \u2013 9)3<\/sup>, then f<\/em>\u20131<\/sup>(x<\/em>) = ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: A function f<\/em> : R<\/em> \u2192 R<\/em> is defined by f<\/em>(x<\/em>) = 6 \u2013 (x<\/em> \u2013 9)3<\/sup>

f<\/em>(x<\/em>)=6-(x<\/em>-9)3\u21d2y<\/em>=6-(x<\/em>-9)3\u21d2(x<\/em>-9)3=6-y<\/em>\u21d2x<\/em>-9=(6-y<\/em>)13\u21d2x<\/em>=9+(6-y<\/em>)13Thus, f<\/em>-1(x<\/em>)=9+(6-x<\/em>)13

Hence, if f<\/em> : R<\/em> \u2192 R<\/em> is defined by f<\/em>(x<\/em>) = 6 \u2013 (x<\/em> \u2013 9)3<\/sup>, then f\u2212<\/sup><\/em>1<\/sup>(x<\/em>) = 9+(6-x<\/em>)13.<\/p>\n\n\n\n

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Question 12:<\/h4>\n\n\n\n

Let A<\/em> = {1, 2, 3, 4} and f<\/em> : A<\/em> \u2192 A <\/em>be given by f<\/em>  = {(1, 4), (2, 3), (3, 2), (4, 1)}. Then f<\/em>\u20131<\/sup> = ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: A function f<\/em> : A<\/em> \u2192 A <\/em>be given by f<\/em>  = {(1, 4), (2, 3), (3, 2), (4, 1)}

f<\/em>={(1, 4), (2, 3), (3, 2), (4, 1)}\u21d2f<\/em>-1={(4, 1), (3, 2), (2, 3), (1, 4)}Thus, f<\/em>-1={(4, 1), (3, 2), (2, 3), (1, 4)}

Hence, f<\/em>-1={(4, 1), (3, 2), (2, 3), (1, 4)}.<\/p>\n\n\n\n

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Question 13:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> be defined by f<\/em>(x<\/em>) = (2 \u2013 x<\/em>5<\/sup>)1\/5<\/sup>, then fof<\/em>(x<\/em>) = ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = (2 \u2013 x<\/em>5<\/sup>)1\/5<\/sup>

f<\/em>o<\/em>f<\/em>(x<\/em>)=f<\/em>(f<\/em>(x<\/em>))         =f<\/em>((2-x<\/em>5)15)         =[2-((2-x<\/em>5)15)5]15         =[2-(2-x<\/em>5)15\u00d75]15         =[2-2+x<\/em>5]15         =[x<\/em>5]15         =x<\/em>5\u00d715         =x<\/em>


Hence, if f<\/em> : R<\/em> \u2192 R<\/em> be defined by f<\/em>(x<\/em>) = (2 \u2013 x<\/em>5<\/sup>)1\/5<\/sup>, then fof<\/em>(x<\/em>) = x<\/em>.<\/p>\n\n\n\n

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Question 14:<\/h4>\n\n\n\n

Let A<\/em> = {1, 2, 3, 4, 5, …, 10} and f <\/em>: A<\/em> \u2192 A <\/em>be an invertible function. Then, \u221110r<\/em>=1(f<\/em>\u22121o<\/em>f<\/em>) (r<\/em>)=___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f <\/em>: A<\/em> \u2192 A <\/em>is an invertible function, where A<\/em> = {1, 2, 3, 4, 5, …, 10}

Since, f<\/em> is invertibleTherefore, f<\/em>-1o<\/em>f<\/em>(x<\/em>)=x<\/em>       …(1)Now,\u221110r<\/em>=1(f<\/em>-1o<\/em>f<\/em>)(r<\/em>)=f<\/em>-1o<\/em>f<\/em>(1)+f<\/em>-1o<\/em>f<\/em>(2)+f<\/em>-1o<\/em>f<\/em>(3)+….+f<\/em>-1o<\/em>f<\/em>(10)                   =1+2+3+….+10             (From (1))                   =10(10+1)2                       (\u2235 1+2+3+…+n<\/em>=n<\/em>(n<\/em>+1)2)                   =5(11)                   =55


Hence, \u221110r<\/em>=1(f<\/em>\u22121o<\/em>f<\/em>) (r<\/em>)= 55.<\/p>\n\n\n\n

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Question 15:<\/h4>\n\n\n\n

Let A<\/em> = {1, 2, 3, 4, 5, 6) and B<\/em> = (2, 4, 6, 8, 10, 12). If f<\/em> : A<\/em> \u2192 B<\/em> is given by f<\/em>(x<\/em>) = 2x<\/em>, then f<\/em>\u20131<\/sup> as set of ordered pairs, is ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: A functionf<\/em> : A<\/em> \u2192 B<\/em> defined as f<\/em>(x<\/em>) = 2x, <\/em>where A<\/em> = {1, 2, 3, 4, 5, 6} and B<\/em> = {2, 4, 6, 8, 10, 12}

Since, f<\/em>(x<\/em>)=2x<\/em>Therefore,f<\/em>={(1, 2), (2, 4), (3, 6), (4, 8), (5, 10), (6, 12)}Hence,f<\/em>-1={(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6)}


Hence, if f<\/em> : A<\/em> \u2192 B<\/em> is given by f<\/em>(x<\/em>) = 2x<\/em>, then f<\/em>\u20131<\/sup> as set of ordered pairs, is {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6)}.<\/p>\n\n\n\n

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Question 16:<\/h4>\n\n\n\n

Let f <\/em> = {(0, \u20131), (\u20131, 3), (2, 3), (3, 5)} be a function from Z<\/em> to Z<\/em> defined by f<\/em>(x<\/em>) = ax<\/em> + b<\/em>. Then, (a<\/em>, b<\/em>) = ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f <\/em> = {(0, \u20131), (\u20131, \u20133), (2, 3), (3, 5)} is a function from Z<\/em> to Z<\/em> defined by f<\/em>(x<\/em>) = ax<\/em> + b<\/em>


f <\/em> = {(0, \u20131), (\u20131, \u20133), (2, 3), (3, 5)} defined by f<\/em>(x<\/em>) = ax<\/em> + b<\/em>

f<\/em>(0)=-1\u21d2a<\/em>(0)+b<\/em>=-1\u21d20+b<\/em>=-1\u21d2b<\/em>=-1          …(1)f<\/em>(2)=3\u21d2a<\/em>(2)+b<\/em>=3\u21d22a<\/em>+b<\/em>=3\u21d22a<\/em>-1=3             (From (1))\u21d22a<\/em>=3+1\u21d22a<\/em>=4\u21d2a<\/em>=2               …(2)Thus,a<\/em>=2 and b<\/em>=-1


Hence, (a<\/em>, b<\/em>) = \u200b(2, \u200b\u20131).


Disclaimer: The function f<\/em> must be equal to f <\/em> =  {(0, \u20131), (\u20131, \u20133), (2, 3), (3, 5)}.<\/p>\n\n\n\n

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Question 17:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em> \u2192 R<\/em> and g<\/em> : R<\/em> \u2192 R<\/em> be functions defined by f<\/em>(x<\/em>) = 5 \u2013 x<\/em>2<\/sup> and g<\/em>(x<\/em>) = 3x<\/em> \u2013 4. Then the value of fog<\/em> (\u20131) is ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = 5 \u2013 x<\/em>2<\/sup> and g<\/em>(x<\/em>) = 3x<\/em> \u2013 4


f<\/em>o<\/em>g<\/em>(-1)=f<\/em>(g<\/em>(-1))            =f<\/em>(3(-1)-4)            =f<\/em>(-3-4)            =f<\/em>(-7)            =5-(-7)2            =5-49            =-44


Hence, the value of fog<\/em> (\u20131) is \u200b\u201344.

 <\/p>\n\n\n\n

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Question 18:<\/h4>\n\n\n\n

Let f<\/em> be the greatest integer function defined as f<\/em>(x<\/em>) = [x<\/em>] and g<\/em> be the modules function defined as g<\/em>(x<\/em>) = |x<\/em>|, then the value of g<\/em>o<\/em>f<\/em>(\u221254) is ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = [x<\/em>] and g<\/em>(x<\/em>) = |x<\/em>|


g<\/em>o<\/em>f<\/em>(-54)=g<\/em>(f<\/em>(-54))              =g<\/em>([-54])              =g<\/em>(-2)              =|-2|              =2


Hence, the value of g<\/em>o<\/em>f<\/em>(\u221254) is \u200b2.
 <\/p>\n\n\n\n

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Question 19:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>) = cos [e<\/em>] x<\/em> + cos [\u2013e<\/em>] x<\/em>, then f<\/em>(\u03c0) = ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = cos[e<\/em>]x<\/em> + cos[\u2013e<\/em>]x<\/em>


f<\/em>(x<\/em>)=cos[e<\/em>]x<\/em>+cos[-e<\/em>]x<\/em>              =cos2x<\/em>+cos(-3x<\/em>)       (\u2235 e<\/em>=2.718 approx)      =cos2x<\/em>+cos3x<\/em>\u21d2f<\/em>(\u03c0)=cos2\u03c0+cos3\u03c0           =1-1           =0


Hence, f<\/em>(\u03c0) = \u200b0.<\/p>\n\n\n\n

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Question 20:<\/h4>\n\n\n\n

Let A<\/em> = {1, 2, 3} and B<\/em> = {a<\/em>, b<\/em>} be two sets. Then the number of constant functions from A<\/em> to B<\/em> is ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: Sets A<\/em> = {1, 2, 3} and B<\/em> = {a<\/em>, b<\/em>}

Two constant functions can be formed from A<\/em> to B<\/em>.
i.e., one is f<\/em>(x<\/em>) = a<\/em> and other is f<\/em>(x<\/em>) = b<\/em>

Hence, the number of constant functions from A<\/em> to B<\/em> is 2.<\/p>\n\n\n\n

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Question 21:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>) = cos [\u03c02<\/sup>] x<\/em> + cos [\u2013\u03c02<\/sup>] x<\/em>, then f<\/em>(\u03c02)=______________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = cos [\u03c02<\/sup>] x<\/em> + cos [\u2013\u03c02<\/sup>] x<\/em>


f<\/em>(x<\/em>)=cos[\u03c02]x<\/em>+cos[-\u03c02]x<\/em>              =cos9x<\/em>+cos(-10x<\/em>)       (\u2235 \u03c02=9.85 approx)      =cos9x<\/em>+cos10x<\/em>\u21d2f<\/em>(\u03c02)=cos9(\u03c02)+cos10(\u03c02)             =cos9\u03c02+cos5\u03c0             =0-1             =-1


Hence, f<\/em>(\u03c02)= \u200b\u20131.<\/p>\n\n\n\n

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Question 22:<\/h4>\n\n\n\n

The number of onto functions from A<\/em> = {a, b<\/em>, c<\/em>} to B<\/em> = {1, 2, 3, 4} is __________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: A function from A<\/em> = {a, b<\/em>, c<\/em>} to B<\/em> = {1, 2, 3, 4}

If a function from A<\/em> to B<\/em> is onto, then number of elements of A<\/em> \u2265 number of elements of B<\/em>.

But here, number of elements of A<\/em> < number of elements of B<\/em>

Thus, no onto function exist from A<\/em> = {a, b<\/em>, c<\/em>} to B<\/em> = {1, 2, 3, 4}.

Hence, the number of onto functions from A<\/em> = {a, b<\/em>, c<\/em>} to B<\/em> = {1, 2, 3, 4} is 0.<\/p>\n\n\n\n

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Question 23:<\/h4>\n\n\n\n

If f<\/em>(0, \u221e) \u2192 R<\/em> is given by f<\/em>(x<\/em>) = log10<\/sub> x<\/em>, then f<\/em>\u20131<\/sup>(x<\/em>) = ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = log10<\/sub>x<\/em>

f<\/em>(x<\/em>)=log10x<\/em>\u21d2y<\/em>=log10x<\/em>\u21d2x<\/em>=10y<\/em>Thus, f<\/em>-1(y<\/em>)=10y<\/em>.

Hence,  f\u2212<\/sup><\/em>1<\/sup>(x<\/em>) = 10x<\/sup><\/em>.<\/p>\n\n\n\n

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Question 24:<\/h4>\n\n\n\n

If f <\/em>: R<\/em>+<\/sup> \u2192 R<\/em> is defined as f<\/em>(x<\/em>) = log3 <\/sub>x<\/em>, then f<\/em>\u20131<\/sup>(x<\/em>) = ______________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = log3<\/sub>x<\/em>

f<\/em>(x<\/em>)=log3x<\/em>\u21d2y<\/em>=log3x<\/em>\u21d2x<\/em>=3y<\/em>Thus, f<\/em>-1(y<\/em>)=3y<\/em>.

Hence, f\u2212<\/sup><\/em>1<\/sup>(x<\/em>) = 3x<\/sup><\/em>.<\/p>\n\n\n\n

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Question 25:<\/h4>\n\n\n\n

If f <\/em>: R<\/em> \u2192 R<\/em>, g<\/em> : R<\/em> \u2192 R<\/em> are defined by f<\/em>(x<\/em>) = 5x<\/em> \u2013 3, g<\/em>(x<\/em>) = x<\/em>2<\/sup> + 3, then (gof<\/em>\u20131<\/sup>) (3) = _______________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = 5x<\/em> \u2013 3 and g<\/em>(x<\/em>) = x<\/em>2<\/sup> + 3

f<\/em>(x<\/em>)=5x<\/em>-3\u21d2y<\/em>=5x<\/em>-3\u21d25x<\/em>=y<\/em>+3\u21d2x<\/em>=y<\/em>+35Thus, f<\/em>-1(y<\/em>)=y<\/em>+35.Now,g<\/em>o<\/em>f<\/em>-1(3)=g<\/em>(f<\/em>-1(3))             =g<\/em>(3+35)             =g<\/em>(65)             =(65)2+3             =3625+3             =36+7525             =11125

Hence, gof\u2212<\/sup><\/em>1<\/sup>(3) = 11125.<\/p>\n\n\n\n

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Question 26:<\/h4>\n\n\n\n

If f <\/em>: R<\/em> \u2192 R<\/em> is given by f<\/em>(x<\/em>) = 2x<\/em> + |x<\/em>|, then f<\/em>(2x<\/em>) + f<\/em>(\u2013x<\/em>) + 4x<\/em> = _______________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = 2x<\/em> + |x<\/em>|

f<\/em>(x<\/em>)=2x<\/em>+|x<\/em>|f<\/em>(2x<\/em>)=2(2x<\/em>)+|2x<\/em>|\u21d2f<\/em>(2x<\/em>)=4x<\/em>+2|x<\/em>|         …(1)f<\/em>(-x<\/em>)=2(-x<\/em>)+|-x<\/em>|\u21d2f<\/em>(-x<\/em>)=-2x<\/em>+|x<\/em>|        …(2)Now,f<\/em>(2x<\/em>)+f<\/em>(-x<\/em>)+4x<\/em>=4x<\/em>+2|x<\/em>|-2x<\/em>+|x<\/em>|+4x<\/em>                           =6x<\/em>+3|x<\/em>|                           =3(2x<\/em>+|x<\/em>|)                           =3f<\/em>(x<\/em>)

Hence, f<\/em>(2x<\/em>) + f<\/em>(\u2013x<\/em>) + 4x<\/em> = 3f<\/em>(x<\/em>).<\/p>\n\n\n\n

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Question 27:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>)=1\u2212x<\/em>1+x<\/em>, then fof<\/em>(cos 2\u03b8) = ______________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>)=1\u2212x<\/em>1+x<\/em>

f<\/em>o<\/em>f<\/em>(cos2\u03b8<\/em>)=f<\/em>(f<\/em>(cos2\u03b8<\/em>))                =f<\/em>(1-cos2\u03b8<\/em>1+cos2\u03b8<\/em>)                =f<\/em>(2sin2\u03b8<\/em>2cos2\u03b8<\/em>)                =f<\/em>(tan2\u03b8<\/em>)                =1-tan2\u03b8<\/em>1+tan2\u03b8<\/em>                =cos2\u03b8<\/em>

Hence, fof<\/em>(cos 2\u03b8) = cos2\u03b8<\/em>.<\/p>\n\n\n\n

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Question 28:<\/h4>\n\n\n\n

Let f<\/em>(x<\/em>)=x<\/em>x<\/em>\u22121andf<\/em>(\u03b1<\/em>)f<\/em>(\u03b1<\/em>+1)=f<\/em>(\u03b1<\/em>k<\/em>), then k<\/em> = ______________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>)=x<\/em>x<\/em>\u22121andf<\/em>(\u03b1<\/em>)f<\/em>(\u03b1<\/em>+1)=f<\/em>(\u03b1<\/em>k<\/em>)

f<\/em>(a<\/em>)f<\/em>(a<\/em>+1)=a<\/em>a<\/em>-1a<\/em>+1a<\/em>+1-1             =a<\/em>a<\/em>-1a<\/em>+1a<\/em>             =a<\/em>2(a<\/em>-1)(a<\/em>+1)             =a<\/em>2a<\/em>2-1             …(1)                It is given that,f<\/em>(a<\/em>)f<\/em>(a<\/em>+1)=f<\/em>(a<\/em>k<\/em>)\u21d2a<\/em>2a<\/em>2-1=a<\/em>k<\/em>a<\/em>k<\/em>-1\u21d2k<\/em>=2

Hence, k<\/em> = 2.<\/p>\n\n\n\n

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Question 29:<\/h4>\n\n\n\n

If f<\/em> (f<\/em>(x<\/em>)) = x<\/em> + 1 for all x<\/em> \u2208 R<\/em> and if f<\/em>(0)=12, then f<\/em>(1) = ____________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em> (f<\/em>(x<\/em>)) = x<\/em> + 1 for all x<\/em> \u2208 R<\/em> and f<\/em>(0)=12

f<\/em>(f<\/em>(x<\/em>))=x<\/em>+1\u21d2f<\/em>(f<\/em>(0))=0+1\u21d2f<\/em>(12)=1          (\u2235 f<\/em>(0)=12)       ….(1)Now,f<\/em>(f<\/em>(12))=12+1\u21d2f<\/em>(1)=1+22     (\u2235 f<\/em>(12)=1)\u21d2f<\/em>(1)=32

Hence, f<\/em>(1) = 32.<\/p>\n\n\n\n

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Question 30:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>) = 3x<\/em> + 10 and g<\/em>(x<\/em>) = x<\/em>2<\/sup> \u20131, then (fog<\/em>)\u20131<\/sup> is equal to ___________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em>(x<\/em>) = 3x<\/em> + 10 and g<\/em>(x<\/em>) = x<\/em>2<\/sup> \u20131

f<\/em>o<\/em>g<\/em>(x<\/em>)=f<\/em>(g<\/em>(x<\/em>))         =f<\/em>(x<\/em>2-1)         =3(x<\/em>2-1)+10         =3x<\/em>2-3+10         =3x<\/em>2+7Thus, f<\/em>o<\/em>g<\/em>(x<\/em>)=3x<\/em>2+7\u21d2y<\/em>=3x<\/em>2+7\u21d23x<\/em>2=y<\/em>-7\u21d2x<\/em>2=y<\/em>-73\u21d2x<\/em>=\u00b1\u221ay<\/em>-73f<\/em>o<\/em>g<\/em>-1(x<\/em>)=\u00b1\u221ax<\/em>-73

Hence, (fog<\/em>)\u20131<\/sup> is equal to \u00b1\u221ax<\/em>-73.<\/p>\n\n\n\n

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Question 31:<\/h4>\n\n\n\n

Let f<\/em> = {(1, 2), (3, 5), (4, 1)) and g<\/em> = {(2, 3), (5, 1), (1, 3)}. Then, gof<\/em> = __________ and fog<\/em> = __________.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given: f<\/em> = {(1, 2), (3, 5), (4, 1)) and g<\/em> = {(2, 3), (5, 1), (1, 3)}

f<\/em>o<\/em>g<\/em>(x<\/em>)=f<\/em>(g<\/em>(x<\/em>))f<\/em>o<\/em>g<\/em>(1)=f<\/em>(g<\/em>(1))         =f<\/em>(3)         =5f<\/em>o<\/em>g<\/em>(2)=f<\/em>(g<\/em>(2))         =f<\/em>(3)         =5f<\/em>o<\/em>g<\/em>(5)=f<\/em>(g<\/em>(5))         =f<\/em>(1)         =2Hence, f<\/em>o<\/em>g<\/em>={(1, 5), (2, 5), (5, 2)}g<\/em>o<\/em>f<\/em>(x<\/em>)=g<\/em>(f<\/em>(x<\/em>))g<\/em>o<\/em>f<\/em>(1)=g<\/em>(f<\/em>(1))         =g<\/em>(2)         =3g<\/em>o<\/em>f<\/em>(3)=g<\/em>(f<\/em>(3))         =g<\/em>(5)         =1g<\/em>o<\/em>f<\/em>(4)=g<\/em>(f<\/em>(4))         =g<\/em>(1)         =3Hence, g<\/em>o<\/em>f<\/em>={(1, 3), (3, 1), (4, 3)}.

Hence, gof<\/em> = {(1, 3), (3, 1), (4, 3)} and fog<\/em> = {(1, 5), (2, 5), (5, 2)}.<\/p>\n\n\n\n

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Question 1:<\/h4>\n\n\n\n

Which one of the following graphs represents a function?

\"\"       \"\"<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

In graph (b), 0 has more than one image, whereas every value of x<\/em> in graph (a) has a unique image.
Thus, graph (a) represents a function.
So, the answer is (a).<\/p>\n\n\n\n

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Question 2:<\/h4>\n\n\n\n

Which of the following graphs represents a one-one function?

\"\" \"\"<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

In the graph of (b), different elements on the x<\/em>-axis have different images on the y<\/em>-axis.But in (a), the graph cuts the x<\/em>-axis at 3 points, which means that 3 points on the x<\/em>-axis have the same image as 0 and hence, it is not one-one.<\/p>\n\n\n\n

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Question 3:<\/h4>\n\n\n\n

If A<\/em> = {1, 2, 3} and B<\/em> = {a<\/em>, b<\/em>}, write the total number of functions from A<\/em> to B<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Formula:
If set A<\/em> has m<\/em> elements and set B<\/em> has n<\/em> elements, then the number of functions from A<\/em> to B<\/em> is nm<\/em>.
Given:
A<\/em> = {1, 2, 3} and B<\/em> = {a<\/em>, b<\/em>}
\u21d2n<\/em>(A<\/em>) <\/em>= 3 and  n<\/em>(B<\/em>) = 2
\u2234 Number of functions from A<\/em> to B =<\/em> 23<\/sup> = 8<\/p>\n\n\n\n

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Question 4:<\/h4>\n\n\n\n

If A<\/em> = {a<\/em>, b<\/em>, c<\/em>} and B<\/em> = {\u22122, \u22121, 0, 1, 2}, write the total number of one-one functions from A<\/em> to B<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>:A<\/em>\u2192B<\/em> be a one-one function.
 Then, f(a) can take 5 values, f(b) can take 4 values and f(c) can take 3 values.<\/em>

Then, the number of one-one functions = 5 \u00d7 4 \u00d7 3 = 60<\/p>\n\n\n\n

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Question 5:<\/h4>\n\n\n\n

Write the total number of one-one functions from set A<\/em> = {1, 2, 3, 4} to set B<\/em> = {a<\/em>, b<\/em>, c<\/em>}.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

A <\/em>has 4 elements and B<\/em> has 3 elements.
Also, one-one function is only possible from A<\/em> to B <\/em>if n<\/em>(A<\/em>)\u2264n<\/em>(B<\/em>).
But, here n<\/em>(A<\/em>)>n<\/em>(B<\/em>).
So, the number of one-one functions from A<\/em> to B <\/em>is 0.<\/p>\n\n\n\n

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Question 6:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> is defined by f<\/em>(x<\/em>) = x<\/em>2<\/sup>, write f<\/em>\u22121<\/sup> (25).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(25)=x<\/em>            … (1)\u21d2f<\/em>(x<\/em>)=25\u21d2x<\/em>2=25\u21d2x<\/em>2-25=0\u21d2(x<\/em>-5)(x<\/em>+5)=0\u21d2x<\/em>=\u00b15\u21d2f<\/em>-1(25)={-5, 5}       [from (1)]<\/p>\n\n\n\n

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Question 7:<\/h4>\n\n\n\n

If f<\/em> : C<\/em> \u2192 C<\/em> is defined by f<\/em>(x<\/em>) = x<\/em>2<\/sup>, write f<\/em>\u22121<\/sup> (\u22124). Here, C<\/em> denotes the set of all complex numbers.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(-4)=x<\/em>                  … (1)\u21d2f<\/em>(x<\/em>)=-4\u21d2x<\/em>2=-4\u21d2x<\/em>2+4=0\u21d2(x<\/em>+2i<\/em>)(x<\/em>-2i<\/em>)=0          [using the identity: a<\/em>2+b<\/em>2=(a<\/em>–i<\/em>b<\/em>)(a<\/em>+i<\/em>b<\/em>)]\u21d2x<\/em>=\u00b12i<\/em>                           [as x<\/em>\u2208C<\/em>]\u21d2f<\/em>-1(25)={-2i<\/em>, 2i<\/em>}         [from (1)]<\/p>\n\n\n\n

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Question 8:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> is given by f<\/em>(x<\/em>) = x<\/em>3<\/sup>, write f<\/em>\u22121<\/sup> (1).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(1)= x<\/em>           … (1)\u21d2f<\/em>(x<\/em>)= 1\u21d2x<\/em>3= 1\u21d2x<\/em>3-1= 0\u21d2(x<\/em>-1)(x<\/em>2+x<\/em>+1)= 0        [using the identity:a<\/em>3-b<\/em>3=(a<\/em>–b<\/em>)(a<\/em>2+a<\/em>b<\/em>+b<\/em>2)]\u21d2x<\/em>=1          ( as x<\/em>\u2208R<\/em>) \u21d2f<\/em>-1(1)= {1}               [from (1)]<\/p>\n\n\n\n

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Question 9:<\/h4>\n\n\n\n

Let C<\/em> denote the set of all complex numbers. A function f<\/em> : C<\/em> \u2192 C<\/em> is defined by f<\/em>(x<\/em>) = x<\/em>3<\/sup>. Write f<\/em>\u22121<\/sup> (1).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(1)=x<\/em>                   … (1)\u21d2f<\/em>(x<\/em>)=1\u21d2x<\/em>3=1\u21d2x<\/em>3-1=0\u21d2(x<\/em>-1)(x<\/em>2+x<\/em>+1)=0        [Using identity: a<\/em>3-b<\/em>3=(a<\/em>–b<\/em>)(a<\/em>2+a<\/em>b<\/em>+b<\/em>2)]\u21d2(x<\/em>-1)(x<\/em>–\u03c9<\/em>)(x<\/em>–\u03c9<\/em>2)=0, where \u03c9<\/em>=1\u00b1i<\/em>\u221a32\u21d2x<\/em>=1, \u03c9<\/em> or \u03c9<\/em>2                 (as x<\/em>\u2208C<\/em>)\u21d2f<\/em>-1(1)={1, \u03c9<\/em>, \u03c9<\/em>2}             [from (1)]<\/p>\n\n\n\n

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Question 10:<\/h4>\n\n\n\n

Let f<\/em>  be a function from C<\/em> (set of all complex numbers) to itself given by f<\/em>(x<\/em>) = x<\/em>3<\/sup>. Write f<\/em>\u22121<\/sup> (\u22121).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(-1)=x<\/em>                          … (1)\u21d2f<\/em>(x<\/em>)=-1\u21d2x<\/em>3=-1\u21d2x<\/em>3+1=0\u21d2(x<\/em>+1)(x<\/em>2-x<\/em>+1)=0        [using the identity: a<\/em>3+b<\/em>3=(a<\/em>+b<\/em>)(a<\/em>2-a<\/em>b<\/em>+b<\/em>2)]\u21d2(x<\/em>+1)(x<\/em>+\u03c9<\/em>)(x<\/em>+\u03c9<\/em>2)=0, where \u03c9<\/em>= 1\u00b1i<\/em>\u221a32    \u21d2x<\/em>=-1, –\u03c9<\/em>, –\u03c9<\/em>2        (as x<\/em>\u2208C<\/em>)\u21d2f<\/em>-1(-1)={-1, –\u03c9<\/em>, –\u03c9<\/em>2}     [from (1)]<\/p>\n\n\n\n

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Question 11:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> be defined by f<\/em>(x<\/em>) = x<\/em>4<\/sup>, write f<\/em>\u22121<\/sup> (1).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(1)=x<\/em>             … (1)\u21d2f<\/em>(x<\/em>)=1\u21d2x<\/em>4=1\u21d2x<\/em>4-1=0\u21d2(x<\/em>2-1)(x<\/em>2+1)=0               [using identity: a<\/em>2-b<\/em>2=(a<\/em>–b<\/em>)(a<\/em>+b<\/em>)]\u21d2(x<\/em>-1)(x<\/em>+1)(x<\/em>2+1)=0         [using identity: a<\/em>2-b<\/em>2=(a<\/em>–b<\/em>)(a<\/em>+b<\/em>)]\u21d2x<\/em>=\u00b11               [as x<\/em>\u2208R<\/em>]\u21d2f<\/em>-1(1)={-1, 1}       [ from (1)]<\/p>\n\n\n\n

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Question 12:<\/h4>\n\n\n\n

If f<\/em> : C<\/em> \u2192 C<\/em> is defined by f<\/em>(x<\/em>) = x<\/em>4<\/sup>, write f<\/em>\u22121<\/sup> (1).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(1)=x<\/em>   … (1)\u21d2f<\/em>(x<\/em>)=1\u21d2x<\/em>4=1\u21d2x<\/em>4-1=0\u21d2(x<\/em>2-1)(x<\/em>2+1)=0                                                   [using identity: a<\/em>2-b<\/em>2=(a<\/em>–b<\/em>)(a<\/em>+b<\/em>)]\u21d2(x<\/em>-1)(x<\/em>+1)(x<\/em>–i<\/em>)(x<\/em>+i<\/em>)=0, where i<\/em>=\u221a-1           [using identity: a<\/em>2-b<\/em>2=(a<\/em>–b<\/em>)(a<\/em>+b<\/em>)]\u21d2x<\/em>=\u00b11, \u00b1i<\/em> \u21d2f<\/em>-1(1)={-1, 1, i<\/em>,-i<\/em>}      [from (1)]<\/p>\n\n\n\n

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Question 13:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> is defined by f<\/em>(x) = x<\/em>2<\/sup>, find f<\/em>\u22121<\/sup> (\u221225).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(-25)=x<\/em>\u21d2f<\/em>(x<\/em>)=-25\u21d2x<\/em>2=-25We cannot find x\u2208R, <\/em>such that x2=-25       (as x2\u22650 for all x\u2208R)So, f<\/em>–<\/em>1<\/em>(–<\/em>25<\/em>)=<\/em>\u03d5<\/em><\/p>\n\n\n\n

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Question 14:<\/h4>\n\n\n\n

If f<\/em> : C<\/em> \u2192 C<\/em> is defined by f<\/em>(x<\/em>) = (x<\/em> \u2212 2)3<\/sup>, write f<\/em>\u22121<\/sup> (\u22121).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(-1)=x<\/em>                              … (1)\u21d2f<\/em>(x<\/em>)=-1\u21d2(x<\/em>-2)3=-1\u21d2x<\/em>-2=-1 or –\u03c9<\/em> or –\u03c9<\/em>2         (as the roots of (-1)13are -1, -\u03c9 and -\u03c92, where \u03c9=1\u00b1i\u221a32)\u21d2x<\/em>=-1+2 or 2-\u03c9<\/em> or 2-\u03c9<\/em>2=1,  2-\u03c9<\/em>, 2-\u03c9<\/em>\u21d2f<\/em>-1(-1)={1,  2-\u03c9<\/em>, 2-\u03c9<\/em>2}            [from (1)]<\/p>\n\n\n\n

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Question 15:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> is defined by f<\/em>(x<\/em>) = 10 x<\/em> \u2212 7, then write f<\/em>\u22121<\/sup> (x<\/em>).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(x<\/em>)=y<\/em>                     … (1)\u21d2f<\/em>(y<\/em>)=x<\/em>\u21d210y<\/em>-7=x<\/em>\u21d210y<\/em>=x<\/em>+7\u21d2y<\/em>=x<\/em>+710\u21d2f<\/em>-1(x<\/em>)=x<\/em>+710             (From (1))<\/p>\n\n\n\n

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Question 16:<\/h4>\n\n\n\n

Let f<\/em> : (-\u03c0<\/em>2, \u03c0<\/em>2)\u2192R<\/em> be a function defined by f<\/em>(x<\/em>) = cos [x<\/em>]. Write range (f<\/em>).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Domain =(-\u03c02, \u03c02)=(-1.57, 1.57)   (as \u03c0=227)So, cos [x<\/em>]=cos (-2)=cos 2   \u2200x<\/em>\u2208(-1.57, 0)Also, cos 0=1 for x<\/em>=0And cos [x<\/em>]=cos 1  \u2200x<\/em>\u2208(0, 1.57)\u2234Range={1, cos 1, cos 2}<\/p>\n\n\n\n

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Question 17:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> defined by f<\/em>(x<\/em>) = 3x<\/em> \u2212 4 is invertible, then write f<\/em>\u22121<\/sup> (x<\/em>).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(x<\/em>)=y<\/em>                 … (1)\u21d2f<\/em>(y<\/em>)=x<\/em>\u21d23y<\/em>-4=x<\/em>\u21d23y<\/em>=x<\/em>+4\u21d2y<\/em>=x<\/em>+43\u21d2f<\/em>-1(x<\/em>)=x<\/em>+43       [from (1)]<\/p>\n\n\n\n

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Question 18:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em>, g<\/em> : R<\/em> \u2192 are given by f<\/em>(x<\/em>) = (x<\/em> + 1)2<\/sup> and g<\/em>(x<\/em>) = x<\/em>2<\/sup> + 1, then write the value of fog<\/em> (\u22123).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(f<\/em>o<\/em>g<\/em>)(-3)=f<\/em> (g<\/em> (-3))=f<\/em>((-3)2+1)=f<\/em>(10)=(10+1)2=121<\/p>\n\n\n\n

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Question 19:<\/h4>\n\n\n\n

Let A<\/em> = {x<\/em> \u2208 R<\/em> : \u22124 \u2264 x<\/em> \u2264 4 and x<\/em> \u2260 0} and f<\/em> : A<\/em> \u2192 R<\/em> be defined by f<\/em>(x<\/em>)=|x<\/em>|x<\/em>. Write the range of f<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

\u2235f<\/em>(x<\/em>)=|x<\/em>|x<\/em>=\u00b1x<\/em>x<\/em>=\u00b11 \u2200x<\/em>\u2208A<\/em>, range of f<\/em>={-1, 1}.<\/p>\n\n\n\n

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Question 20:<\/h4>\n\n\n\n

Let f<\/em> : [-\u03c0<\/em>2, \u03c0<\/em>2]\u2192 A<\/em> be defined by f<\/em>(x<\/em>) = sin x<\/em>. If f<\/em> is a bijection, write set A<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

\u2235f<\/em> is a bijection,
co-domain of f<\/em> = range of f<\/em>
As -1\u2264sin x\u22641,
-1\u2264y\u22641<\/em>
So, A = <\/em>[-1, 1]<\/p>\n\n\n\n

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Question 21:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em> \u2192 R<\/em>+<\/sup> be defined by f<\/em>(x<\/em>) = ax<\/sup><\/em>, a<\/em> > 0 and a<\/em> \u2260 1. Write f<\/em>\u22121<\/sup> (x<\/em>).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(x<\/em>)=y<\/em>                    … (1)\u21d2f<\/em>(y<\/em>)=x<\/em>\u21d2a<\/em>y<\/em>=x<\/em>\u21d2y<\/em>=loga<\/em> x<\/em>\u21d2f<\/em>-1(x<\/em>)=log a<\/em> x<\/em>            [from (1)]<\/p>\n\n\n\n

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Question 22:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em> \u2212 {\u22121} \u2192 R<\/em> \u2212 {1} be given by f<\/em>(x<\/em>)=x<\/em>x<\/em>+1. Write f<\/em>-1 (x<\/em>).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(x<\/em>)=y<\/em>                … (1)\u21d2f<\/em>(y<\/em>)=x<\/em>\u21d2y<\/em>y<\/em>+1=x<\/em>\u21d2y<\/em>=x<\/em>y<\/em>+x<\/em>\u21d2y<\/em>–x<\/em>y<\/em>=x<\/em>\u21d2y<\/em>(1-x<\/em>)=x<\/em>\u21d2y<\/em>=x<\/em>1-x<\/em>\u21d2f<\/em>-1(x<\/em>)=x<\/em>1-x<\/em>            [from (1)]<\/p>\n\n\n\n

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Question 23:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em>-{-35}\u2192R<\/em> be a function defined as f<\/em>(x<\/em>)=2x<\/em>5x<\/em>+3.

Write f<\/em>-1 : Range of f<\/em>\u2192R<\/em>-{-35}.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(x<\/em>)=y<\/em>              … (1)\u21d2f<\/em>(y<\/em>)=x<\/em>\u21d22y<\/em>5y<\/em>+3=x<\/em>\u21d22y<\/em>=5x<\/em>y<\/em>+3x<\/em>\u21d22y<\/em>-5x<\/em>y<\/em>=3x<\/em>\u21d2y<\/em>(2-5x<\/em>)=3x<\/em>\u21d2y<\/em>=3x<\/em>2-5x<\/em>\u21d2f<\/em>-1(x<\/em>)=3x<\/em>2-5x<\/em>   [from (1)]<\/p>\n\n\n\n

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Question 24:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em> \u2192 R<\/em>, g<\/em> : R<\/em> \u2192 R<\/em> be two functions defined by f<\/em>(x<\/em>) = x<\/em>2<\/sup> + x<\/em> + 1 and g<\/em>(x<\/em>) = 1 \u2212 x<\/em>2<\/sup>. Write fog<\/em> (\u22122).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(f<\/em>o<\/em>g<\/em>)(-2)=f<\/em> (g<\/em> (-2))=f<\/em>(1-(-2)2)=f<\/em>(-3)=(-3)2+(-3)+1=9-3+1=7<\/p>\n\n\n\n

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Question 25:<\/h4>\n\n\n\n

Let f<\/em> : R<\/em> \u2192 R<\/em> be defined as f<\/em>(x<\/em>)=2x<\/em>-34. Write f<\/em>o<\/em>f<\/em>-1 (1).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Let f<\/em>-1(x<\/em>)=y<\/em>                …(1)\u21d2f<\/em>(y<\/em>)=x<\/em>\u21d22y<\/em>-34=x<\/em>\u21d22y<\/em>-3=4x<\/em>\u21d22y<\/em>=4x<\/em>+3\u21d2y<\/em>=4x<\/em>+32\u21d2f<\/em>-1(x<\/em>)=4x<\/em>+32      [from (1)]\u21d2f<\/em>-1(x<\/em>)=4x<\/em>+32\u2234(f<\/em>o<\/em>f<\/em>-1)(1)=f<\/em>(4(1)+32)=f<\/em>(72)=2(72)-34=7-34=44=1<\/p>\n\n\n\n

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Question 26:<\/h4>\n\n\n\n

Let f<\/em> be an invertible real function. Write

(f<\/em>-1 o<\/em>f<\/em>) (1) +(f<\/em>-1 o<\/em>f<\/em>) (2)+…+(f<\/em>-1o<\/em>f<\/em>) (100).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Given that f<\/em>  is an invertible real function.
f<\/em>-1o<\/em> f<\/em>=I<\/em>,where I is an identity function.So,(f<\/em>-1o<\/em> f<\/em>)(1)+(f<\/em>-1o<\/em> f<\/em>)(2)+…+(f<\/em>-1o<\/em> f<\/em>)(100)=I<\/em>(1)+I<\/em>(2)+… +I<\/em>(100)=1+2+…+100 (As I<\/em>(x<\/em>)=x<\/em>, \u2200x<\/em>\u2208R<\/em>)=100(100+1)2[Sum of first n natural numbers=n<\/em>(n<\/em>+1)2]=5050<\/p>\n\n\n\n

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Question 27:<\/h4>\n\n\n\n

Let A<\/em> = {1, 2, 3, 4} and B<\/em> = {a<\/em>, b<\/em>} be two sets. Write the total number of onto functions from A<\/em> to B<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Formula:

When two sets A<\/em> and B <\/em>have m<\/em> and n <\/em>elements respectively, then the number of onto functions from A<\/em> to B<\/em> is
{\u2211n<\/em>r<\/em>=1 (-1)r<\/em> n<\/em>C<\/em>r<\/em> r<\/em>m<\/em>, if m<\/em>\u2265n<\/em>o<\/em>, if m<\/em><n<\/em>

Here, number of elements in A<\/em> = 4 = m<\/em>
Number of elements in B<\/em> = 2 = n<\/em>
So, m<\/em> > n
Number of onto functions
=\u22112r<\/em>=1 (-1)r<\/em> 2C<\/em>r<\/em> r<\/em>4=(-1)1 2C<\/em>1 14+(-1)2 2C<\/em>2 24=-2+16=14<\/p>\n\n\n\n

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Question 28:<\/h4>\n\n\n\n

Write the domain of the real function f<\/em>(x<\/em>)=\u221ax<\/em>-[x<\/em>].<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

[x] is the greatest integral function.
So, 0\u2264x<\/em>-[x<\/em>]<1\u21d2\u221ax<\/em>-[x<\/em>] exists for every x<\/em>\u2208R<\/em>.\u21d2Domain =R<\/em><\/p>\n\n\n\n

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Question 29:<\/h4>\n\n\n\n

Write the domain of the real function f<\/em>(x<\/em>)=\u221a[x<\/em>]-x<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

[x<\/em>] is the greatest integer function.
[x<\/em>]\u2264x<\/em>, \u2200x<\/em>\u2208R<\/em>\u21d2[x<\/em>]-x<\/em>\u22640,\u2200x<\/em>\u2208R<\/em>\u21d2\u221a[x<\/em>]-x<\/em> does not exist for any x<\/em>\u2208R<\/em>.Domain =\u03d5<\/em><\/p>\n\n\n\n

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Question 30:<\/h4>\n\n\n\n

Write the domain of the real function f<\/em>(x<\/em>)=1\u221a|x<\/em>|-x<\/em>.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

Case-1: When x<\/em>>0|x<\/em>|=x<\/em>\u21d21\u221a|x<\/em>|-x<\/em>=1\u221ax<\/em>–x<\/em>=10=\u221eCase-2: When x<\/em><0|x<\/em>|=-x<\/em>\u21d21\u221a|x<\/em>|-x<\/em>=1\u221a-x<\/em>–x<\/em>=1\u221a-2x<\/em> (exists because when x<0, -2x>0)\u21d2f<\/em>(x<\/em>) is defined when x<\/em><0So, domain =(-\u221e,0)<\/p>\n\n\n\n

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Question 31:<\/h4>\n\n\n\n

Write whether f<\/em> : R<\/em> \u2192 R,<\/em> given by f<\/em>(x<\/em>)=x<\/em>+\u221ax<\/em>2, is one-one, many-one, onto or into.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em>(x<\/em>)=x<\/em>+\u221ax<\/em>2=x<\/em>\u00b1x<\/em>=0 o<\/em>r<\/em> 2x<\/em>So, each element x<\/em> in the domain may contain 2 images.For example,f<\/em>(0)=0+\u221a02=0f<\/em>(-1)=-1+\u221a(-1)2=-1+\u221a1=-1+1=0Here, the image of 0 and -1 is 0.
Hence, f <\/em>is may-one.<\/p>\n\n\n\n

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Question 32:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>) = x<\/em> + 7 and g<\/em>(x<\/em>) = x<\/em> \u2212 7, x<\/em> \u2208 R<\/em>, write fog<\/em> (7).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(f<\/em>o<\/em>g<\/em>)(7)=f<\/em> ( g<\/em>(7))=f<\/em>(7-7)=f<\/em> (0)=0+7=7<\/p>\n\n\n\n

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Question 33:<\/h4>\n\n\n\n

What is the range of the function f<\/em>(x<\/em>)=|x<\/em>-1|x<\/em>-1?<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em>(x<\/em>)=|x<\/em>-1|x<\/em>-1=\u00b1(x<\/em>-1)x<\/em>-1=\u00b11Range of f<\/em>={-1, 1}<\/p>\n\n\n\n

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Question 34:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> be defined by f<\/em>(x<\/em>) = (3 \u2212 x<\/em>3<\/sup>)1\/3, then find fof<\/em> (x<\/em>).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

(f<\/em>o<\/em>f<\/em>) (x<\/em>)=f<\/em> (f<\/em> (x<\/em>))=f<\/em> ((3-x<\/em>3)13)=[3-((3-x<\/em>3)13)3]13=[3-(3-x<\/em>3)]13=(x<\/em>3)13=x<\/em><\/p>\n\n\n\n

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Question 35:<\/h4>\n\n\n\n

If f<\/em> : R<\/em> \u2192 R<\/em> is defined by f<\/em>(x<\/em>) = 3x<\/em> + 2, find f<\/em> (f<\/em> (x<\/em>)).<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em>(f<\/em> (x<\/em>))=f<\/em> (3x<\/em>+2)=3 (3x<\/em>+2)+2=9x<\/em>+6+2=9x<\/em>+8<\/p>\n\n\n\n

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Question 36:<\/h4>\n\n\n\n

Let A<\/em> = {1, 2, 3}, B<\/em> = {4, 5, 6, 7} and let f<\/em> = {(1, 4), (2, 5), (3, 6)} be a function from A<\/em> to B<\/em>. State whether f<\/em> is one-one or not.<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

f<\/em> = {(1, 4), (2, 5), (3, 6)}

Here, different elements of the domain have different images in the co-domain.
So, f<\/em> is one-one.<\/p>\n\n\n\n

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Question 37:<\/h4>\n\n\n\n

If f<\/em> : {5, 6} \u2192 {2, 3} and g<\/em> : {2, 3} \u2192 {5, 6} are given by f<\/em> = {(5, 2), (6, 3)} and g<\/em> = {(2, 5), (3, 6)}, then find fog<\/em>.    [NCERT EXEMPLAR]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,
f<\/em> : {5, 6} \u2192 {2, 3} and g<\/em> : {2, 3} \u2192 {5, 6} are given by f<\/em> = {(5, 2), (6, 3)} and g<\/em> = {(2, 5), (3, 6)}

As,
fog<\/em>(2) = f<\/em>(g<\/em>(2)) = f<\/em>(5) = 2,
fog<\/em>(3) = f<\/em>(g<\/em>(3)) = f<\/em>(6) = 3,

So,
fog<\/em> : {2, 3} \u2192 {2, 3} is defined as
fog<\/em> = {(2, 2), (3, 3)}<\/p>\n\n\n\n

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Question 38:<\/h4>\n\n\n\n

Let f<\/em> : R<\/strong> \u2192 R<\/strong> be the function defined by f<\/em>(x<\/em>) = 4x<\/em> – 3 for all x<\/em> \u2208 R<\/strong>. Then write f <\/em>-1<\/sup>.                                                 [NCERT EXEMPLAR]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,
f<\/em> : R<\/strong> \u2192 R<\/strong> is the function defined by f<\/em>(x<\/em>) = 4x<\/em> – 3 for all x<\/em> \u2208 R<\/strong>

Let f<\/em>(x<\/em>)=y<\/em>. Then,y<\/em>=4x<\/em>-3\u21d24x<\/em>=y<\/em>+3\u21d2x<\/em>=y<\/em>+34So, f<\/em>-1(y<\/em>)=y<\/em>+34or, f<\/em>-1(x<\/em>)=x<\/em>+34<\/p>\n\n\n\n

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Question 39:<\/h4>\n\n\n\n

Which one the following relations on A<\/em> = {1, 2, 3} is a function?
f<\/em> = {(1, 3), (2, 3), (3, 2)}, g<\/em> = {(1, 2), (1, 3), (3, 1)}                                                                                                         [NCERT EXEMPLAR]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

As, each element of the domain set has unique image in the relation f<\/em> = {(1, 3), (2, 3), (3, 2)}

So, f<\/em> is a function.

Also, the element 1 of the domain set has two images 2 and 3 of the range set in the relation g<\/em> = {(1, 2), (1, 3), (3, 1)}

So, g<\/em> is not a function.<\/p>\n\n\n\n

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Question 40:<\/h4>\n\n\n\n

Write the domain of the real function f<\/em> defined by f<\/em>(x<\/em>) = \u221a25-x<\/em>2.                                                                                 [NCERT EXEMPLAR]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,f<\/em>(x<\/em>)=\u221a25-x<\/em>2The function is defined only when 25-x<\/em>2\u22650\u21d2x<\/em>2-25\u22640\u21d2(x<\/em>+5)(x<\/em>-5)\u22640\u21d2x<\/em>\u2208[-5, 5]So, the domain of the given function is [-5, 5].<\/p>\n\n\n\n

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Question 41:<\/h4>\n\n\n\n

Let A<\/em> = {a<\/em>, b<\/em>, c<\/em>, d<\/em>} and f<\/em> : A<\/em> \u2192 A<\/em> be given by f<\/em> = {(a<\/em>, b<\/em>), (b<\/em>, d<\/em>), (c<\/em>, a<\/em>), (d<\/em>, c<\/em>)}. Write f <\/em>-1<\/sup>.                                           [NCERT EXEMPLAR]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,

A<\/em> = {a<\/em>, b<\/em>, c<\/em>, d<\/em>} and f<\/em> : A<\/em> \u2192 A<\/em> be given by f<\/em> = {(a<\/em>, b<\/em>), (b<\/em>, d<\/em>), (c<\/em>, a<\/em>), (d<\/em>, c<\/em>)}

Since, the elements of a function when interchanged gives inverse function.

So, f <\/em>-1<\/sup> = {(b<\/em>, a<\/em>), (d<\/em>, b<\/em>), (a<\/em>, c<\/em>), (c<\/em>, d<\/em>)}<\/p>\n\n\n\n

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Question 42:<\/h4>\n\n\n\n

Let f<\/em>, g<\/em> : R<\/strong> \u2192 R<\/strong> be defined by f<\/em>(x<\/em>) = 2x<\/em> + l and g<\/em>(x<\/em>) = x<\/em>2<\/sup> – 2 for all x<\/em> \u2208 R<\/strong>, respectively. Then, find gof<\/em>.                 [NCERT EXEMPLAR]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,

 f<\/em>, g<\/em> : R<\/strong> \u2192 R<\/strong> are defined by f<\/em>(x<\/em>) = 2x<\/em> + l and g<\/em>(x<\/em>) = x<\/em>2<\/sup> – 2 for all x<\/em> \u2208 R<\/strong>, respectively

Now,g<\/em>o<\/em>f<\/em>(x<\/em>)=g<\/em>(f<\/em>(x<\/em>))=g<\/em>(2x<\/em>+1)=(2x<\/em>+1)2-2=4x<\/em>2+4x<\/em>+1-2=4x<\/em>2+4x<\/em>-1<\/p>\n\n\n\n

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Question 43:<\/h4>\n\n\n\n

If the mapping f<\/em> : {1, 3, 4} \u2192 {1, 2, 5} and g<\/em> : {1, 2, 5} \u2192 {1, 3}, given by f<\/em> = {(1, 2), (3, 5), (4, 1)} and g<\/em> = {(2, 3), (5, 1), (1, 3)}, then write fog<\/em>.                                                                                                                                                                           [NCERT EXEMPLAR]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,

 f<\/em> : {1, 3, 4} \u2192 {1, 2, 5} and g<\/em> : {1, 2, 5} \u2192 {1, 3}, are given by f<\/em> = {(1, 2), (3, 5), (4, 1)} and g<\/em> = {(2, 3), (5, 1), (1, 3)}, respectively


As,

f<\/em>o<\/em>g<\/em>(2)=f<\/em>(g<\/em>(2))=f<\/em>(3)=5,f<\/em>o<\/em>g<\/em>(5)=f<\/em>(g<\/em>(5))=f<\/em>(1)=2,f<\/em>o<\/em>g<\/em>(1)=f<\/em>(g<\/em>(1))=f<\/em>(3)=5,So,f<\/em>o<\/em>g<\/em> : {1,2,5}\u2192{1,2,5} is given byf<\/em>o<\/em>g<\/em>={(2,5),(5,2),(1,5)}<\/p>\n\n\n\n

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Question 44:<\/h4>\n\n\n\n

If a function g<\/em> = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by g<\/em>(x<\/em>) = \u03b1<\/em>x<\/em>+\u03b2<\/em>, then find the values of \u03b1<\/em> and \u03b2<\/em>.                [NCERT EXEMPLAR]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,

A function g<\/em> = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by g<\/em>(x<\/em>) = \u03b1<\/em>x<\/em>+\u03b2<\/em>

As, g<\/em>(1)=1 and g<\/em>(2)=3So, \u03b1<\/em>(1)+\u03b2<\/em>=1\u21d2\u03b1<\/em>+\u03b2<\/em>=1         …..(i)and \u03b1<\/em>(2)+\u03b2<\/em>=3\u21d22\u03b1<\/em>+\u03b2<\/em>=3       …..(ii)(ii)-(i), we get2\u03b1<\/em>–\u03b1<\/em>=2\u21d2\u03b1<\/em>=2Substituting \u03b1<\/em>=2 in (i), we get2+\u03b2<\/em>=1\u21d2\u03b2<\/em>=-1<\/p>\n\n\n\n

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Question 45:<\/h4>\n\n\n\n

If f<\/em>(x<\/em>) = 4 – (x<\/em> – 7)3<\/sup>, then write f<\/em> -1<\/sup>(x<\/em>).                                                                                                                        [NCERT EXEMPLAR]<\/p>\n\n\n\n

ANSWER:<\/h4>\n\n\n\n

We have,f<\/em>(x<\/em>)=4-(x<\/em>-7)3Let y<\/em>=4-(x<\/em>-7)3\u21d2(x<\/em>-7)3=4-y<\/em>\u21d2x<\/em>-7=3\u221a4-y<\/em>\u21d2x<\/em>=7+3\u221a4-y<\/em>\u21d2f<\/em>-1(y<\/em>)=7+3\u221a4-y<\/em>\u2234 f<\/em>-1(x<\/em>)=7+3\u221a4-x<\/em><\/p>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 2: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 2\u2013Functions<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 12 Maths Chapter 2\u2013Functions PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 12 Maths :<\/strong><\/h2>\n\n\n\n