RD Sharma Solutions for Class 12 Maths Chapter 2–Functions

Class 12: Maths Chapter 1 solutions. Complete Class 12 Maths Chapter 1 Notes.

RD Sharma Solutions for Class 12 Maths Chapter 2–Functions

RD Sharma 12th Maths Chapter 1, Class 12 Maths Chapter 1 solutions

Page No 2.31:

Question 1:

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

ANSWER:

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

f: Z → Z given by f(x)=3x+2

Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
 f(x)=f(y)
⇒3x + 2 =3y + 2
⇒3x = 3y
⇒x = y
⇒f(x) = f(y) ⇒x = y
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
⇒3x + 2 = y
⇒3x = y – 2

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

(ii) which is not one-one but onto.
f: Z → N ∪ {0} given by f(x) = |x|

Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
⇒|x| = |y|
⇒x= ± y
So, different elements of domain f may give the same image.
So, f is not one-one.

Surjectivity:
Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
⇒|x| = y
⇒x = ± y, which is an element in Z (domain).
So, for every element in the co-domain, there exists a pre-image in the domain.
Thus, f is onto.

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

f: Z → Z given by f(x) = 2x² + 1

Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y)⇒2x2+1 = 2y2+1⇒2x2 = 2y2⇒x2 = y2⇒x = ±y
So, different elements of domain f may give the same image.
Thus, f is not one-one.

Surjectivity:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
⇒2x2+1=y⇒2x2=y-1⇒x2=y-12⇒x=±√y-12, ∉ Z always.For example, if we take, y = 4,x=±√y-12=±√4-12=±√32, ∉ ZSo, x may not be in Z (domain).

Thus, f is not onto.

Page No 2.31:

Question 2:

Which of the following functions from A to B are one-one and onto?
(i) f₁ = {(1, 3), (2, 5), (3, 7)} ; A = {1, 2, 3}, B = {3, 5, 7}
(ii) f₂ = {(2, a), (3, b), (4, c)} ; A = {2, 3, 4}, B = {a, b, c}
(iii) f₃ = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}

ANSWER:

(i) f₁ = {(1, 3), (2, 5), (3, 7)} ; A = {1, 2, 3}, B = {3, 5, 7}

Injectivity:
f₁ (1) = 3
f₁ (2) = 5
f₁ (3) = 7
⇒Every element of A has different images in B.
So, f₁ is one-one.

Surjectivity:
Co-domain of f₁ = {3, 5, 7}
Range of f₁ =set of images  =  {3, 5, 7}
⇒Co-domain = range
So, f₁ is onto.

(ii) f₂ = {(2, a), (3, b), (4, c)} ; A = {2, 3, 4}, B = {a, b, c}

Injectivity:
f₂ (2) = a
f₂ (3) = b
f₂ (4) = c
⇒Every element of A has different images in B.
So, f₂ is one-one.

Surjectivity:
Co-domain of f₂ = {a, b, c}
Range of f₂ = set of images = {a, b, c}
⇒Co-domain = range
So, f₂ is onto.

(iii) f₃ = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}

Injectivity:
f₃ (a) = x
f₃ (b) = x
f₃ (c) = z
f₃ (d) = z
⇒a and b have the same image x. (Also c and d have the same image z)
So, f₃ is not one-one.

Surjectivity:
Co-domain of f₁ ={x, y, z}
Range of f₁ =set of images  =  {x, z}
So, the co-domain  is not same as the range.
So, f₃ is not onto.

Page No 2.31:

Question 3:

Prove that the function f : N → N, defined by f(x) = x² + x + 1, is one-one but not onto.

ANSWER:

f : N → N, defined by f(x) = x² + x + 1

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

Surjectivity:
The minimum number in N is 1.When x=1,x2+x+1=1+1+1=3⇒x2+x+1≥3, for every x in N.⇒f(x) will not assume the values 1 and 2.So, f is not onto.

Page No 2.31:

Question 4:

Let A = {−1, 0, 1} and f = {(x, x²) : x ∈ A}. Show that f : A → A is neither one-one nor onto.

ANSWER:

A = {−1, 0, 1} and f = {(x, x²) : x ∈ A}
Given, f(x) = x²

Injectivity:
f(1) = 1²=1 and
f(-1)=(-1)²=1

⇒1 and -1 have the same images.
So, f  is not one-one.

Surjectivity:
Co-domain of  f  = {-1, 0, 1}

f(1) = 1² = 1,
f(-1) = (-1)² = 1 and
f(0) = 0
⇒Range of f  = {0, 1}
So, both are not same.
Hence, f  is not onto.

Page No 2.31:

Question 5:

Classify the following functions as injection, surjection or bijection :
(i) f : N → N given by f(x) = x²
(ii) f : Z → Z given by f(x) = x²
(iii) f : N → N given by f(x) = x³
(iv) f : Z → Z given by f(x) = x³
(v) f : R → R, defined by f(x) = |x|
(vi) f : Z → Z, defined by f(x) = x² + x
(vii) f : Z → Z, defined by f(x) = x − 5
(viii) f : R → R, defined by f(x) = sinx
(ix) f : R → R, defined by f(x) = x³ + 1
(x) f : R → R, defined by f(x) = x³ − x
(xi) f : R → R, defined by f(x) = sin²x + cos²x
(xii) f : Q − {3} → Q, defined by f(x)=2x+3x-3
(xiii) f : Q → Q, defined by f(x) = x³ + 1
(xiv) f : R → R, defined by f(x) = 5x³ + 4
(xv) f : R → R, defined by f(x) = 3 − 4x
(xvi) f : R → R, defined by f(x) = 1 + x²
(xvii) f : R → R, defined by f(x) = xx2+1                                                                                                                    [NCERT EXEMPLAR]

ANSWER:

(i) f : N → N, given by f(x) = x²

Injection test:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x)=f(y)

x2=y2x=y       (We do not get ± because x and y are in N)

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

x2=yx=√y, which may not be in N.For example, if y=3,x=√3 is not in N.

So, f is not a surjection.

So, f is not a bijection.

(ii) f : Z → Z, given by f(x) = x²

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

x2=y2x=±y

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x2=yx=±√y which may not be in Z.For example, if y=3,x=±√3 is not in Z.

So, f is not a surjection.

So, f is not a bijection.

(iii) f : N → N, given by f(x) = x³

Injection test:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x) = f(y)

x3=y3x=y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

x3=yx=3√ywhich may not be in N.For example, if y=3,x=3√3 is not in N.

So, f is not a surjection and  f is not a bijection.

(iv) f : Z → Z, given by f(x) = x³

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y)

f(x) = f(y)

x3=y3x=y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x3=yx=3√y which may not be in Z.For example, if y=3,x=3√3 is not in Z.

So, f is not a surjection and f is not a bijection.

(v) f : R → R, defined by f(x) = |x|

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y)

f(x) = f(y)

|x|=|y|x=±y

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

|x|=yx=±y∈Z

So, f is a surjection and  f is not a bijection.

(vi) f : Z → Z, defined by f(x) = x² + x

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

x2+x=y2+yHere, we cannot say that x = y.For example, x = 2 and y = – 3Then, x2+x=22+2= 6y2+y=(-3)2-3= 6So, we have two numbers 2 and -3 in the domain Z whose image is same as 6.

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x2+x=yHere, we cannot say x∈Z.For example, y =-4.x2+x=-4x2+x+4=0x=-1±√-152=-1±i√152 which is not in Z.

So, f is not a surjection and  f is not a bijection.

(vii) f : Z → Z, defined by f(x) = x − 5

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

x – 5 = y – 5

x = y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x – 5 = y

x = y + 5, which is in Z.

So, f is a surjection and f is a bijection.

(viii) f : R → R, defined by f(x) = sinx

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

sinx=sinyHere, x may not be equal to y because sin0=sinπ.So, 0 and π have the same image 0.

So, f is not an injection .

Surjection test:

Range of f = [-1, 1]

Co-domain of f = R

Both are not same.

So, f is not a surjection and f is not a bijection.

(ix) f : R → R, defined by f(x) = x³ + 1

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

x3+1=y3+1x3=y3x=y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

x3+1=yx=3√y-1∈R

So, f is a surjection.

So, f is a bijection.

(x) f : R → R, defined by f(x) = x³ − x

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

x3-x=y3-yHere, we cannot say x=y.For example, x=1 and y=-1x3-x=1-1=0y3-y=(-1)3-(-1)-1+1=0So, 1 and -1 have the same image 0.

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

x3-x=yBy observation we can say that there exist some x in R, such that x3-x=y.

So, f is a surjection and  f is not a bijection.

(xi) f : R → R, defined by f(x) = sin²x + cos²x

f(x) = sin²x + cos²x = 1

So, f(x) = 1 for every x in R.

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

So, f is not an injection.

Range of f = {1}

Co-domain of f = R

Both are not same.

So, f is not a surjection and  f is not a bijection.

(xii) f : Q − {3} → Q, defined by f(x)=2x+3x-3

Injection test:

Let x and y be any two elements in the domain (Q − {3}), such that f(x) = f(y).

f(x) = f(y)

2x+3x-3=2y+3y-3(2x+3)(y-3)=(2y+3)(x-3)2xy-6x+3y-9=2xy-6y+3x-99x=9yx=y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Q − {3}), such that f(x) = y for some element x in Q (domain).

f(x) = y

2x+3x-3=y2x+3=xy-3y2x–xy=-3y-3x(2-y)=-3(y+1)x=3(y+1)y-2, which is not defined at y=2.

So, f is not a surjection and f is not a bijection.

(xiii) f : Q → Q, defined by f(x) = x³ + 1

Injection test:

Let x and y be any two elements in the domain (Q), such that f(x) = f(y).

f(x) = f(y)

x3+1=y3+1x3=y3x=y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (Q), such that f(x) = y for some element x in Q (domain).

f(x) = y

x3+1=yx=3√y-1, which may not be in Q.For example, if y= 8,x3+1= 8x3=7x=3√7, which is not in Q.

So, f is not a surjection and f is not a bijection.

So, f is a surjection and f is a bijection.

(xiv) f : R → R, defined by f(x) = 5x³ + 4

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

5x3+4 = 5y3+45x3= 5y3x3= y3x=y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

5x3+4=y5x3=y-4x3=y-45x=3√y-45∈R

So, f is a surjection and f is a bijection.

(xv) f : R → R, defined by f(x) = 3 − 4x

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

3-4x=3-4y-4x=-4yx= y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

3-4x=y4x=3-yx=3-y4∈R

So, f is a surjection and f is a bijection.

(xvi) f : R → R, defined by f(x) = 1 + x²

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

1+x2=1+y2x2=y2x= ±y

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

1+x2=yx2=y-1x=±√y-1 which may not be in RFor example, if y=0,x=±√-1=±i is not in R.

So, f is not a surjection and f is not a bijection.

(xvii)  f : R → R, defined by f(x) = xx2+1

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

xx2+1=yy2+1xy2+x=x2y+yxy2-x2y+x–y=0-xy(-y+x)+1(x–y)=0(x–y)(1-xy)=0x=y or x=1y

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

xx2+1=yyx2-x+y=0x=-(-1)±√1-4y22y, if y≠0=1±√1-4y22y, which may not be in RFor example, if y=1, thenx=1±√1-42=1±i√32, which is not in RSo, f is not surjection and f is not bijection.

So, f is not a surjection and f is not a bijection.

Page No 2.31:

Question 6:

If f : A → B is an injection, such that range of f = {a}, determine the number of elements in A.

ANSWER:

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

Page No 2.31:

Question 7:

Show that the function f : R − {3} → R − {2} given by f(x)=x-2x-3 is a bijection.

ANSWER:

f : R − {3} → R − {2} given by
f(x)=x-2x-3
Injectivity:
Let x and y be any two elements in the domain (R − {3}), such that f(x) = f(y).
 f(x) = f(y)
⇒x-2x-3=y-2y-3⇒(x-2)(y-3)=(y-2)(x-3)⇒xy-3x-2y+6=xy-3y-2x+6⇒x=y
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (R − {2}), such that f(x) = y for some element x in R − {3} (domain).
f(x) = y
⇒x-2x-3=y⇒x-2=xy-3y⇒xy–x=3y-2⇒x(y-1)=3y-2⇒x=3y-2y-1, which is in R-{3}
So, for every element in the co-domain, there exists some pre-image in the domain.
⇒f  is onto.
Since, f  is both one-one and onto, it is a bijection.

Page No 2.32:

Question 8:

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

(i) f(x) = x2                                (ii) g(x) = |x|                                (iii) h(x) = x²                                                               [NCERT EXEMPLAR]

ANSWER:

(i) f : A → A, given by f(x) = x2

Injection test:

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

f(x) = f(y)

x2 = y2

x = y

So, f is one-one.

Surjection test:

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

f(x) = y

x2 = y

x = 2y, which may not be in A.

For example, if y = 1, then

x = 2, which is not in A.

So, f is not onto.

So, f is not bijective.

(ii) g(x) = |x|

Injection test:

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

f(x) = f(y)

|x| = |y|

x = ±y

So, f is not one-one.

Surjection test:

For y = -1, there is no value of x in A.

So, f is not onto.

So, f is not bijective.

(iii) h(x) = x²

Injection test:

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

f(x) = f(y)

x² = y²

x = ±y

So, f is not one-one.

Surjection test:

For y = -1, there is no value of x in A.

So, f is not onto.

So, f is not bijective.

 

Page No 2.32:

Question 9:

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

(i) {(x, y) : x is a person, y is the mother of x}
(ii) {(a, b) : a is a person, b is an ancestor of a}                                                                                                               [NCERT EXEMPLAR]

ANSWER:

(i) f = {(x, y) : x is a person, y is the mother of x}

As, for each element x in domain set, there is a unique related element y in co-domain set.

So, f is the function.

Injection test:

As, y can be mother of two or more persons

So, f is not injective.

Surjection test:

For every mother y defined by (x, y), there exists a person x for whom y is mother.

So, f is surjective.

Therefore, f is surjective function.

(ii) g = {(a, b) : a is a person, b is an ancestor of a}

Since, the ordered map (a, b) does not map ‘a‘ – a person to a living person.
So, g is not a function.

Page No 2.32:

Question 10:

Let A = {1, 2, 3}. Write all one-one from A to itself.

ANSWER:

A ={1, 2, 3}
Number of elements in  A = 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 )}

Page No 2.32:

Question 11:

If f : R → R be the function defined by f(x) = 4x³ + 7, show that f is a bijection.

ANSWER:

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
⇒4x3+7=4y3+7⇒4x3=4y3⇒x3=y3⇒x=y
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
⇒4x3+7=y⇒4x3=y-7⇒x3=y-74⇒x=3√y-74∈R
So, for every element in the co-domain, there exists some pre-image in the domain.
⇒f is onto.
Since, f is both one-to-one and onto, it is a bijection.

Page No 2.32:

Question 12:

Show that the exponential function f : R → R, given by f(x) = e^x, is one-one but not onto. What happens if the co-domain is replaced by R+0 (set of all positive real numbers)?

ANSWER:

f : R → R, given by f(x) = e^x

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
 f(x)=f(y)
⇒ex=ey⇒x=y
So, f is one-one.

Surjectivity:
We know that range of e^x is (0, ∞) = R⁺
Co-domain = R
Both are not same.
So, f is not onto.

If the co-domain is replaced by R⁺, then the co-domain and range become the same and in that case, f is onto and hence, it is a bijection.

Page No 2.32:

Question 13:

Show that the logarithmic function f : R+0→R given by f(x)=loga x, a>0 is a bijection.

ANSWER:

f:R⁺→R given by f(x)= loga x, a>0
Injectivity:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
 f(x) = f(y)
loga x=loga y⇒x=y
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R⁺ (domain).
f(x) = y
loga x=y⇒x=ay ∈R+
So, for every element in the co-domain, there exists some pre-image in the domain.
⇒f is onto.
Since f is one-one and onto, it is a bijection.

Page No 2.32:

Question 14:

If A = {1, 2, 3}, show that a one-one function f : A → A must be onto.

ANSWER:

A ={1, 2, 3}
Number of elements in  A = 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.

Page No 2.32:

Question 15:

If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.

ANSWER:

A ={1, 2, 3}
Possible onto functions from A to A 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.

Page No 2.32:

Question 16:

Find the number of all onto functions from the set A = {1, 2, 3, …, n} to itself.

ANSWER:

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

Page No 2.32:

Question 17:

Give examples of two one-one functions f₁ and f₂ from R to R, such that f₁ + f₂ : R → R. defined by (f₁ + f₂) (x) = f₁ (x) + f₂ (x) is not one-one.

ANSWER:

We know that f₁: R → R, given by f₁(x)=x, and f₂(x)=-x are one-one.
Proving f₁ is one-one:
Let f1(x)=f1(y)⇒x=y
So, f₁ is one-one.

Proving f₂ is one-one:
Let f2(x)=f2(y)⇒-x=-y⇒x=y
So, f₂  is one-one.

Proving (f₁ + f₂) is not one-one:
Given:
(f₁ + f₂) (x) = f₁ (x) + f₂ (x)= x + (-x) =0
So, for every real number x, (f₁ + f₂) (x)=0
So, the image of ever number in the domain is same as 0.
Thus, (f₁ + f₂) is not one-one.

Page No 2.32:

Question 18:

Give examples of two surjective functions f₁ and f₂ from Z to Z such that f₁ + f₂ is not surjective.

ANSWER:

We know that f₁: R → R, given by f₁(x) = x, and f₂(x) = -x are surjective functions.
Proving f₁ is surjective :
Let y be an element in the co-domain (R), such that f₁(x) = y.
f₁(x) = y
⇒x = y, which is in R.
So, for every element in the co-domain, there exists some pre-image in the domain.1(x)=f1(y)x=y
So, f₁is surjective .

Proving f₂ is surjective :Let f2(x)=f2(y)−x=−yx=y
Let y be an element in the co domain (R) such that f₂(x) = y.
 f₂(x) = y
⇒x = y, which is in R.
So, for every element in the co-domain, there exists some pre-image in the domain.1(x)=f1(y)x=y
So, f₂ is surjective .

Proving (f₁ + f₂) is not surjective :
Given:
(f₁ + f₂) (x) = f₁ (x) + f₂ (x)= x + (-x) =0
So, for every real number x, (f₁ + f₂) (x)=0
So, the image of every number in the domain is same as 0.
⇒Range = {0}
Co-domain = R
So, both are not same.
So, f₁ + f₂ is not surjective.

Page No 2.32:

Question 19:

Show that if f₁ and f₂ are one-one maps from R to R, then the product f1×f2 : R→R defined by (f1×f2) (x)=f1 (x) f2 (x) need not be one-one.

ANSWER:

We know that f₁: R → R, given by f₁(x) = x, and f₂(x) = x are one-one.
Proving f₁ is one-one:
Let x and y be two elements in the domain R, such that
f₁(x) = f₁(y)
⇒x = yet f1(x)=f1(y)x=y
So, f₁ is one-one.

Proving f₂ is one-one:
Let x and y be two elements in the domain R, such that
f₂(x) = f₂(y)
⇒x = yet f1(x)=f1(y)x=y
So, f₂ is one-one.

Proving f1 × f2 is not one-one:
Given:
(f1 × f2)(x)=f1 (x) × f2 (x)=x × x=x2Let x and y be two elements in the domain R, such that(f1 × f2)(x)=(f1 × f2)(y)⇒x2 = y2⇒x=± ySo, (f1 × f2) is not one-one.f1×f2

Page No 2.32:

Question 20:

Suppose f₁ and f₂ are non-zero one-one functions from R to R. Is f1f2 necessarily one-one? Justify your answer. Here, f1f2:R→R is given by (f1f2) (x)=f1 (x)f2 (x) for all x∈R.

ANSWER:

We know that f₁: R → R, given by f₁(x)=x³ and f₂(x)=x are one-one.
Injectivity of f₁:
Let x and y be two elements in the domain R, such that
f1(x)=f2(y)⇒x3=y⇒x=3√y∈RLet f1(x)=f1(y)x=y
So, f₁ is one-one.

Injectivity of f₂:
Let x and y be two elements in the domain R, such that
f2(x)=f2(y)⇒x=y ⇒x∈R.Let f2(x)=f2(y)−x=−yx=y
So, f₂  is one-one.

Proving f1f2is not one-one:
Given that f1f2(x)=f1(x)f2(x)=x3x=x2
Let x and y be two elements in the domain R, such that

f1f2(x)=f1f2(y)⇒x2=y2⇒x=±y
So, f1f2 is not one-one.

Page No 2.32:

Question 21:

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

(i) an injective map from A to B
(ii) a mapping from A to B which is not injective
(iii) a mapping from A to B.

ANSWER:

(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.

Page No 2.32:

Question 22:

Show that f : R → R, given by f(x) = x – [x], is neither one-one nor onto.

ANSWER:

We have, f(x) = x – [x]

Injection test:

f(x) = 0 for all x ∈ Z

So, f is a many-one function.

Surjection test:

Range (f) = [0, 1) ≠ R.

So, f is an into function.

Therefore, f is neither one-one nor onto.

Page No 2.32:

Question 23:

Let f : N → N be defined by

f(n)={n+1, if n is oddn-1, if n is even

Show that f is a bijection.                                                                                                                                                   [CBSE 2012, NCERT]

ANSWER:

We have,f(n)={n+1, if n is oddn-1, if n is evenInjection test:Case I: If n is odd,Let x, y∈N such that f(x)=f(y)As, f(x)=f(y)⇒x+1=y+1⇒x=yCase II: If n is even,Let x, y∈N such that f(x)=f(y)As, f(x)=f(y)⇒x-1=y-1⇒x=ySo, f is injective.Surjection test:Case I: If n is odd,As, for every n∈N, there exists y=n-1 in N such thatf(y)=f(n-1)=n-1+1=nCase II: If n is even,As, for every n∈N, there exists y=n+1 in N such thatf(y)=f(n+1)=n+1-1=nSo, f is surjective.So, f is a bijection.

Page No 2.46:

Question 1:

Find gof and fog when f : R → R and g : R → R are defined by
(i) f(x) = 2x + 3                and       g(x) = x² + 5
(ii) f(x) = 2x + x²             and       g(x) = x³
(iii) f(x) = x² + 8              and       g(x) = 3x³ + 1
(iv) f(x) = x                       and       g(x) = |x|
(v) f(x) = x² + 2x − 3       and       g(x) = 3x − 4
(vi) f(x) = 8x³                   and       g(x) = x^1/3

ANSWER:

Given, f : R → R and g : R → R
So, gof : R → R  and fog : R → R

(i) f(x) = 2x + 3  and g(x) = x² + 5
Now, (gof) (x)
= g (f (x))
= g (2x +3)
= (2x + 3)² + 5
= 4x²+ 9 + 12x +5
=4x²+  12x + 14

(fog) (x)
=f (g (x))
= f (x² + 5)
= 2 (x² + 5) +3
= 2 x²+ 10 + 3
= 2x² + 13

(ii) f(x) = 2x + x² and g(x) = x^{3
(gof) (x)=g (f (x))=g (2x+x2)=(2x+x2)3(fog) (x)=f (g (x))=f (x3)=2 (x3)+(x3)2=2x3+x6}

(iii) f(x) = x² + 8  and g(x) = 3x³ + 1
(gof) (x)=g (f(x))=g (x2+8)=3 (x2+8)3+1(fog) (x)=f (g (x))=f (3x3+1)=(3x3+1)2+8=9x6+6x3+1+8=9x6+6x3+9

(iv) f(x) = x and g(x) = |x|
(gof) (x)=g (f(x))=g (x)=|x|(fog) (x)=f (g (x))=f (|x|)=|x|

(v) f(x) = x² + 2x − 3 and g(x) = 3x − 4
(gof) (x)=g (f(x))=g (x2+2x-3)=3 (x2+2x-3)-4=3x2+6x-9-4=3x2+6x-13(fog) (x)=f (g (x))=f (3x-4)=(3x-4)2+2 (3x-4)-3=9x2+16-24x+6x-8-3=9x2-18x+5

(vi) f(x) = 8x³ and g(x) = x^1/3
(gof) (x)=g (f (x))=g (8x3)=(8x3)13=[(2x)3]13=2x(fog) (x)=f (g (x))=f (x13)=8 (x13)3=8x

Page No 2.46:

Question 2:

Let f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}. Show that gof and fog are both defined. Also, find fog and gof.

ANSWER:

f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}

f : {3, 9, 12} → {1, 3,4} and g : {1, 3, 4, 5} → {3, 9}

Co-domain of f is a subset of the domain of g.
So, gof exists and gof : {3, 9, 12} → {3, 9}
(gof) (3)=g (f (3))=g (1)=3(gof) (9)=g (f (9))=g (3)=3(gof) (12)=g (f (12))=g (4)=9⇒gof ={(3, 3), (9, 3), (12, 9)}
Co-domain of g is a subset of the domain of f.
So, fog exists and fog : {1, 3, 4, 5} → {3, 9, 12}
(fog) (1)=f (g (1))=f (3)=1(fog) (3)=f (g (3))=f (3)=1(fog) (4)=f (g (4))=f (9)=3(fog) (5)=f (g (5))=f (9)=3⇒fog={(1, 1), (3, 1), (4, 3), (5, 3)}

Page No 2.46:

Question 3:

Let f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.

ANSWER:

f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}
f : {1, 4, 9, 16} → {-1, -2, -3, 4} and g : {-1, -2, -3, 4} → {-2, -4, -6, 8}
Co-domain of f = domain of g
So, gof exists and gof : {1, 4, 9, 16} → {-2, -4, -6, 8}
(gof) (1)=g (f (1))=g (-1)=-2(gof) (4)=g (f (4))=g (-2)=-4(gof) (9)=g (f (9))=g (-3)=-6(gof) (16)=g (f (16))=g (4)=8So, gof={(1, -2), (4, -4), (9, -6), (16, 8)}

But the co-domain of g is not same as the domain of f.
So, fog does not exist.

Page No 2.46:

Question 4:

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

    f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.

Show that f and g both are bijections and find fog and gof.

ANSWER:

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

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

Finding  fog:
Co-domain of g is same as the domain of f.
So, fog exists and fog : {u v, w} → {u v, w}
(fog) (u)=f (g (u))=f (b)=u(fog) (v)=f (g (v))=f (a)=v(fog) (w)=f (g (w))=f (c)=wSo, fog ={(u, u), (v, v), (w, w)}

Finding gof:
Co-domain of f is same as the domain of g.
So, fog exists and gof : {a, b, c} → {a, b, c}
(gof) (a)=g (f (a))=g (v)=a(gof) (b)=g (f (b))=g (u)=b(gof) (c)=g (f (c))=g (w)=cSo, gof={(a, a), (b, b), (c, c)}

Page No 2.46:

Question 5:

Find  fog (2) and gof (1) when : f : R → R ; f(x) = x² + 8 and g : R → R; g(x) = 3x³ + 1.

ANSWER:

(fog) (2)=f (g (2))=f(3×23+1)=f(25)=252+8=633(gof) (1)=g (f (1))=g (12+8)=g (9)=3×93+1=2188

Page No 2.46:

Question 6:

Let R⁺ be the set of all non-negative real numbers. If f : R⁺ → R⁺ and g : R⁺ → R⁺ are defined as f(x)=x2 and g(x)=+√x, find fog and gof. Are they equal functions?

ANSWER:

Given,  f : R⁺ → R⁺ and g : R⁺ → R⁺
So,  fog : R⁺ → R⁺  and gof : R⁺ → R⁺
Domains of fog  and gof  are the same.
(fog) (x)=f (g (x))=f (√x)=(√x)2=x(gof) (x)=g (f (x))=g (x2)=√x2=xSo, (fog) (x)=(gof) (x),∀x∈R+
Hence, fog = gof

Page No 2.46:

Question 7:

Let f : R → R and g : R → R be defined by f(x) = x² and g(x) = x + 1. Show that fog ≠ gof.

ANSWER:

Given,  f : R → R and g : R → R.
So, the domains of f and g are the same.

(fog) (x)=f (g (x))=f (x+1)=(x+1)2=x2+1+2x(gof) (x)=g (f (x))=g (x2)=x2+1
So,  fog ≠ gof

Page No 2.46:

Question 8:

Let f : R → R and g : R → R be defined by f(x) = x + 1 and g(x) = x − 1. Show that fog = gof = I_R.

ANSWER:

Given,  f : R → R and g : R → R
⇒fog :  R → R and gof : R → R (Also, we know that I_R : R → R)
So, the domains of all fog, gof and I_R are the same.
(fog) (x)=f (g (x))=f (x-1)=x-1+1=x=IR (x)      … (1)(gof) (x)=g (f (x))=g (x+1)=x+1-1=x=IR (x)      … (2)From (1) and (2), (fog) (x)=(gof) (x)=IR (x), ∀x∈RHence, fog=gof=IR

Page No 2.46:

Question 9:

Verify associativity for the following three mappings : f : N → Z₀ (the set of non-zero integers), g : Z₀ → Q and h : Q → R given by f(x) = 2x, g(x) = 1/x and h(x) = e^x.

ANSWER:

Given that f : N → Z₀ , g : Z₀ → Q and h : Q → R .
gof : N → Q  and hog : Z₀ → R
⇒h o (gof ) : N → R and (hog) o f: N → R
So, both have the same domains.
(gof) (x)=g (f (x))=g (2x)=12x         …(1)(hog) (x)=h (g (x))=h (1x)=e1x       …(2)Now,(h o(gof)) (x)=h((gof) (x))=h (12x)=e12x           [from (1)]((hog) o f)(x)=(hog) (f (x))= (hog) (2x)=e12x    [from (2)]⇒(h o(gof)) (x)=((hog) o f)(x), ∀x∈NSo, h o(gof)=(hog) o f
Hence, the associative property has been verified.

Page No 2.46:

Question 10:

Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and h(z) = sin z for all x, y, z ∈ N. Show that ho (gof) = (hog) of.

ANSWER:

Given, f : N → N, g : N → N and h : N → R
⇒gof : N → N and hog : N → R
⇒ho (gof) : N → R and (hog) of : N → R
So, both have the same domains.
(gof) (x)=g (f (x))=g (2x)=3 (2x)+4=6x+4  …(1)(hog) (x)=h(g (x))=h (3x+4)=sin (3x+4)        … (2)Now,(h o (gof)) (x)=h ((gof) (x))=h(6x+4) = sin (6x+4)    [from (1)]((hog) o f) (x)=(hog) (f (x))=(hog) (2x)=sin (6x+4)    [from (2)]So, (h o (gof)) (x)=((hog) o f) (x), ∀x∈NHence, h o (gof)=(hog) o f

Page No 2.46:

Question 11:

Give examples of two functions f : N → N and g : N → N, such that gof is onto but f is not onto.

ANSWER:

Let us consider a function f : N → N given by f(x) = x +1 , which is not onto.
[This not onto because if we take 0 in N (co-domain), then,
0=x+1
⇒x=-1∉N]

Let us consider g : N → N  given by
g (x)={x-1, if x>11, if x=1Now, let us find (gof) (x)Case 1: x>1(gof) (x)=g (f (x))=g (x+1)=x+1-1=xCase 2: x=1(gof) (x)=g (f (x))=g (x+1)=1From case-1 and case-2, (gof) (x)=x, ∀x∈N, which is an identity function and, hence, it is onto.

Page No 2.46:

Question 12:

Give examples of two functions f : N → Z and g : Z → Z, such that gof is injective but g is not injective.

ANSWER:

Let f : N → Z be given by f (x) = x, which is injective.
(If we take f(x) = f(y), then it gives x = y)

Let g : Z → Z be given by g (x) = |x|, which is not injective.
If we take f(x) = f(y), we get:
|x| = |y|
⇒x = ± y

Now, gof : N → Z.
(gof) (x)=g (f (x))=g (x)=|x|
Let us take two elements x and y in the domain of gof , such that
(gof) (x)=(gof) (y)⇒|x|=|y|⇒x=y (We don’t get ± here because x, y ∈N)
So, gof is injective.
 

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Question 13:

If f : A → B and g : B → C are one-one functions, show that gof is a one-one function.

ANSWER:

Given,  f : A → B and g : B → C are one – one.
Then, gof : A → B
Let us take two elements x and y from A, such that
(gof) (x)=(gof) (y)⇒g (f (x))=g (f (y))⇒f (x)=f (y) (As, g is one-one)⇒x=y (As, f is one-one)
Hence, gof is one-one.

Page No 2.46:

Question 14:

If f : A → B and g : B → C are onto functions, show that gof is a onto function.

ANSWER:

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

Page No 2.54:

Question 1:

Find fog and gof  if
(i) f(x)=ex, g(x)=loge x
(ii) f(x)=x2, g(x)=cos x
(iii) f(x)=|x|, g (x)=sin x
(iv) f(x)=x+1, g(x)=ex
(v) f(x)=sin-1 x, g(x)=x2
(vi) f(x)=x+1, g(x)=sin x
(vii) f(x)=x+1, g(x)=2x+3
(viii) f(x)=c, c ∈ R, g(x)=sin x2
(ix) f(x)=x2+2, g(x)=1-11-x

ANSWER:

(i) f (x)=ex, g(x)=loge xf:R→(0,∞); g:(0,∞)→RComputing fog:Clearly, the range of g is a subset of the domain of f.fog : (0,∞)→R(fog) (x)=f (g (x))=f (loge x)=loge ex=xComputing gof:Clearly, the range of f is a subset of the domain of g.⇒fog : R→R(gof) (x)=g (f (x))=g (ex)=loge ex=x

(ii) f (x)=x2, g(x)=cos xf:R→[0, ∞) ; g:R→[-1, 1]Computing fog:Clearly, the range of g is not a subset of the domain of f.⇒Domain (fog)={x: x∈domain of g and g(x)∈domain of f}⇒Domain (fog)=x: x∈R and cos x ∈R}⇒Domain of (fog)=Rfog: R→R(fog) (x)=f (g (x))=f (cos x)=cos2xComputing gof:Clearly, the range of f is a subset of the domain of g.⇒fog : R→R(gof) (x)=g (f (x))=g (x2)=cos (x2)


(iii) f (x)=|x|, g(x)=sin xf:R→(0, ∞); g:R→[-1, 1]Computing fog:Clearly, the range of g is a subset of the domain of f.⇒fog : R→R(fog) (x)=f (g (x))=f (sin x)=|sin x|Computing gof:Clearly, the range of f is a subset of the domain of g.⇒fog : R→R(gof) (x)=g (f (x))=g (|x|)=sin |x|

(iv) f (x)=x+1, g(x)=exf:R→R; g:R→[1, ∞)Computing fog:Clearly, range of g is a subset of domain of f.⇒fog : R→R(fog) (x)=f (g (x))=f (ex)=ex+1Computing gof:Clearly, range of f is a subset of domain of g.⇒fog : R→R(gof) (x)=g (f (x))=g (x+1)=ex+1

(v) f (x)=sin-1x, g(x)=x2f:[-1,1]→[-π2,π2] ; g:R→[0, ∞)Computing fog:Clearly, the range of g is not a subset of the domain of f.Domain (fog)={x: x∈domain of g and g(x)∈domain of f}Domain (fog)={x: x∈R and x2∈[-1,1]}Domain (fog)={x: x∈R and x∈[-1,1]}Domain of (fog)=[-1,1]fog: [-1,1]→R(fog) (x)=f (g (x))=f (x2)=sin-1 (x2)Computing gof:Clearly, the range of f is a subset of the domain of g.fog : [-1,1]→R(gof) (x)=g (f (x))=g (sin-1x)=(sin-1 x)2

(vi) f(x)=x+1, g(x)=sin xf:R→R ; g:R→[-1, 1]Computing fog:Clearly, the range of g is a subset of the domain of f.⇒fog: R→R(fog) (x)=f (g (x))=f (sin x)=sin x+1Computing gof:Clearly, the range of f is a subset of the domain of g.⇒fog : R→R(gof) (x)=g (f (x))=g (x+1)=sin (x+1)

(vii) f (x)=x+1, g(x)=2x+3f:R→R ; g:R→RComputing fog:Clearly, the range of g is a subset of the domain of f.⇒fog: R→R(fog) (x)=f (g (x))=f (2x+3)=2x+3+1=2x+4Computing gof:Clearly, the range of f is a subset of the domain of g.⇒fog : R→R(gof) (x)=g (f (x))=g (x+1)=2 (x+1)+3=2x+5

(viii) f (x)=c, g(x)=sin x2f:R→{c} ; g:R→[0, 1]Computing fog:Clearly, the range of g is a subset of the domain of f.fog: R→R(fog) (x)=f (g (x))=f (sin x2)=cComputing gof:Clearly, the range of f is a subset of the domain of g.⇒fog : R→R(gof) (x)=g (f (x))=g (c)=sin c2

(ix) f(x)=x2+2f:R→[2,∞)  g(x)=1-11-xFor domain of g: 1-x≠0 ⇒x≠1⇒Domain of g=R-{1}g(x)=1-11-x=1-x-11-x=-x1-xFor range of g:y=-x1-x⇒y–xy=-x⇒y=xy–x⇒y=x(y-1)⇒x=yy-1Range of g =R-{1}So, g: R-{1}→R-{1}Computing fog:Clearly, the range of g is a subset of the domain of f.⇒fog: R-{1}→R(fog) (x)=f (g (x))=f (-xx-1)=(-xx-1)2+2=x2+2x2+2-4x(1-x)2=3x2-4x+2(1-x)2Computing gof:Clearly, the range of f is a subset of the domain of g.⇒gof : R→R(gof) (x)=g (f (x))=g (x2+2)=1-11-(x2+2)=1-1-(x2+1)=x2+2x2+1

Page No 2.54:

Question 2:

Let f(x) = x² + x + 1 and g(x) = sin x. Show that fog ≠ gof.

ANSWER:

(fog) (x)=f (g (x))=f(sin x)=sin2x+sin x+1and (gof) (x)=g (f (x))=g (x2+x+1)= sin (x2+x+1)So, fog≠gof.

Page No 2.54:

Question 3:

If f(x) = |x|, prove that fof = f.

ANSWER:

Domains of  f and fof are same as R.
(fof) (x)=f (f (x))=f (|x|)=| |x| |=|x|=f (x)So,(fof) (x)=f (x), ∀x∈RHence, fof=f

Page No 2.54:

Question 4:

If f(x) = 2x + 5 and g(x) = x² + 1 be two real functions, then describe each of the following functions:
(i) fog
(ii) gof
(iii) fof
(iv) f²

Also, show that fof ≠ f²

ANSWER:

f(x) and g(x) are polynomials.
⇒f : R → R and g : R → R.
So, fog : R → R  and gof : R → 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)×f(x)=(2x+5)(2x+5)=(2x+5)2=4×2+20x+25


→→  →

Page No 2.54:

Question 5:

If f(x) = sin x and g(x) = 2x be two real functions, then describe gof and fog. Are these equal functions?
 

ANSWER:

We know thatf:R→[-1, 1] and g: R→RClearly, the range of f is a subset of the domain of g.gof:R→R(gof) (x)=g (f (x))=g (sin x)=2sin x

Clearly, the range of g is a subset of the domain of f.fog:R→RSo, (fog) (x)=f (g (x))=f (2x)=sin (2x)

Clearly, fog≠gof

Page No 2.54:

Question 6:

Let f, g, h be real functions given by f(x) = sin x, g (x) = 2x and h (x) = cos x. Prove that fog = go (fh).

ANSWER:

We know that f:R→[-1, 1] and g:R→RClearly, the range of g is a subset of the domain of f.fog:R→RNow, (fh) (x)=f(x)h(x)=(sin x) (cos x)=12 sin (2x)Domain of fh is R.Since range of sin x is [-1,1],-1≤sin 2x≤1⇒-12≤sin x2≤12Range of fh  =[-12, 12]So, (fh):R→[-12, 12]Clearly, range of fh is a subset of g.⇒go(fh):R→R⇒domains of fog and go(fh) are the same.So, (fog) (x)=f (g (x))=f (2x)=sin (2x)and (go(fh))(x)= g ((fh) (x))=g (sinx cos x)=2sin x cos x=sin (2x)⇒(fog) (x)=( go(fh))(x), ∀x∈RHence, fog = go(fh)

Page No 2.54:

Question 7:

Let  f  be any real function and let g be a function given by g(x) = 2x. Prove that gof = f + f.

ANSWER:

Given, f:R→RSince g(x)=2x is a polynomial, g:R→RClearly, gof:R→R and f+f:R→RSo, domains of gof and f+f are the same.(gof) (x)=g (f (x))=2 f(x)(f+f) (x)=f(x)+f(x)=2 f(x)⇒(gof) (x)=(f+f) (x), ∀x∈RHence, gof =f+f

Page No 2.54:

Question 8:

If f(x)=√1-x and g(x)=loge x are two real functions, then describe functions fog and gof.

ANSWER:

 f(x)=√1-xFor domain, 1-x≥0⇒x≤1⇒domain of f =(-∞, 1]⇒f:(-∞, 1]→(0,∞) g(x)=loge xClearly, g : (0, ∞)→RComputation of fog:Clearly, the range of g is not a subset of the domain of f.So,we need to compute the domain of fog.⇒Domain (fog)={x : x∈Domain (g) and g(x)∈Domain of f}⇒Domain (fog)={x: x∈(0, ∞) and loge x ∈ (-∞, 1]}⇒Domain (fog)={x:x∈(0, ∞) and x∈ (0, e]}⇒Domain (fog)={x: x ∈(0, e]}⇒Domain (fog)=(0, e]⇒fog: (0, e)→RSo, (fog) (x)=f (g (x))=f (loge x)=√1-loge x Computation of gof:Clearly, the range of f is  a subset of the domain of g.⇒gof:(-∞,1]→R⇒(gof) (x)=g (f (x))=g (√1-x)=loge√1-x=loge (1-x)12=12loge (1-x)

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Question 9:

If f:(-π2,π2)→R and g:[-1, 1]→R be defined as
f(x)=tan x and g(x)=√1-x2 respectively, describe fog and gof.

ANSWER:

g (x)=√1-x2⇒x2≥0, ∀x∈[-1, 1]⇒-x2≤0, ∀x∈[-1, 1]⇒1-x2≤1, ∀x∈[-1, 1]We know that 1-x2≥0⇒0≤1-x2≤1⇒Range of g(x)=[0, 1]So, f:(–π2, π2)→R and g:[–1, 1]→ [0, 1]Computation of fog:Clearly, the range of g is a subset of the domain of f.So, fog: [-1, 1]→R(fog) (x)=f (g (x))=f (√1-x2)=tan √1-x2Computation of gof:Clearly, the range of f is not a subset of the domain of g.⇒Domain (gof)={x∈domain of f and f(x)∈domain of g}⇒Domain (gof)={x∈(–π2, π2) and tan x ∈[–1,1]}⇒Domain (gof)={x∈(–π2, π2) and x ∈(–π4, π4)}⇒Domain (gof)={x∈(–π4, π4)}Now, gof:(–π4, π4)→RSo, (gof) (x)=g (f (x))=g (tan x)=√1-tan2x 

Page No 2.54:

Question 10:

If f(x)=√x+3 and g(x)=x2+1 be two real functions, then find fog and gof.

ANSWER:

f(x)=√x+3For domain,x+3≥0⇒x≥-3Domain of f =[-3, ∞)Since f is a square root function, range of f =[0, ∞)f: [-3, ∞)→[0, ∞)g(x)=x2+1 is a polynomial.⇒g:R→RComputation of fog:Range of g  is not a subset of the domain of f.and domain (fog)={x: x∈domain of g and g(x)∈domain of f(x)}⇒Domain (fog)={x:x∈R and  x2+1∈[-3, ∞)}⇒Domain (fog)={x:x∈R and  x2+1≥-3}⇒Domain (fog)={x:x∈R and  x2+4≥0}⇒Domain (fog)={x:x∈R and x∈R}⇒Domain (fog)=Rfog:R→R(fog) (x)=f(g (x))=f (x2+1)=√x2+1+3=√x2+4Computation of gof:Range of f  is a subset of the domain of g.gof: [-3, ∞)→R⇒(gof) (x)=g (f (x))=g (√x+3)=(√x+3)2+1=x+3+1=x+4

Page No 2.54:

Question 11:

Let f be a real function given by f(x)=√x-2.
Find each of the following:
(i) fof
(ii) fofof
(iii) (fofof) (38)
(iv) f²

Also, show that fof ≠ f² .