{"id":543129,"date":"2021-09-29T09:56:14","date_gmt":"2021-09-29T09:56:14","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=543129"},"modified":"2022-12-26T10:00:31","modified_gmt":"2022-12-26T10:00:31","slug":"rd-sharma-solutions-for-class-11-maths-chapter-3-functions","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-3-functions\/","title":{"rendered":"RD Sharma Solutions for Class 11 Maths Chapter 3\u2013Functions"},"content":{"rendered":"\n

Class 11: Maths Chapter 3 solutions. Complete Class 11 Maths Chapter 3 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 3\u2013Functions<\/h2>\n\n\n\n

RD Sharma 11th Maths Chapter 3, Class 11 Maths Chapter 3 solutions<\/p>\n\n\n\n

EXERCISE 3.1 PAGE NO: 3.7<\/p>\n\n\n\n

1. Define a function as a set of ordered pairs.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let A and B be two non-empty sets. A relation from A to B, i.e., a subset of A\u00d7B, is called a function (or a mapping) from A to B, if<\/p>\n\n\n\n

(i) for each a \u2208 A there exists b \u2208 B such that (a, b) \u2208 f<\/p>\n\n\n\n

(ii) (a, b) \u2208 f and (a, c) \u2208 f \u21d2 b = c<\/p>\n\n\n\n

2. Define a function as a correspondence between two sets.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let A and B be two non-empty sets. Then a function \u2018f\u2019 from set A to B is a rule or method or correspondence which associates elements of set A to elements of set B such that:<\/p>\n\n\n\n

(i) all elements of set A are associated to elements in set B.<\/p>\n\n\n\n

(ii) an element of set A is associated to a unique element in set B.<\/p>\n\n\n\n

3. What is the fundamental difference between a relation and a function? Is every relation a function?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let \u2018f\u2019 be a function and R be a relation defined from set X to set Y.<\/p>\n\n\n\n

The domain of the relation R might be a subset of the set X, but the domain of the function f must be equal to X. This is because each element of the domain of a function must have an element associated with it, whereas this is not necessary for a relation.<\/p>\n\n\n\n

In relation, one element of X might be associated with one or more elements of Y, while it must be associated with only one element of Y in a function.<\/p>\n\n\n\n

Thus, not every relation is a function. However, every function is necessarily a relation.<\/p>\n\n\n\n

4. Let A = {\u20132, \u20131, 0, 1, 2} and f: A \u2192 Z be a function defined by f(x) = x2<\/sup> \u2013 2x \u2013 3. Find:
(i) range of f i.e. f (A)
(ii) pre-images of 6, \u20133 and 5<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

A = {\u20132, \u20131, 0, 1, 2}<\/p>\n\n\n\n

f : A \u2192 Z such that f(x) = x2<\/sup> \u2013 2x \u2013 3<\/p>\n\n\n\n

(i)<\/strong> Range of f i.e. f (A)<\/p>\n\n\n\n

A is the domain of the function f. Hence, range is the set of elements f(x) for all x \u2208 A.<\/p>\n\n\n\n

Substituting x = \u20132 in f(x), we get<\/p>\n\n\n\n

f(\u20132) = (\u20132)2<\/sup> \u2013 2(\u20132) \u2013 3<\/p>\n\n\n\n

= 4 + 4 \u2013 3<\/p>\n\n\n\n

= 5<\/p>\n\n\n\n

Substituting x = \u20131 in f(x), we get<\/p>\n\n\n\n

f(\u20131) = (\u20131)2<\/sup> \u2013 2(\u20131) \u2013 3<\/p>\n\n\n\n

= 1 + 2 \u2013 3<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Substituting x = 0 in f(x), we get<\/p>\n\n\n\n

f(0) = (0)2<\/sup> \u2013 2(0) \u2013 3<\/p>\n\n\n\n

= 0 \u2013 0 \u2013 3<\/p>\n\n\n\n

= \u2013 3<\/p>\n\n\n\n

Substituting x = 1 in f(x), we get<\/p>\n\n\n\n

f(1) = 12<\/sup> \u2013 2(1) \u2013 3<\/p>\n\n\n\n

= 1 \u2013 2 \u2013 3<\/p>\n\n\n\n

= \u2013 4<\/p>\n\n\n\n

Substituting x = 2 in f(x), we get<\/p>\n\n\n\n

f(2) = 22<\/sup> \u2013 2(2) \u2013 3<\/p>\n\n\n\n

= 4 \u2013 4 \u2013 3<\/p>\n\n\n\n

= \u20133<\/p>\n\n\n\n

Thus, the range of f is {-4, -3, 0, 5}.<\/p>\n\n\n\n

(ii)<\/strong> pre-images of 6, \u20133 and 5<\/p>\n\n\n\n

Let x be the pre-image of 6 \u21d2 f(x) = 6<\/p>\n\n\n\n

x2<\/sup> \u2013 2x \u2013 3 = 6<\/p>\n\n\n\n

x2<\/sup> \u2013 2x \u2013 9 = 0<\/p>\n\n\n\n

x = [-(-2) \u00b1 \u221a<\/strong> ((-2)2<\/sup> \u2013 4(1) (-9))] \/ 2(1)<\/p>\n\n\n\n

= [2 \u00b1 \u221a<\/strong> (4+36)] \/ 2<\/p>\n\n\n\n

= [2 \u00b1 \u221a<\/strong>40] \/ 2<\/p>\n\n\n\n

= 1 \u00b1 \u221a<\/strong>10<\/p>\n\n\n\n

However, 1 \u00b1 \u221a<\/strong>10 \u2209 A<\/p>\n\n\n\n

Thus, there exists no pre-image of 6.<\/p>\n\n\n\n

Now, let x be the pre-image of \u20133 \u21d2 f(x) = \u20133<\/p>\n\n\n\n

x2<\/sup> \u2013 2x \u2013 3 = \u20133<\/p>\n\n\n\n

x2<\/sup> \u2013 2x = 0<\/p>\n\n\n\n

x(x \u2013 2) = 0<\/p>\n\n\n\n

x = 0 or 2<\/p>\n\n\n\n

Clearly, both 0 and 2 are elements of A.<\/p>\n\n\n\n

Thus, 0 and 2 are the pre-images of \u20133.<\/p>\n\n\n\n

Now, let x be the pre-image of 5 \u21d2 f(x) = 5<\/p>\n\n\n\n

x2<\/sup> \u2013 2x \u2013 3 = 5<\/p>\n\n\n\n

x2<\/sup> \u2013 2x \u2013 8= 0<\/p>\n\n\n\n

x2<\/sup> \u2013 4x + 2x \u2013 8= 0<\/p>\n\n\n\n

x(x \u2013 4) + 2(x \u2013 4) = 0<\/p>\n\n\n\n

(x + 2)(x \u2013 4) = 0<\/p>\n\n\n\n

x = \u20132 or 4<\/p>\n\n\n\n

However, 4 \u2209 A but \u20132 \u2208 A<\/p>\n\n\n\n

Thus, \u20132 is the pre-images of 5.<\/p>\n\n\n\n

\u2234 \u00d8, {0, 2}, -2 are the pre-images of 6, -3, 5<\/p>\n\n\n\n

5. If a function f: R \u2192 R be defined by

Find: f (1), f (\u20131), f (0), f (2).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Let us find f (1), f (\u20131), f (0) and f (2).<\/p>\n\n\n\n

When x > 0, f (x) = 4x + 1<\/p>\n\n\n\n

Substituting x = 1 in the above equation, we get<\/p>\n\n\n\n

f (1) = 4(1) + 1<\/p>\n\n\n\n

= 4 + 1<\/p>\n\n\n\n

= 5<\/p>\n\n\n\n

When x < 0, f(x) = 3x \u2013 2<\/p>\n\n\n\n

Substituting x = \u20131 in the above equation, we get<\/p>\n\n\n\n

f (\u20131) = 3(\u20131) \u2013 2<\/p>\n\n\n\n

= \u20133 \u2013 2<\/p>\n\n\n\n

= \u20135<\/p>\n\n\n\n

When x = 0, f(x) = 1<\/p>\n\n\n\n

Substituting x = 0 in the above equation, we get<\/p>\n\n\n\n

f (0) = 1<\/p>\n\n\n\n

When x > 0, f(x) = 4x + 1<\/p>\n\n\n\n

Substituting x = 2 in the above equation, we get<\/p>\n\n\n\n

f (2) = 4(2) + 1<\/p>\n\n\n\n

= 8 + 1<\/p>\n\n\n\n

= 9<\/p>\n\n\n\n

\u2234 f (1) = 5, f (\u20131) = \u20135, f (0) = 1 and f (2) = 9.<\/p>\n\n\n\n

6. A function f: R \u2192 R is defined by f(x) = x2<\/sup>. Determine
(i) range of f
(ii) {x: f(x) = 4}
(iii) {y: f(y) = \u20131}<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

f : R \u2192 R and f(x) = x2<\/sup>.<\/p>\n\n\n\n

(i)<\/strong> range of f<\/p>\n\n\n\n

Domain of f = R (set of real numbers)<\/p>\n\n\n\n

We know that the square of a real number is always positive or equal to zero.<\/p>\n\n\n\n

\u2234 range of f = R+<\/sup>\u222a {0}<\/p>\n\n\n\n

(ii)<\/strong> {x: f(x) = 4}<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

f(x) = 4<\/p>\n\n\n\n

we know, x2<\/sup> = 4<\/p>\n\n\n\n

x2<\/sup> \u2013 4 = 0<\/p>\n\n\n\n

(x \u2013 2)(x + 2) = 0<\/p>\n\n\n\n

\u2234 x = \u00b1 2<\/p>\n\n\n\n

\u2234 {x: f(x) = 4} = {\u20132, 2}<\/p>\n\n\n\n

(iii)<\/strong> {y: f(y) = \u20131}<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

f(y) = \u20131<\/p>\n\n\n\n

y2<\/sup> = \u20131<\/p>\n\n\n\n

However, the domain of f is R, and for every real number y, the value of y2<\/sup> is non-negative.<\/p>\n\n\n\n

Hence, there exists no real y for which y2<\/sup> = \u20131.<\/p>\n\n\n\n

\u2234{y: f(y) = \u20131} = \u2205<\/p>\n\n\n\n

7.<\/strong> Let f: R+<\/sup>\u2192 R, where R+<\/sup> is the set of all positive real numbers, be such that f(x) = loge <\/sub>x. Determine
(i) the image set of the domain of f
(ii) {x: f (x) = \u20132}
(iii) whether f (xy) = f (x) + f (y) holds.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f: R+<\/sup>\u2192 R and f(x) = loge <\/sub>x.<\/p>\n\n\n\n

(i)<\/strong> the image set of the domain of f<\/p>\n\n\n\n

Domain of f = R+<\/sup> (set of positive real numbers)<\/p>\n\n\n\n

We know the value of logarithm to the base e (natural logarithm) can take all possible real values.<\/p>\n\n\n\n

\u2234 The image set of f = R<\/p>\n\n\n\n

(ii)<\/strong> {x: f(x) = \u20132}<\/p>\n\n\n\n

Given f(x) = \u20132<\/p>\n\n\n\n

loge <\/sub>x = \u20132<\/p>\n\n\n\n

\u2234 x = e-2<\/sup> [since, logb <\/sub>a = c \u21d2 a = bc<\/sup>]<\/p>\n\n\n\n

\u2234 {x: f(x) = \u20132} = {e\u20132<\/sup>}<\/p>\n\n\n\n

(iii)<\/strong> Whether f (xy) = f (x) + f (y) holds.<\/p>\n\n\n\n

We have f (x) = loge <\/sub>x \u21d2 f (y) = loge <\/sub>y<\/p>\n\n\n\n

Now, let us consider f (xy)<\/p>\n\n\n\n

F (xy) = loge <\/sub>(xy)<\/p>\n\n\n\n

f (xy) = loge <\/sub>(x \u00d7 y) [since, logb <\/sub>(a\u00d7c) = logb <\/sub>a + logb <\/sub>c]<\/p>\n\n\n\n

f (xy) = loge <\/sub>x + loge <\/sub>y<\/p>\n\n\n\n

f (xy) = f (x) + f (y)<\/p>\n\n\n\n

\u2234 the equation f (xy) = f (x) + f (y) holds.<\/p>\n\n\n\n

8. Write the following relations as sets of ordered pairs and find which of them are functions:
(i) {(x, y): y = 3x, x \u2208 {1, 2, 3}, y \u2208 {3, 6, 9, 12}}
(ii) {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}
(iii) {(x, y): x + y = 3, x, y \u2208 {0, 1, 2, 3}}<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> {(x, y): y = 3x, x \u2208 {1, 2, 3}, y \u2208 {3, 6, 9, 12}}<\/p>\n\n\n\n

When x = 1, y = 3(1) = 3<\/p>\n\n\n\n

When x = 2, y = 3(2) = 6<\/p>\n\n\n\n

When x = 3, y = 3(3) = 9<\/p>\n\n\n\n

\u2234 R = {(1, 3), (2, 6), (3, 9)}<\/p>\n\n\n\n

Hence, the given relation R is a function.<\/p>\n\n\n\n

(ii)<\/strong> {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}<\/p>\n\n\n\n

When x = 1, y > 1 + 1 or y > 2 \u21d2 y = {4, 6}<\/p>\n\n\n\n

When x = 2, y > 2 + 1 or y > 3 \u21d2 y = {4, 6}<\/p>\n\n\n\n

\u2234 R = {(1, 4), (1, 6), (2, 4), (2, 6)}<\/p>\n\n\n\n

Hence, the given relation R is not a function.<\/p>\n\n\n\n

(iii)<\/strong> {(x, y): x + y = 3, x, y \u2208 {0, 1, 2, 3}}<\/p>\n\n\n\n

When x = 0, 0 + y = 3 \u21d2 y = 3<\/p>\n\n\n\n

When x = 1, 1 + y = 3 \u21d2 y = 2<\/p>\n\n\n\n

When x = 2, 2 + y = 3 \u21d2 y = 1<\/p>\n\n\n\n

When x = 3, 3 + y = 3 \u21d2 y = 0<\/p>\n\n\n\n

\u2234 R = {(0, 3), (1, 2), (2, 1), (3, 0)}<\/p>\n\n\n\n

Hence, the given relation R is a function.<\/p>\n\n\n\n

9. Let f: R \u2192 R and g: C \u2192 C be two functions defined as f(x) = x2<\/sup> and g(x) = x2<\/sup>. Are they equal functions?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

f: R \u2192 R \u2208 f(x) = x2<\/sup> and g : R \u2192 R \u2208 g(x) = x2<\/sup><\/p>\n\n\n\n

f is defined from R to R, the domain of f = R.<\/p>\n\n\n\n

g is defined from C to C, the domain of g = C.<\/p>\n\n\n\n

Two functions are equal only when the domain and codomain of both the functions are equal.<\/p>\n\n\n\n

In this case, the domain of f \u2260 domain of g.<\/p>\n\n\n\n

\u2234 f and g are not equal functions.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 3.2 PAGE NO: 3.11<\/p>\n\n\n\n

1. If f (x) = x2<\/sup> \u2013 3x + 4, then find the values of x satisfying the equation f (x) = f (2x + 1).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

f(x) = x2<\/sup> \u2013 3x + 4.<\/p>\n\n\n\n

Let us find x satisfying f (x) = f (2x + 1).<\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

f (2x + 1) = (2x + 1)2<\/sup> \u2013 3(2x + 1) + 4<\/p>\n\n\n\n

= (2x) 2<\/sup> + 2(2x) (1) + 12<\/sup> \u2013 6x \u2013 3 + 4<\/p>\n\n\n\n

= 4x2<\/sup> + 4x + 1 \u2013 6x + 1<\/p>\n\n\n\n

= 4x2<\/sup> \u2013 2x + 2<\/p>\n\n\n\n

Now, f (x) = f (2x + 1)<\/p>\n\n\n\n

x2<\/sup> \u2013 3x + 4 = 4x2<\/sup> \u2013 2x + 2<\/p>\n\n\n\n

4x2<\/sup> \u2013 2x + 2 \u2013 x2<\/sup> + 3x \u2013 4 = 0<\/p>\n\n\n\n

3x2<\/sup> + x \u2013 2 = 0<\/p>\n\n\n\n

3x2<\/sup> + 3x \u2013 2x \u2013 2 = 0<\/p>\n\n\n\n

3x(x + 1) \u2013 2(x + 1) = 0<\/p>\n\n\n\n

(x + 1)(3x \u2013 2) = 0<\/p>\n\n\n\n

x + 1 = 0 or 3x \u2013 2 = 0<\/p>\n\n\n\n

x = \u20131 or 3x = 2<\/p>\n\n\n\n

x = \u20131 or 2\/3<\/p>\n\n\n\n

\u2234 The values of x are \u20131 and 2\/3.<\/p>\n\n\n\n

2. If f (x) = (x \u2013 a)2 <\/sup>(x \u2013 b)2<\/sup>, find f (a + b).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

F (x) = (x \u2013 a)2<\/sup>(x \u2013 b)2<\/sup><\/p>\n\n\n\n

Let us find f (a + b).<\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

f (a + b) = (a + b \u2013 a)2 <\/sup>(a + b \u2013 b)2<\/sup><\/p>\n\n\n\n

f (a + b) = (b)2<\/sup> (a)2<\/sup><\/p>\n\n\n\n

\u2234 f (a + b) = a2<\/sup>b2<\/sup><\/p>\n\n\n\n

3. If y = f (x) = (ax \u2013 b) \/ (bx \u2013 a), show that x = f (y).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

y = f (x) = (ax \u2013 b) \/ (bx \u2013 a) \u21d2 f (y) = (ay \u2013 b) \/ (by \u2013 a)<\/p>\n\n\n\n

Let us prove that x = f (y).<\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

y = (ax \u2013 b) \/ (bx \u2013 a)<\/p>\n\n\n\n

By cross-multiplying,<\/p>\n\n\n\n

y(bx \u2013 a) = ax \u2013 b<\/p>\n\n\n\n

bxy \u2013 ay = ax \u2013 b<\/p>\n\n\n\n

bxy \u2013 ax = ay \u2013 b<\/p>\n\n\n\n

x(by \u2013 a) = ay \u2013 b<\/p>\n\n\n\n

x = (ay \u2013 b) \/ (by \u2013 a) = f (y)<\/p>\n\n\n\n

\u2234 x = f (y)<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

4. If f (x) = 1 \/ (1 \u2013 x), show that f [f {f (x)}] = x.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

f (x) = 1 \/ (1 \u2013 x)<\/p>\n\n\n\n

Let us prove that f [f {f (x)}] = x.<\/p>\n\n\n\n

Firstly, let us solve for f {f (x)}.<\/p>\n\n\n\n

f {f (x)} = f {1\/(1 \u2013 x)}<\/p>\n\n\n\n

= 1 \/ 1 \u2013 (1\/(1 \u2013 x))<\/p>\n\n\n\n

= 1 \/ [(1 \u2013 x \u2013 1)\/(1 \u2013 x)]<\/p>\n\n\n\n

= 1 \/ (-x\/(1 \u2013 x))<\/p>\n\n\n\n

= (1 \u2013 x) \/ -x<\/p>\n\n\n\n

= (x \u2013 1) \/ x<\/p>\n\n\n\n

\u2234 f {f (x)} = (x \u2013 1) \/ x<\/p>\n\n\n\n

Now, we shall solve for f [f {f (x)}]<\/p>\n\n\n\n

f [f {f (x)}] = f [(x-1)\/x]<\/p>\n\n\n\n

= 1 \/ [1 \u2013 (x-1)\/x]<\/p>\n\n\n\n

= 1 \/ [(x \u2013 (x-1))\/x]<\/p>\n\n\n\n

= 1 \/ [(x \u2013 x + 1)\/x]<\/p>\n\n\n\n

= 1 \/ (1\/x)<\/p>\n\n\n\n

\u2234 f [f {f (x)}] = x<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

5. If f (x) = (x + 1) \/ (x \u2013 1), show that f [f (x)] = x.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

f (x) = (x + 1) \/ (x \u2013 1)<\/p>\n\n\n\n

Let us prove that f [f (x)] = x.<\/p>\n\n\n\n

f [f (x)] = f [(x+1)\/(x-1)]<\/p>\n\n\n\n

= [(x+1)\/(x-1) + 1] \/ [(x+1)\/(x-1) \u2013 1]<\/p>\n\n\n\n

= [[(x+1) + (x-1)]\/(x-1)] \/ [[(x+1) \u2013 (x-1)]\/(x-1)]<\/p>\n\n\n\n

= [(x+1) + (x-1)] \/ [(x+1) \u2013 (x-1)]<\/p>\n\n\n\n

= (x+1+x-1)\/(x+1-x+1)<\/p>\n\n\n\n

= 2x\/2<\/p>\n\n\n\n

= x<\/p>\n\n\n\n

\u2234 f [f (x)] = x<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

6. If<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Find:<\/strong><\/p>\n\n\n\n

(i) f (1\/2)<\/strong><\/p>\n\n\n\n

(ii) f (-2)<\/strong><\/p>\n\n\n\n

(iii) f (1)<\/strong><\/p>\n\n\n\n

(iv) f (\u221a3)<\/strong><\/p>\n\n\n\n

(v) f (\u221a-3)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> f (1\/2)<\/p>\n\n\n\n

When, 0 \u2264 x \u2264 1, f(x) = x<\/p>\n\n\n\n

\u2234 f (1\/2) = \u00bd<\/p>\n\n\n\n

(ii)<\/strong> f (-2)<\/p>\n\n\n\n

When, x < 0, f(x) = x2<\/sup><\/p>\n\n\n\n

f (\u20132) = (\u20132)2<\/sup><\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

\u2234 f (\u20132) = 4<\/p>\n\n\n\n

(iii)<\/strong> f (1)<\/p>\n\n\n\n

When, x \u2265 1, f (x) = 1\/x<\/p>\n\n\n\n

f (1) = 1\/1<\/p>\n\n\n\n

\u2234 f(1) = 1<\/p>\n\n\n\n

(iv)<\/strong> f (\u221a3)<\/p>\n\n\n\n

We have \u221a3 = 1.732 > 1<\/p>\n\n\n\n

When, x \u2265 1, f (x) = 1\/x<\/p>\n\n\n\n

\u2234 f (\u221a3) = 1\/\u221a3<\/p>\n\n\n\n

(v)<\/strong> f (\u221a-3)<\/p>\n\n\n\n

We know \u221a-3 is not a real number and the function f(x) is defined only when x \u2208 R.<\/p>\n\n\n\n

\u2234 f (\u221a-3) does not exist.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 3.3 PAGE NO: 3.18<\/p>\n\n\n\n

1. Find the domain of each of the following real valued functions of real variable:<\/strong><\/p>\n\n\n\n

(i) f (x) = 1\/x<\/strong><\/p>\n\n\n\n

(ii) f (x) = 1\/(x-7)<\/strong><\/p>\n\n\n\n

(iii) f (x) = (3x-2)\/(x+1)<\/strong><\/p>\n\n\n\n

(iv) f (x) = (2x+1)\/(x2<\/sup>-9)<\/strong><\/p>\n\n\n\n

(v) f (x) = (x2<\/sup>+2x+1)\/(x2<\/sup>-8x+12)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>f (x) = 1\/x<\/p>\n\n\n\n

We know, f (x) is defined for all real values of x, except for the case when x = 0.<\/p>\n\n\n\n

\u2234 Domain of f = R \u2013 {0}<\/p>\n\n\n\n

(ii)<\/strong> f (x) = 1\/(x-7)<\/p>\n\n\n\n

We know, f (x) is defined for all real values of x, except for the case when x \u2013 7 = 0 or x = 7.<\/p>\n\n\n\n

\u2234 Domain of f = R \u2013 {7}<\/p>\n\n\n\n

(iii)<\/strong> f (x) = (3x-2)\/(x+1)<\/p>\n\n\n\n

We know, f(x) is defined for all real values of x, except for the case when x + 1 = 0 or x = \u20131.<\/p>\n\n\n\n

\u2234 Domain of f = R \u2013 {\u20131}<\/p>\n\n\n\n

(iv) <\/strong>f (x) = (2x+1)\/(x2<\/sup>-9)<\/p>\n\n\n\n

We know, f (x) is defined for all real values of x, except for the case when x2<\/sup> \u2013 9 = 0.<\/p>\n\n\n\n

x2<\/sup> \u2013 9 = 0<\/p>\n\n\n\n

x2<\/sup> \u2013 32<\/sup> = 0<\/p>\n\n\n\n

(x + 3)(x \u2013 3) = 0<\/p>\n\n\n\n

x + 3 = 0 or x \u2013 3 = 0<\/p>\n\n\n\n

x = \u00b1 3<\/p>\n\n\n\n

\u2234 Domain of f = R \u2013 {\u20133, 3}<\/p>\n\n\n\n

(v)<\/strong> f (x) = (x2<\/sup>+2x+1)\/(x2<\/sup>-8x+12)<\/p>\n\n\n\n

We know, f(x) is defined for all real values of x, except for the case when x2<\/sup> \u2013 8x + 12 = 0.<\/p>\n\n\n\n

x2<\/sup> \u2013 8x + 12 = 0<\/p>\n\n\n\n

x2<\/sup> \u2013 2x \u2013 6x + 12 = 0<\/p>\n\n\n\n

x(x \u2013 2) \u2013 6(x \u2013 2) = 0<\/p>\n\n\n\n

(x \u2013 2)(x \u2013 6) = 0<\/p>\n\n\n\n

x \u2013 2 = 0 or x \u2013 6 = 0<\/p>\n\n\n\n

x = 2 or 6<\/p>\n\n\n\n

\u2234 Domain of f = R \u2013 {2, 6}<\/p>\n\n\n\n

2. Find the domain of each of the following real valued functions of real variable:<\/strong><\/p>\n\n\n\n

(i) f (x) = \u221a(x-2)<\/strong><\/p>\n\n\n\n

(ii) f (x) = 1\/(\u221a(x2<\/sup>-1))<\/strong><\/p>\n\n\n\n

(iii) f (x) = \u221a(9-x2<\/sup>)<\/strong><\/p>\n\n\n\n

(iv) f (x) = \u221a(x-2)\/(3-x)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>f (x) = \u221a(x-2)<\/p>\n\n\n\n

We know the square of a real number is never negative.<\/p>\n\n\n\n

f (x) takes real values only when x \u2013 2 \u2265 0<\/p>\n\n\n\n

x \u2265 2<\/p>\n\n\n\n

\u2234 x \u2208 [2, \u221e)<\/p>\n\n\n\n

\u2234 Domain (f) = [2, \u221e)<\/p>\n\n\n\n

(ii)<\/strong> f (x) = 1\/(\u221a(x2<\/sup>-1))<\/p>\n\n\n\n

We know the square of a real number is never negative.<\/p>\n\n\n\n

f (x) takes real values only when x2<\/sup> \u2013 1 \u2265 0<\/p>\n\n\n\n

x2<\/sup> \u2013 12<\/sup> \u2265 0<\/p>\n\n\n\n

(x + 1) (x \u2013 1) \u2265 0<\/p>\n\n\n\n

x \u2264 \u20131 or x \u2265 1<\/p>\n\n\n\n

\u2234 x \u2208 (\u2013\u221e, \u20131] \u222a [1, \u221e)<\/p>\n\n\n\n

In addition, f (x) is also undefined when x2<\/sup> \u2013 1 = 0 because denominator will be zero and the result will be indeterminate.<\/p>\n\n\n\n

x2<\/sup> \u2013 1 = 0 \u21d2 x = \u00b1 1<\/p>\n\n\n\n

So, x \u2208 (\u2013\u221e, \u20131] \u222a [1, \u221e) \u2013 {\u20131, 1}<\/p>\n\n\n\n

x \u2208 (\u2013\u221e, \u20131) \u222a (1, \u221e)<\/p>\n\n\n\n

\u2234 Domain (f) = (\u2013\u221e, \u20131) \u222a (1, \u221e)<\/p>\n\n\n\n

(iii)<\/strong> f (x) = \u221a(9-x2<\/sup>)<\/p>\n\n\n\n

We know the square of a real number is never negative.<\/p>\n\n\n\n

f (x) takes real values only when 9 \u2013 x2<\/sup> \u2265 0<\/p>\n\n\n\n

9 \u2265 x2<\/sup><\/p>\n\n\n\n

x2<\/sup> \u2264 9<\/p>\n\n\n\n

x2<\/sup> \u2013 9 \u2264 0<\/p>\n\n\n\n

x2<\/sup> \u2013 32<\/sup> \u2264 0<\/p>\n\n\n\n

(x + 3)(x \u2013 3) \u2264 0<\/p>\n\n\n\n

x \u2265 \u20133 and x \u2264 3<\/p>\n\n\n\n

x \u2208 [\u20133, 3]<\/p>\n\n\n\n

\u2234 Domain (f) = [\u20133, 3]<\/p>\n\n\n\n

(iv)<\/strong> f (x) = \u221a(x-2)\/(3-x)<\/p>\n\n\n\n

We know the square root of a real number is never negative.<\/p>\n\n\n\n

f (x) takes real values only when x \u2013 2 and 3 \u2013 x are both positive and negative.<\/p>\n\n\n\n

(a)<\/strong> Both x \u2013 2 and 3 \u2013 x are positive<\/p>\n\n\n\n

x \u2013 2 \u2265 0<\/p>\n\n\n\n

x \u2265 2<\/p>\n\n\n\n

3 \u2013 x \u2265 0<\/p>\n\n\n\n

x \u2264 3<\/p>\n\n\n\n

Hence, x \u2265 2 and x \u2264 3<\/p>\n\n\n\n

\u2234 x \u2208 [2, 3]<\/p>\n\n\n\n

(b)<\/strong> Both x \u2013 2 and 3 \u2013 x are negative<\/p>\n\n\n\n

x \u2013 2 \u2264 0<\/p>\n\n\n\n

x \u2264 2<\/p>\n\n\n\n

3 \u2013 x \u2264 0<\/p>\n\n\n\n

x \u2265 3<\/p>\n\n\n\n

Hence, x \u2264 2 and x \u2265 3<\/p>\n\n\n\n

However, the intersection of these sets is null set. Thus, this case is not possible.<\/p>\n\n\n\n

Hence, x \u2208 [2, 3] \u2013 {3}<\/p>\n\n\n\n

x \u2208 [2, 3]<\/p>\n\n\n\n

\u2234 Domain (f) = [2, 3]<\/p>\n\n\n\n

3. Find the domain and range of each of the following real valued functions:<\/strong><\/p>\n\n\n\n

(i) f (x) = (ax+b)\/(bx-a)<\/strong><\/p>\n\n\n\n

(ii) f (x) = (ax-b)\/(cx-d)<\/strong><\/p>\n\n\n\n

(iii) f (x) = \u221a(x-1)<\/strong><\/p>\n\n\n\n

(iv) f (x) = \u221a(x-3)<\/strong><\/p>\n\n\n\n

(v) f (x) = (x-2)\/(2-x)<\/strong><\/p>\n\n\n\n

(vi) f (x) = |x-1|<\/strong><\/p>\n\n\n\n

(vii) f (x) = -|x|<\/strong><\/p>\n\n\n\n

(viii) f (x) = \u221a(9-x2<\/sup>)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>f (x) = (ax+b)\/(bx-a)<\/p>\n\n\n\n

f(x) is defined for all real values of x, except for the case when bx \u2013 a = 0 or x = a\/b.<\/p>\n\n\n\n

Domain (f) = R \u2013 (a\/b)<\/p>\n\n\n\n

Let f (x) = y<\/p>\n\n\n\n

(ax+b)\/(bx-a) = y<\/p>\n\n\n\n

ax + b = y(bx \u2013 a)<\/p>\n\n\n\n

ax + b = bxy \u2013 ay<\/p>\n\n\n\n

ax \u2013 bxy = \u2013ay \u2013 b<\/p>\n\n\n\n

x(a \u2013 by) = \u2013(ay + b)<\/p>\n\n\n\n

\u2234 x = \u2013 (ay+b)\/(a-by)<\/p>\n\n\n\n

When a \u2013 by = 0 or y = a\/b<\/p>\n\n\n\n

Hence, f(x) cannot take the value a\/b.<\/p>\n\n\n\n

\u2234 Range (f) = R \u2013 (a\/b)<\/p>\n\n\n\n

(ii) <\/strong>f (x) = (ax-b)\/(cx-d)<\/p>\n\n\n\n

f(x) is defined for all real values of x, except for the case when cx \u2013 d = 0 or x = d\/c. Domain (f) = R \u2013 (d\/c)<\/p>\n\n\n\n

Let f (x) = y<\/p>\n\n\n\n

(ax-b)\/(cx-d) = y<\/p>\n\n\n\n

ax \u2013 b = y(cx \u2013 d)<\/p>\n\n\n\n

ax \u2013 b = cxy \u2013 dy<\/p>\n\n\n\n

ax \u2013 cxy = b \u2013 dy<\/p>\n\n\n\n

x(a \u2013 cy) = b \u2013 dy<\/p>\n\n\n\n

\u2234 x = (b-dy)\/(a-cy)<\/p>\n\n\n\n

When a \u2013 cy = 0 or y = a\/c,<\/p>\n\n\n\n

Hence, f(x) cannot take the value a\/c.<\/p>\n\n\n\n

\u2234 Range (f) = R \u2013 (a\/c)<\/p>\n\n\n\n

(iii)<\/strong> f (x) = \u221a(x-1)<\/p>\n\n\n\n

We know the square of a real number is never negative.<\/p>\n\n\n\n

f(x) takes real values only when x \u2013 1 \u2265 0<\/p>\n\n\n\n

x \u2265 1<\/p>\n\n\n\n

\u2234 x \u2208 [1, \u221e)<\/p>\n\n\n\n

Thus, domain (f) = [1, \u221e)<\/p>\n\n\n\n

When x \u2265 1, we have x \u2013 1 \u2265 0<\/p>\n\n\n\n

Hence, \u221a(x-1) \u2265 0 \u21d2 f (x) \u2265 0<\/p>\n\n\n\n

f(x) \u2208 [0, \u221e)<\/p>\n\n\n\n

\u2234 Range (f) = [0, \u221e)<\/p>\n\n\n\n

(iv)<\/strong> f (x) = \u221a(x-3)<\/p>\n\n\n\n

We know the square of a real number is never negative.<\/p>\n\n\n\n

f (x) takes real values only when x \u2013 3 \u2265 0<\/p>\n\n\n\n

x \u2265 3<\/p>\n\n\n\n

\u2234 x \u2208 [3, \u221e)<\/p>\n\n\n\n

Domain (f) = [3, \u221e)<\/p>\n\n\n\n

When x \u2265 3, we have x \u2013 3 \u2265 0<\/p>\n\n\n\n

Hence, \u221a(x-3) \u2265 0 \u21d2 f (x) \u2265 0<\/p>\n\n\n\n

f(x) \u2208 [0, \u221e)<\/p>\n\n\n\n

\u2234 Range (f) = [0, \u221e)<\/p>\n\n\n\n

(v)<\/strong> f (x) = (x-2)\/(2-x)<\/p>\n\n\n\n

f(x) is defined for all real values of x, except for the case when 2 \u2013 x = 0 or x = 2.<\/p>\n\n\n\n

Domain (f) = R \u2013 {2}<\/p>\n\n\n\n

We have, f (x) = (x-2)\/(2-x)<\/p>\n\n\n\n

f (x) = -(2-x)\/(2-x)<\/p>\n\n\n\n

= \u20131<\/p>\n\n\n\n

When x \u2260 2, f(x) = \u20131<\/p>\n\n\n\n

\u2234 Range (f) = {\u20131}<\/p>\n\n\n\n

(vi)<\/strong> f (x) = |x-1|<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now we have,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Hence, f(x) is defined for all real numbers x.<\/p>\n\n\n\n

Domain (f) = R<\/p>\n\n\n\n

When, x < 1, we have x \u2013 1 < 0 or 1 \u2013 x > 0.<\/p>\n\n\n\n

|x \u2013 1| > 0 \u21d2 f(x) > 0<\/p>\n\n\n\n

When, x \u2265 1, we have x \u2013 1 \u2265 0.<\/p>\n\n\n\n

|x \u2013 1| \u2265 0 \u21d2 f(x) \u2265 0<\/p>\n\n\n\n

\u2234 f(x) \u2265 0 or f(x) \u2208 [0, \u221e)<\/p>\n\n\n\n

Range (f) = [0, \u221e)<\/p>\n\n\n\n

(vii)<\/strong> f (x) = -|x|<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now we have,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Hence, f(x) is defined for all real numbers x.<\/p>\n\n\n\n

Domain (f) = R<\/p>\n\n\n\n

When, x < 0, we have \u2013|x| < 0<\/p>\n\n\n\n

f (x) < 0<\/p>\n\n\n\n

When, x \u2265 0, we have \u2013x \u2264 0.<\/p>\n\n\n\n

\u2013|x| \u2264 0 \u21d2 f (x) \u2264 0<\/p>\n\n\n\n

\u2234 f (x) \u2264 0 or f (x) \u2208 (\u2013\u221e, 0]<\/p>\n\n\n\n

Range (f) = (\u2013\u221e, 0]<\/p>\n\n\n\n

(viii) <\/strong>f (x) = \u221a(9-x2<\/sup>)<\/p>\n\n\n\n

We know the square of a real number is never negative.<\/p>\n\n\n\n

f(x) takes real values only when 9 \u2013 x2<\/sup> \u2265 0<\/p>\n\n\n\n

9 \u2265 x2<\/sup><\/p>\n\n\n\n

x2<\/sup> \u2264 9<\/p>\n\n\n\n

x2<\/sup> \u2013 9 \u2264 0<\/p>\n\n\n\n

x2<\/sup> \u2013 32<\/sup> \u2264 0<\/p>\n\n\n\n

(x + 3)(x \u2013 3) \u2264 0<\/p>\n\n\n\n

x \u2265 \u20133 and x \u2264 3<\/p>\n\n\n\n

\u2234 x \u2208 [\u20133, 3]<\/p>\n\n\n\n

Domain (f) = [\u20133, 3]<\/p>\n\n\n\n

When, x \u2208 [\u20133, 3], we have 0 \u2264 9 \u2013 x2<\/sup> \u2264 9<\/p>\n\n\n\n

0 \u2264 \u221a(9-x2<\/sup>) \u2264 3 \u21d2 0 \u2264 f (x) \u2264 3<\/p>\n\n\n\n

\u2234 f(x) \u2208 [0, 3]<\/p>\n\n\n\n

Range (f) = [0, 3]<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 3.4 PAGE NO: 3.38<\/p>\n\n\n\n

1. Find f + g, f \u2013 g, cf (c \u2208 R, c \u2260 0), fg, 1\/f and f\/g in each of the following:
(i) f (x) = x3<\/sup> + 1 and g (x) = x + 1<\/strong><\/p>\n\n\n\n

(ii) f (x) = <\/strong>\u221a(x-1) and g (x) = \u221a(x+1)<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) <\/strong>f (x) = x3<\/sup> + 1 and g(x) = x + 1<\/p>\n\n\n\n

We have f(x): R \u2192 R and g(x): R \u2192 R<\/p>\n\n\n\n

(a) f + g<\/p>\n\n\n\n

We know, (f + g) (x) = f(x) + g(x)<\/p>\n\n\n\n

(f + g) (x) = x3<\/sup> + 1 + x + 1<\/p>\n\n\n\n

= x3<\/sup> + x + 2<\/p>\n\n\n\n

So, (f + g) (x): R \u2192 R<\/p>\n\n\n\n

\u2234 f + g: R \u2192 R is given by (f + g) (x) = x3<\/sup> + x + 2<\/p>\n\n\n\n

(b) f \u2013 g<\/p>\n\n\n\n

We know, (f \u2013 g) (x) = f(x) \u2013 g(x)<\/p>\n\n\n\n

(f \u2013 g) (x) = x3<\/sup> + 1 \u2013 (x + 1)<\/p>\n\n\n\n

= x3<\/sup> + 1 \u2013 x \u2013 1<\/p>\n\n\n\n

= x3<\/sup> \u2013 x<\/p>\n\n\n\n

So, (f \u2013 g) (x): R \u2192 R<\/p>\n\n\n\n

\u2234 f \u2013 g: R \u2192 R is given by (f \u2013 g) (x) = x3<\/sup> \u2013 x<\/p>\n\n\n\n

(c) cf (c \u2208 R, c \u2260 0)<\/p>\n\n\n\n

We know, (cf) (x) = c \u00d7 f(x)<\/p>\n\n\n\n

(cf)(x) = c(x3<\/sup> + 1)<\/p>\n\n\n\n

= cx3<\/sup> + c<\/p>\n\n\n\n

So, (cf) (x) : R \u2192 R<\/p>\n\n\n\n

\u2234 cf: R \u2192 R is given by (cf) (x) = cx3<\/sup> + c<\/p>\n\n\n\n

(d) fg<\/p>\n\n\n\n

We know, (fg) (x) = f(x) g(x)<\/p>\n\n\n\n

(fg) (x) = (x3<\/sup> + 1) (x + 1)<\/p>\n\n\n\n

= (x + 1) (x2<\/sup> \u2013 x + 1) (x + 1)<\/p>\n\n\n\n

= (x + 1)2 <\/sup>(x2<\/sup> \u2013 x + 1)<\/p>\n\n\n\n

So, (fg) (x): R \u2192 R<\/p>\n\n\n\n

\u2234 fg: R \u2192 R is given by (fg) (x) = (x + 1)2<\/sup>(x2<\/sup> \u2013 x + 1)<\/p>\n\n\n\n

(e) 1\/f<\/p>\n\n\n\n

We know, (1\/f) (x) = 1\/f (x)<\/p>\n\n\n\n

1\/f (x) = 1 \/ (x3<\/sup> + 1)<\/p>\n\n\n\n

Observe that 1\/f(x) is undefined when f(x) = 0 or when x = \u2013 1.<\/p>\n\n\n\n

So, 1\/f: R \u2013 {\u20131} \u2192 R is given by 1\/f (x) = 1 \/ (x3<\/sup> + 1)<\/p>\n\n\n\n

(f) f\/g<\/p>\n\n\n\n

We know, (f\/g) (x) = f(x)\/g(x)<\/p>\n\n\n\n

(f\/g) (x) = (x3<\/sup> + 1) \/ (x + 1)<\/p>\n\n\n\n

Observe that (x3<\/sup> + 1) \/ (x + 1) is undefined when g(x) = 0 or when x = \u20131.<\/p>\n\n\n\n

Using x3<\/sup> + 1 = (x + 1) (x2<\/sup> \u2013 x + 1), we have<\/p>\n\n\n\n

(f\/g) (x) = [(x+1) (x2<\/sup>\u2013 x+1)\/(x+1)]<\/p>\n\n\n\n

= x2<\/sup> \u2013 x + 1<\/p>\n\n\n\n

\u2234 f\/g: R \u2013 {\u20131} \u2192 R is given by (f\/g) (x) = x2<\/sup> \u2013 x + 1<\/p>\n\n\n\n

(ii) <\/strong>f (x) = \u221a(x-1) and g (x) = \u221a(x+1)<\/p>\n\n\n\n

We have f(x): [1, \u221e) \u2192 R+<\/sup> and g(x): [\u20131, \u221e) \u2192 R+<\/sup> as real square root is defined only for non-negative numbers.<\/p>\n\n\n\n

(a) f + g<\/p>\n\n\n\n

We know, (f + g) (x) = f(x) + g(x)<\/p>\n\n\n\n

(f+g) (x) = \u221a(x-1) + \u221a(x+1)<\/p>\n\n\n\n

Domain of (f + g) = Domain of f \u2229 Domain of g<\/p>\n\n\n\n

Domain of (f + g) = [1, \u221e) \u2229 [\u20131, \u221e)<\/p>\n\n\n\n

Domain of (f + g) = [1, \u221e)<\/p>\n\n\n\n

\u2234 f + g: [1, \u221e) \u2192 R is given by (f+g) (x) = \u221a(x-1) + \u221a(x+1)<\/p>\n\n\n\n

(b) f \u2013 g<\/p>\n\n\n\n

We know, (f \u2013 g) (x) = f(x) \u2013 g(x)<\/p>\n\n\n\n

(f-g) (x) = \u221a(x-1) \u2013 \u221a(x+1)<\/p>\n\n\n\n

Domain of (f \u2013 g) = Domain of f \u2229 Domain of g<\/p>\n\n\n\n

Domain of (f \u2013 g) = [1, \u221e) \u2229 [\u20131, \u221e)<\/p>\n\n\n\n

Domain of (f \u2013 g) = [1, \u221e)<\/p>\n\n\n\n

\u2234 f \u2013 g: [1, \u221e) \u2192 R is given by (f-g) (x) = \u221a(x-1) \u2013 \u221a(x+1)<\/p>\n\n\n\n

(c) cf (c \u2208 R, c \u2260 0)<\/p>\n\n\n\n

We know, (cf) (x) = c \u00d7 f(x)<\/p>\n\n\n\n

(cf) (x) = c\u221a(x-1)<\/p>\n\n\n\n

Domain of (cf) = Domain of f<\/p>\n\n\n\n

Domain of (cf) = [1, \u221e)<\/p>\n\n\n\n

\u2234 cf: [1, \u221e) \u2192 R is given by (cf) (x) = c\u221a(x-1)<\/p>\n\n\n\n

(d) fg<\/p>\n\n\n\n

We know, (fg) (x) = f(x) g(x)<\/p>\n\n\n\n

(fg) (x) = \u221a(x-1) \u221a(x+1)<\/p>\n\n\n\n

= \u221a(x2<\/sup> -1)<\/p>\n\n\n\n

Domain of (fg) = Domain of f \u2229 Domain of g<\/p>\n\n\n\n

Domain of (fg) = [1, \u221e) \u2229 [\u20131, \u221e)<\/p>\n\n\n\n

Domain of (fg) = [1, \u221e)<\/p>\n\n\n\n

\u2234 fg: [1, \u221e) \u2192 R is given by (fg) (x) = \u221a(x2<\/sup> -1)<\/p>\n\n\n\n

(e) 1\/f<\/p>\n\n\n\n

We know, (1\/f) (x) = 1\/f(x)<\/p>\n\n\n\n

(1\/f) (x) = 1\/\u221a(x-1)<\/p>\n\n\n\n

Domain of (1\/f) = Domain of f<\/p>\n\n\n\n

Domain of (1\/f) = [1, \u221e)<\/p>\n\n\n\n

Observe that 1\/\u221a(x-1) is also undefined when x \u2013 1 = 0 or x = 1.<\/p>\n\n\n\n

\u2234 1\/f: (1, \u221e) \u2192 R is given by (1\/f) (x) = 1\/\u221a(x-1)<\/p>\n\n\n\n

(f) f\/g<\/p>\n\n\n\n

We know, (f\/g) (x) = f(x)\/g(x)<\/p>\n\n\n\n

(f\/g) (x) = \u221a(x-1)\/\u221a(x+1)<\/p>\n\n\n\n

(f\/g) (x) = \u221a[(x-1)\/(x+1)]<\/p>\n\n\n\n

Domain of (f\/g) = Domain of f \u2229 Domain of g<\/p>\n\n\n\n

Domain of (f\/g) = [1, \u221e) \u2229 [\u20131, \u221e)<\/p>\n\n\n\n

Domain of (f\/g) = [1, \u221e)<\/p>\n\n\n\n

\u2234 f\/g: [1, \u221e) \u2192 R is given by (f\/g) (x) = \u221a[(x-1)\/(x+1)]<\/p>\n\n\n\n

2.<\/strong> Let f(x) = 2x + 5 and g(x) = x2<\/sup> + x. Describe<\/strong><\/p>\n\n\n\n

(i) f + g
(ii) f \u2013 g
(iii) fg
(iv) f\/g
Find the domain in each case.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

f(x) = 2x + 5 and g(x) = x2<\/sup> + x<\/p>\n\n\n\n

Both f(x) and g(x) are defined for all x \u2208 R.<\/p>\n\n\n\n

So, domain of f = domain of g = R<\/p>\n\n\n\n

(i)<\/strong> f + g<\/p>\n\n\n\n

We know, (f + g)(x) = f(x) + g(x)<\/p>\n\n\n\n

(f + g)(x) = 2x + 5 + x2<\/sup> + x<\/p>\n\n\n\n

= x2<\/sup> + 3x + 5<\/p>\n\n\n\n

(f + g)(x) Is defined for all real numbers x.<\/p>\n\n\n\n

\u2234 The domain of (f + g) is R<\/p>\n\n\n\n

(ii)<\/strong> f \u2013 g<\/p>\n\n\n\n

We know, (f \u2013 g)(x) = f(x) \u2013 g(x)<\/p>\n\n\n\n

(f \u2013 g)(x) = 2x + 5 \u2013 (x2<\/sup> + x)<\/p>\n\n\n\n

= 2x + 5 \u2013 x2<\/sup> \u2013 x<\/p>\n\n\n\n

= 5 + x \u2013 x2<\/sup><\/p>\n\n\n\n

(f \u2013 g)(x) is defined for all real numbers x.<\/p>\n\n\n\n

\u2234 The domain of (f \u2013 g) is R<\/p>\n\n\n\n

(iii)<\/strong> fg<\/p>\n\n\n\n

We know, (fg)(x) = f(x)g(x)<\/p>\n\n\n\n

(fg)(x) = (2x + 5)(x2<\/sup> + x)<\/p>\n\n\n\n

= 2x(x2<\/sup> + x) + 5(x2<\/sup> + x)<\/p>\n\n\n\n

= 2x3<\/sup> + 2x2<\/sup> + 5x2<\/sup> + 5x<\/p>\n\n\n\n

= 2x3<\/sup> + 7x2<\/sup> + 5x<\/p>\n\n\n\n

(fg)(x) is defined for all real numbers x.<\/p>\n\n\n\n

\u2234 The domain of fg is R<\/p>\n\n\n\n

(iv)<\/strong> f\/g<\/p>\n\n\n\n

We know, (f\/g) (x) = f(x)\/g(x)<\/p>\n\n\n\n

(f\/g) (x) = (2x+5)\/(x2<\/sup>+x)<\/p>\n\n\n\n

(f\/g) (x) is defined for all real values of x, except for the case when x2<\/sup> + x = 0.<\/p>\n\n\n\n

x2<\/sup> + x = 0<\/p>\n\n\n\n

x(x + 1) = 0<\/p>\n\n\n\n

x = 0 or x + 1 = 0<\/p>\n\n\n\n

x = 0 or \u20131<\/p>\n\n\n\n

When x = 0 or \u20131, (f\/g) (x) will be undefined as the division result will be indeterminate.<\/p>\n\n\n\n

\u2234 The domain of f\/g = R \u2013 {\u20131, 0}<\/p>\n\n\n\n

3.<\/strong> If f(x) be defined on [\u20132, 2] and is given by<\/strong>
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

  and g(x) = f(|x|) + |f(x)|. Find g(x).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

4. Let f, g be two real functions defined by f(x) = \u221a(x+1) and g(x) = \u221a(9-x2<\/sup>). Then, describe each of the following functions.
(i) f + g
(ii) g \u2013 f
(iii) fg
(iv) f\/g
(v) g\/f
(vi) 2f \u2013 \u221a5g
(vii) f2<\/sup> + 7f
(viii) 5\/g <\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

f(x) = \u221a(x+1) and g(x) = \u221a(9-x2<\/sup>)<\/p>\n\n\n\n

We know the square of a real number is never negative.<\/p>\n\n\n\n

So, f(x) takes real values only when x + 1 \u2265 0<\/p>\n\n\n\n

x \u2265 \u20131, x \u2208 [\u20131, \u221e)<\/p>\n\n\n\n

Domain of f = [\u20131, \u221e)<\/p>\n\n\n\n

Similarly, g(x) takes real values only when 9 \u2013 x2<\/sup> \u2265 0<\/p>\n\n\n\n

9 \u2265 x2<\/sup><\/p>\n\n\n\n

x2<\/sup> \u2264 9<\/p>\n\n\n\n

x2<\/sup> \u2013 9 \u2264 0<\/p>\n\n\n\n

x2<\/sup> \u2013 32<\/sup> \u2264 0<\/p>\n\n\n\n

(x + 3)(x \u2013 3) \u2264 0<\/p>\n\n\n\n

x \u2265 \u20133 and x \u2264 3<\/p>\n\n\n\n

\u2234 x \u2208 [\u20133, 3]<\/p>\n\n\n\n

Domain of g = [\u20133, 3]<\/p>\n\n\n\n

(i)<\/strong> f + g<\/p>\n\n\n\n

We know, (f + g)(x) = f(x) + g(x)<\/p>\n\n\n\n

(f + g) (x) = \u221a(x+1) + \u221a(9-x2<\/sup>)<\/p>\n\n\n\n

Domain of f + g = Domain of f \u2229 Domain of g<\/p>\n\n\n\n

= [\u20131, \u221e) \u2229 [\u20133, 3]<\/p>\n\n\n\n

= [\u20131, 3]<\/p>\n\n\n\n

\u2234 f + g: [\u20131, 3] \u2192 R is given by (f + g) (x) = f(x) + g(x) = \u221a(x+1) + \u221a(9-x2<\/sup>)<\/p>\n\n\n\n

(ii)<\/strong> g \u2013 f<\/p>\n\n\n\n

We know, (g \u2013 f)(x) = g(x) \u2013 f(x)<\/p>\n\n\n\n

(g \u2013 f) (x) = \u221a(9-x2<\/sup>) \u2013 \u221a(x+1)<\/p>\n\n\n\n

Domain of g \u2013 f = Domain of g \u2229 Domain of f<\/p>\n\n\n\n

= [\u20133, 3] \u2229 [\u20131, \u221e)<\/p>\n\n\n\n

= [\u20131, 3]<\/p>\n\n\n\n

\u2234 g \u2013 f: [\u20131, 3] \u2192 R is given by (g \u2013 f) (x) = g(x) \u2013 f(x) = \u221a(9-x2<\/sup>) \u2013 \u221a(x+1)<\/p>\n\n\n\n

(iii)<\/strong> fg<\/p>\n\n\n\n

We know, (fg) (x) = f(x)g(x)<\/p>\n\n\n\n

(fg) (x) = \u221a(x+1) \u221a(9-x2<\/sup>)<\/p>\n\n\n\n

= \u221a[(x+1) (9-x2<\/sup>)]<\/p>\n\n\n\n

= \u221a[x(9-x2<\/sup>) + (9-x2<\/sup>)]<\/p>\n\n\n\n

= \u221a(9x-x3<\/sup>+9-x2<\/sup>)<\/p>\n\n\n\n

= \u221a(9+9x-x2<\/sup>-x3<\/sup>)<\/p>\n\n\n\n

Domain of fg = Domain of f \u2229 Domain of g<\/p>\n\n\n\n

= [\u20131, \u221e) \u2229 [\u20133, 3]<\/p>\n\n\n\n

= [\u20131, 3]<\/p>\n\n\n\n

\u2234 fg: [\u20131, 3] \u2192 R is given by (fg) (x) = f(x) g(x) = \u221a(x+1) \u221a(9-x2<\/sup>) = \u221a(9+9x-x2<\/sup>-x3<\/sup>)<\/p>\n\n\n\n

(iv)<\/strong> f\/g<\/p>\n\n\n\n

We know, (f\/g) (x) = f(x)\/g(x)<\/p>\n\n\n\n

(f\/g) (x) = \u221a(x+1) \/ \u221a(9-x2<\/sup>)<\/p>\n\n\n\n

= \u221a[(x+1) \/ (9-x2<\/sup>)]<\/p>\n\n\n\n

Domain of f\/g = Domain of f \u2229 Domain of g<\/p>\n\n\n\n

= [\u20131, \u221e) \u2229 [\u20133, 3]<\/p>\n\n\n\n

= [\u20131, 3]<\/p>\n\n\n\n

However, (f\/g) (x) is defined for all real values of x \u2208 [\u20131, 3], except for the case when 9 \u2013 x2<\/sup> = 0 or x = \u00b1 3<\/p>\n\n\n\n

When x = \u00b13, (f\/g) (x) will be undefined as the division result will be indeterminate.<\/p>\n\n\n\n

Domain of f\/g = [\u20131, 3] \u2013 {\u20133, 3}<\/p>\n\n\n\n

Domain of f\/g = [\u20131, 3)<\/p>\n\n\n\n

\u2234 f\/g: [\u20131, 3) \u2192 R is given by (f\/g) (x) = f(x)\/g(x) = \u221a(x+1) \/ \u221a(9-x2<\/sup>)<\/p>\n\n\n\n

(v)<\/strong> g\/f<\/p>\n\n\n\n

We know, (g\/f) (x) = g(x)\/f(x)<\/p>\n\n\n\n

(g\/f) (x) = \u221a(9-x2<\/sup>) \/ \u221a(x+1)<\/p>\n\n\n\n

= \u221a[(9-x2<\/sup>) \/ (x+1)]<\/p>\n\n\n\n

Domain of g\/f = Domain of f \u2229 Domain of g<\/p>\n\n\n\n

= [\u20131, \u221e) \u2229 [\u20133, 3]<\/p>\n\n\n\n

= [\u20131, 3]<\/p>\n\n\n\n

However, (g\/f) (x) is defined for all real values of x \u2208 [\u20131, 3], except for the case when x + 1 = 0 or x = \u20131<\/p>\n\n\n\n

When x = \u20131, (g\/f) (x) will be undefined as the division result will be indeterminate.<\/p>\n\n\n\n

Domain of g\/f = [\u20131, 3] \u2013 {\u20131}<\/p>\n\n\n\n

Domain of g\/f = (\u20131, 3]<\/p>\n\n\n\n

\u2234 g\/f: (\u20131, 3] \u2192 R is given by (g\/f) (x) = g(x)\/f(x) = \u221a(9-x2<\/sup>) \/ \u221a(x+1)<\/p>\n\n\n\n

(vi)<\/strong> 2f \u2013 \u221a5g <\/strong><\/p>\n\n\n\n

We know, (2f \u2013 \u221a5g) <\/strong>(x) = 2f(x) \u2013 \u221a5g(x)<\/p>\n\n\n\n

(2f \u2013 \u221a5g) (x) = 2f (x) \u2013 \u221a5g (x)<\/p>\n\n\n\n

= 2\u221a(x+1) \u2013 \u221a5\u221a(9-x2<\/sup>)<\/p>\n\n\n\n

= 2\u221a(x+1) \u2013 \u221a(45- 5x2<\/sup>)<\/p>\n\n\n\n

Domain of 2f \u2013 \u221a5g <\/strong>= Domain of f \u2229 Domain of g<\/p>\n\n\n\n

= [\u20131, \u221e) \u2229 [\u20133, 3]<\/p>\n\n\n\n

= [\u20131, 3]<\/p>\n\n\n\n

\u2234 2f \u2013 \u221a5g: [\u20131, 3] \u2192 R is given by (2f \u2013 \u221a5g) (x) = 2f (x) \u2013 \u221a5g (x) = 2\u221a(x+1) \u2013 \u221a(45- 5x2<\/sup>)<\/p>\n\n\n\n

(vii)<\/strong> f2<\/sup> + 7f<\/p>\n\n\n\n

We know, (f2<\/sup> + 7f) (x) = f2<\/sup>(x) + (7f)(x)<\/p>\n\n\n\n

(f2<\/sup> + 7f) (x) = f(x) f(x) + 7f(x)<\/p>\n\n\n\n

= \u221a(x+1) \u221a(x+1) + 7\u221a(x+1)<\/p>\n\n\n\n

= x + 1 + 7\u221a(x+1)<\/p>\n\n\n\n

Domain of f2<\/sup> + 7f is same as domain of f.<\/p>\n\n\n\n

Domain of f2<\/sup> + 7f = [\u20131, \u221e)<\/p>\n\n\n\n

\u2234 f2<\/sup> + 7f: [\u20131, \u221e) \u2192 R is given by (f2<\/sup> + 7f) (x) = f(x) f(x) + 7f(x) = x + 1 + 7\u221a(x+1)<\/p>\n\n\n\n

(viii)<\/strong> 5\/g<\/p>\n\n\n\n

We know, (5\/g) (x) = 5\/g(x)<\/p>\n\n\n\n

(5\/g) (x) = 5\/\u221a(9-x2<\/sup>)<\/p>\n\n\n\n

Domain of 5\/g = Domain of g = [\u20133, 3]<\/p>\n\n\n\n

However, (5\/g) (x) is defined for all real values of x \u2208 [\u20133, 3], except for the case when 9 \u2013 x2<\/sup> = 0 or x = \u00b1 3<\/p>\n\n\n\n

When x = \u00b13, (5\/g) (x) will be undefined as the division result will be indeterminate.<\/p>\n\n\n\n

Domain of 5\/g = [\u20133, 3] \u2013 {\u20133, 3}<\/p>\n\n\n\n

= (\u20133, 3)<\/p>\n\n\n\n

\u2234 5\/g: (\u20133, 3) \u2192 R is given by (5\/g) (x) = 5\/g(x) = 5\/\u221a(9-x2<\/sup>)<\/p>\n\n\n\n

5. If f(x) = loge <\/sub>(1 \u2013 x) and g(x) = [x], then determine each of the following functions:
(i) f + g
(ii) fg<\/strong><\/p>\n\n\n\n

(iii) f\/g
(iv) g\/f <\/strong><\/p>\n\n\n\n

Also, find (f + g) (\u20131), (fg) (0), (f\/g) (1\/2) and (g\/f) (1\/2).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

f(x) = loge <\/sub>(1 \u2013 x) and g(x) = [x]<\/p>\n\n\n\n

We know, f(x) takes real values only when 1 \u2013 x > 0<\/p>\n\n\n\n

1 > x<\/p>\n\n\n\n

x < 1, \u2234 x \u2208 (\u2013\u221e, 1)<\/p>\n\n\n\n

Domain of f = (\u2013\u221e, 1)<\/p>\n\n\n\n

Similarly, g(x) is defined for all real numbers x.<\/p>\n\n\n\n

Domain of g = [x], x \u2208 R<\/p>\n\n\n\n

= R<\/p>\n\n\n\n

(i)<\/strong> f + g<\/p>\n\n\n\n

We know, (f + g) (x) = f(x) + g(x)<\/p>\n\n\n\n

(f + g) (x) = loge <\/sub>(1 \u2013 x) + [x]<\/p>\n\n\n\n

Domain of f + g = Domain of f \u2229 Domain of g<\/p>\n\n\n\n

Domain of f + g = (\u2013\u221e, 1) \u2229 R<\/p>\n\n\n\n

= (\u2013\u221e, 1)<\/p>\n\n\n\n

\u2234 f + g: (\u2013\u221e, 1) \u2192 R is given by (f + g) (x) = loge <\/sub>(1 \u2013 x) + [x]<\/p>\n\n\n\n

(ii)<\/strong> fg<\/p>\n\n\n\n

We know, (fg) (x) = f(x) g(x)<\/p>\n\n\n\n

(fg) (x) = loge <\/sub>(1 \u2013 x) \u00d7 [x]<\/p>\n\n\n\n

= [x] loge <\/sub>(1 \u2013 x)<\/p>\n\n\n\n

Domain of fg = Domain of f \u2229 Domain of g<\/p>\n\n\n\n

= (\u2013\u221e, 1) \u2229 R<\/p>\n\n\n\n

= (\u2013\u221e, 1)<\/p>\n\n\n\n

\u2234 fg: (\u2013\u221e, 1) \u2192 R is given by (fg) (x) = [x] loge <\/sub>(1 \u2013 x)<\/p>\n\n\n\n

(iii)<\/strong> f\/g<\/p>\n\n\n\n

We know, (f\/g) (x) = f(x)\/g(x)<\/p>\n\n\n\n

(f\/g) (x) = loge <\/sub>(1 \u2013 x) \/ [x]<\/p>\n\n\n\n

Domain of f\/g = Domain of f \u2229 Domain of g<\/p>\n\n\n\n

= (\u2013\u221e, 1) \u2229 R<\/p>\n\n\n\n

= (\u2013\u221e, 1)<\/p>\n\n\n\n

However, (f\/g) (x) is defined for all real values of x \u2208 (\u2013\u221e, 1), except for the case when [x] = 0.<\/p>\n\n\n\n

We have, [x] = 0 when 0 \u2264 x < 1 or x \u2208 [0, 1)<\/p>\n\n\n\n

When 0 \u2264 x < 1, (f\/g) (x) will be undefined as the division result will be indeterminate.<\/p>\n\n\n\n

Domain of f\/g = (\u2013\u221e, 1) \u2013 [0, 1)<\/p>\n\n\n\n

= (\u2013\u221e, 0)<\/p>\n\n\n\n

\u2234 f\/g: (\u2013\u221e, 0) \u2192 R is given by (f\/g) (x) = loge <\/sub>(1 \u2013 x) \/ [x]<\/p>\n\n\n\n

(iv)<\/strong> g\/f<\/p>\n\n\n\n

We know, (g\/f) (x) = g(x)\/f(x)<\/p>\n\n\n\n

(g\/f) (x) = [x] \/ loge <\/sub>(1 \u2013 x)<\/p>\n\n\n\n

However, (g\/f) (x) is defined for all real values of x \u2208 (\u2013\u221e, 1), except for the case when loge <\/sub>(1 \u2013 x) = 0.<\/p>\n\n\n\n

loge <\/sub>(1 \u2013 x) = 0 \u21d2 1 \u2013 x = 1 or x = 0<\/p>\n\n\n\n

When x = 0, (g\/f) (x) will be undefined as the division result will be indeterminate.<\/p>\n\n\n\n

Domain of g\/f = (\u2013\u221e, 1) \u2013 {0}<\/p>\n\n\n\n

= (\u2013\u221e, 0) \u222a (0, 1)<\/p>\n\n\n\n

\u2234 g\/f: (\u2013\u221e, 0) \u222a (0, 1) \u2192 R is given by (g\/f) (x) = [x] \/ loge <\/sub>(1 \u2013 x)<\/p>\n\n\n\n

(a) We need to find (f + g) (\u20131).<\/p>\n\n\n\n

We have, (f + g) (x) = loge <\/sub>(1 \u2013 x) + [x], x \u2208 (\u2013\u221e, 1)<\/p>\n\n\n\n

Substituting x = \u20131 in the above equation, we get<\/p>\n\n\n\n

(f + g)(\u20131) = loge <\/sub>(1 \u2013 (\u20131)) + [\u20131]<\/p>\n\n\n\n

= loge <\/sub>(1 + 1) + (\u20131)<\/p>\n\n\n\n

= loge<\/sub>2 \u2013 1<\/p>\n\n\n\n

\u2234 (f + g) (\u20131) = loge<\/sub>2 \u2013 1<\/p>\n\n\n\n

(b) We need to find (fg) (0).<\/p>\n\n\n\n

We have, (fg) (x) = [x] loge <\/sub>(1 \u2013 x), x \u2208 (\u2013\u221e, 1)<\/p>\n\n\n\n

Substituting x = 0 in the above equation, we get<\/p>\n\n\n\n

(fg) (0) = [0] loge <\/sub>(1 \u2013 0)<\/p>\n\n\n\n

= 0 \u00d7 loge<\/sub>1<\/p>\n\n\n\n

\u2234 (fg) (0) = 0<\/p>\n\n\n\n

(c) We need to find (f\/g) (1\/2)<\/p>\n\n\n\n

We have, (f\/g) (x) = loge <\/sub>(1 \u2013 x) \/ [x], x \u2208 (\u2013\u221e, 0)<\/p>\n\n\n\n

However, 1\/2 is not in the domain of f\/g.<\/p>\n\n\n\n

\u2234 (f\/g) (1\/2) does not exist.<\/p>\n\n\n\n

(d) We need to find (g\/f) (1\/2)<\/p>\n\n\n\n

We have, (g\/f) (x) = [x] \/ loge <\/sub>(1 \u2013 x), x \u2208 (\u2013\u221e, 0) \u222a (0, \u221e)<\/p>\n\n\n\n

Substituting x=1\/2 in the above equation, we get<\/p>\n\n\n\n

(g\/f) (1\/2) = [x] \/ loge <\/sub>(1 \u2013 x)<\/p>\n\n\n\n

= (1\/2)\/ loge <\/sub>(1 \u2013 1\/2)<\/p>\n\n\n\n

= 0.5\/ loge <\/sub>(1\/2)<\/p>\n\n\n\n

= 0 \/ loge <\/sub>(1\/2)<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

\u2234 (g\/f) (1\/2) = 0<\/p>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 3: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 11 Maths Chapter 3\u2013Functions<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 11 Maths Chapter 3\u2013Functions PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 11 Maths :<\/strong><\/h2>\n\n\n\n