Let f : R → R : f ( x ) = 1 2 ( 3 x + 1 ) . Show that f is invertible and find f - 1 .

We have to show that f : R → R : f ( x ) = 1 2 ( 3 x + 1 ) is an invertible function. We know that a function is invertible if and only if it is a bijective function (i.e. one-one and onto) So, we need to prove that f is one-one and onto. Let x 1 , x 2 ∈ R such that f x 1 = f x 2 ⇒ 3 x 1 + 1 2 = 3 x 2 + 1 2 ⇒ 3 x 1 + 1 = 3 x 2 + 1 ⇒ x 1 = x 2 ∴ f ( x ) is a one-one Now, y = f ( x ) ⇒ y = ( 3 x + 1 ) 2 ⇒ x = 2 y - 1 3 For any y ∈ R there exists x = 2 y - 1 3 ∈ R such that f ( x ) = f ( 2 y - 1 3 ) = 1 2 ( 3 × 2 y - 1 3 + 1 ) = 2 y - 1 + 1 2 = y So, range of f ( x ) = co-domain of f ( x ) = R ∴ f ( x ) is an onto function Thus, f is an one-one, onto function and hence it is invertible Inverse of f ( x ) is given by f - 1 : R → R : f - 1 ( y ) = 2 y - 1 3 .

If f ( x ) = ( 4 x + 3 ) ( 6 x - 4 ) , x ≠ 2 3 show that ( f o f ) ( x ) = x for all x ≠ 2 3 . Hence, find f - 1 .

To Show: that f o f ( x ) = x Consider ( f o f ) ( x ) f f x = 4 4 x + 3 6 x - 4 + 3 6 4 x + 3 6 x - 4 - 4 = 4 ( 4 x + 3 ) + 3 ( 6 x - 4 ) 6 ( 4 x + 3 ) - 4 ( 6 x - 4 ) = 16 x + 12 + 18 x - 12 24 x + 18 - 24 x + 16 = 34 x 34 = x Finding f - 1 A function is invertible only when it is bijective (both one-one and onto) Proof to whether the function is one-one A function is said to be one-one if f x 1 = f x 2 ⇒ x 1 = x 2 ⇒ 4 x 1 + 3 6 x 1 - 4 = 4 x 2 + 3 6 x 2 - 4 ⇒ 4 x 1 + 3 6 x 2 - 4 = 4 x 2 + 3 6 x 1 - 4 ⇒ 24 x 1 x 2 + 18 x 2 - 16 x 1 - 12 = 24 x 1 x 2 + 18 x 1 - 16 x 2 - 12 ⇒ 34 x 1 = 34 x 2 ⇒ x 1 = x 2 Hence, the function is one-one. For onto, Range of f ( x ) = R excluding x = 2 3 Let f ( x ) = y . ⇒ 6 x y - 4 y = 4 x + 3 ⇒ 6 x y - 4 x = 4 y + 3 ⇒ x ( 6 y - 4 ) = 4 y + 3 ⇒ x = 4 y + 3 6 y - 4 So Domain of y is R(excluding 2 3 ) = Range of f ( x ) Hence, Range of f ( x ) = co-domain of f ( x ) = R (excluding 2 3 ) So, f ( x ) is onto function As it is bijective function. So it is invertible Inverse of f ( x ) is f - 1 ( y ) = 4 y + 3 6 y - 4

Show that the function f on A = R - 2 3 defined as f ( x ) = 4 x + 3 6 x - 4 is one-one and onto. Hence, find f - 1 .

We know that any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)] So, We need to prove that the given function is one-one and onto. Let x 1 , x 2 ∈ Q and f ( x ) = ( 4 x + 3 ) ( 6 x - 4 ) So f x 1 = f x 2 on solving ⇒ 4 x 1 + 3 6 x 1 - 4 = 4 x 2 + 3 6 x 2 - 4 ⇒ 24 x 1 x 2 - 16 x 1 + 18 x 2 - 12 = 24 x 1 x 2 + 18 x 1 - 16 x 2 - 12 ⇒ 34 x 1 = 34 x 2 we get x 1 = x 2 So f x 1 = f x 2 ⇒ x 1 = x 2 , ∴ f ( x ) is one-one Let y be any arbitrary element of A . Then f ( x ) = y ⇒ ( 4 x + 3 ) ( 6 x - 4 ) = y ⇒ ( 6 x - 4 ) y = 4 x + 3 ⇒ 6 x y - 4 y = 4 x + 3 ⇒ x = 4 y + 3 6 y - 4 Thus, for each y ∈ A , there exist x = 4 y + 3 6 y - 4 such that f ( x ) = f 4 y + 3 6 y - 4 = 4 · 4 y + 3 6 y - 4 + 3 6 · 4 y + 3 6 y - 4 - 4 = 16 y + 12 + 18 y - 12 24 y + 18 - 24 y + 16 = 34 y 34 = y So, f ( x ) is onto function Now f ( x ) = y ⇒ x = f - 1 ( y ) ∴ f - 1 ( y ) = 4 y + 3 6 y - 4 , for all y ∈ A .

Show that the function f on A = R - - 4 3 into itself, defined by f ( x ) = 4 x ( 3 x + 4 ) is one-one and onto. Hence, find f - 1

We know that any function is invertible if and only if it is bijective functions (i.e. one-one and onto) So, We need to prove that the given function is one-one and onto. Let x 1 , x 2 ∈ R and f ( x ) = 4 x ( 3 x + 4 ) So f x 1 = f x 2 ⇒ 4 x 1 3 x 1 + 4 = 4 x 2 3 x 2 + 4 on solving we get x 1 = x 2 So f x 1 = f x 2 ⇒ x 1 = x 2 , f ( x ) is one-one. Given co-domain of f ( x ) is R except 3 x + 4 = 0 . y = f ( x ) = ( 4 x ) ( 3 x + 4 ) ⇒ 3 x y + 4 y = 4 x ⇒ 3 x y - 4 x = - 4 y ⇒ x 3 y - 4 = - 4 y ⇒ x = - 4 y 3 y - 4 So, [ Range of f ( x ) = Domain of y ] So Domain of y is R = Range of f ( x ) Hence, Range of f ( x ) = co-domain of f ( x ) = R except x = - 4 3 So, f ( x ) is onto function As it is bijective function. So it is invertible Inverse of f ( x ) is f - 1 ( y ) = 4 y 4 - 3 y i.e, f - 1 x = 4 x 4 - 3 x

Let R + be the set of all positive real numbers. Show that the function f : R + → [ - 5 , ∞ ) : f ( x ) = 9 x 2 + 6 x - 5 is invertible. Find f - 1 .

We know that any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)] So, We need to prove that the given function is one-one and onto. Let x 1 , x 2 ∈ R and f ( x ) = 9 x 2 + 6 x - 5 So f x 1 = f x 2 9 x 1 2 + 6 x 1 - 5 = 9 x 2 2 + 6 x 2 - 5 on solving we get x 1 = x 2 So f x 1 = f x 2 ⇒ x 1 = x 2 Therefore, f x is one-one. Given co-domain of f ( x ) is [ - 5 , ∞ ) Let y = f ( x ) = 9 x 2 + 6 x - 5 So x = - 1 + y + 6 3 [Range of f ( x ) = Domain of y ] So Domain of y = Range of f ( x ) = [ - 5 , ∞ ) Hence, Range of f ( x ) = co-domain of f ( x ) = [ - 5 , ∞ ) So, f ( x ) is onto function As it is bijective function. So it is invertible Inverse of f ( x ) is f - 1 ( y ) = - 1 + y + 6 3 .

Let f : N → R : f ( x ) = 4 x 2 + 12 x + 15 . Show that f : N → range ( f ) is invertible. Find f - 1

We know that any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)] So, We need to prove that the given function is one-one and onto. Let x 1 , x 2 ∈ R and f ( x ) = 4 x 2 + 12 x + 15 So f x 1 = f x 2 ⇒ 4 x 1 2 + 12 x 1 + 15 = 4 x 2 2 + 12 x 2 + 15 on solving, we get → x 1 = x 2 So f x 1 = f x 2 ⇒ x 1 = x 2 Therefore, f ( x ) is one-one Given co-domain of f ( x ) is Range ( f ) . Let y = f ( x ) = 4 x 2 + 12 x + 15 So x = - 3 + y - 6 2 [ Range of f ( x ) = Domain of y] So Domain of y = Range of f ( x ) = [ 6 , ∞ ) Hence, Range of f ( x ) = co-domain of f ( x ) = [ 6 , ∞ ) So, f ( x ) is onto function . As it is bijective function, So it is invertible Inverse of f ( x ) is f - 1 ( y ) = - 3 + y - 6 2

Let A = R – 2 and B = R – 1 . If f : A → B : f ( x ) = x - 1 x - 2 , show that f is one-one and onto. Hence, find f - 1

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)] So, We need to prove that the given function is one-one and onto. Let x 1 , x 2 ∈ Q and f ( x ) = x - 1 x - 2 So f x 1 = f x 2 x 1 - 1 x 1 - 2 = x 2 - 1 x 2 - 2 on solving we get, x 1 = x 2 So f x 1 = f x 2 ⇒ x 1 = x 2 Therefore, f ( x ) is one-one Given co-domain of f ( x ) is R - { 1 } Let y = f ( x ) = x - 1 x - 2 , So, x = 2 y - 1 y - 1 [ Range of f ( x ) = Domain of y ] So Domain of y = Range of f ( x ) = R – 1 Hence, Range of f ( x ) = R – 1 co-domain of f ( x ) = R – 1 . So, f ( x ) is onto function As it is a bijective function. So it is invertible Inverse of f ( x ) is f - 1 ( y ) = 2 y - 1 y - 1

E M B I B E

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Relations and Functions>Chapter 2 - Functions>EXERCISE

R S Aggarwal Solutions for Exercise 4: EXERCISE

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 4: EXERCISE

Attempt the free practice questions from Exercise 4: EXERCISE with hints and solutions to strengthen your understanding. SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12 solutions are prepared by Experienced Embibe Experts.

Questions from R S Aggarwal Solutions for Exercise 4: EXERCISE with Hints & Solutions

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 2 - Functions>EXERCISE>Q 4

Let $f : R \to R : f (x) = \frac{1}{2} (3 x + 1)$ . Show that $f$ is invertible and find $f^{- 1}$ .

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 2 - Functions>EXERCISE>Q 5

If $f (x) = \frac{(4 x + 3)}{(6 x - 4)}, x \neq \frac{2}{3}$ show that $(f o f) (x) = x$ for all $x \neq \frac{2}{3}$ . Hence, find $f^{- 1}$ .

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 2 - Functions>EXERCISE>Q 6

Show that the function $f$ on $A = R - \{\frac{2}{3}\}$ defined as $f (x) = \frac{4 x + 3}{6 x - 4}$ is one-one and onto. Hence, find $f^{- 1}$ .

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 2 - Functions>EXERCISE>Q 7

Show that the function $f$ on $A = R - \{\frac{- 4}{3}\}$ into itself, defined by $f (x) = \frac{4 x}{(3 x + 4)}$ is one-one and onto. Hence, find $f^{- 1}$

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 2 - Functions>EXERCISE>Q 8

Let $R^{+}$ be the set of all positive real numbers. Show that the function $f : R^{+} \to [- 5, \infty) : f (x) = (9 x^{2} + 6 x - 5)$ is invertible. Find $f^{- 1}$ .

EASY

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 2 - Functions>EXERCISE>Q 9

Let $f : N \to R : f (x) = 4 x^{2} + 12 x + 15$ . Show that $f : N \to$ range $(f)$ is invertible. Find $f^{- 1}$

EASY

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 2 - Functions>EXERCISE>Q 10

Let $A = R - \{2\}$ and $B = R - \{1\}$ . If $f : A \to B : f (x) = \frac{x - 1}{x - 2}$ , show that f is one-one and onto. Hence, find $f^{- 1}$

EASY

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 2 - Functions>EXERCISE>Q 11

Let $f$ and $g$ be two functions from $R$ into $R$ , defined by $f (x) = | x | + x$ and $g (x) = | x | - x$ for all $x \in R$ . Find $f \circ g$ and $g \circ f$ .

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