NCERT Solutions For Class 12 Maths Chapter 1 Relations and Functions Ex 1.3

NCERT Solutions For Class 12 Chapter 1 Maths Relations and Functions Ex 1.3

Get Free NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Relations and Functions Exercise 1.3 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 1 Class 12 Relations and Functions Ex 1.3 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

Topics and Sub Topics in Class 11 Maths Chapter 1 Relations and Functions Ex 1.3:

1 – Relations and Functions
1.1 Introduction
1.2 Types of Relations
1.3 Types of Functions
1.4 Composition of Functions and Invertible Function
1.5 Binary Operations

NCERT Solutions for Class 12 Maths – Chapter 1 – Relations and Functions Ex 1.3

Ex 1.3 Class 12 Maths Question 1.
Let f: {1,3,4} –> {1,2, 5} and g : {1, 2,5} –> {1,3} be given by f = {(1, 2), (3,5), (4,1) and g = {(1,3), (2,3), (5,1)}. Write down g of.
Solution:
f= {(1,2),(3,5),(4,1)}
g= {(1,3),(2,3),(5,1)}
f(1) = 2, g(2) = 3 => gof(1) = 3
f(3) = 5, g(5)= 1 =>gof(3) = 1
f(4) = 1, g(1) = 3 => gof(4) = 3
=> gof= {(1,3), (3,1), (4,3)}

Ex 1.3 Class 12 Maths Question 2.
Let f, g and h be functions from R to R. Show that (f + g) oh = foh + goh, (f • g) oh = (foh) • (goh)
Solution:
f + R –> R, g: R –> R, h: R –> R
(i) (f+g)oh(x)=(f+g)[h(x)]
= f[h(x)]+g[h(x)]
={foh} (x)+ {goh} (x)
=>(f + g) oh = foh + goh
(ii) (f • g) oh (x) = (f • g) [h (x)]
= f[h (x)] • g [h (x)]
= {foh} (x) • {goh} (x)
=> (f • g) oh = (foh) • (goh)

Ex 1.3 Class 12 Maths Question 3.
Find gof and fog, if
(i) f (x) = |x| and g (x) = |5x – 2|
(ii) f (x) = 8x³ and g (x) = ${ x }^{ 1/3 }$ .
Solution:
(i) f(x) = |x|, g(x) = |5x – 2|
(a) gof(x) = g(f(x)) = g|x|= |5| x | – 2|
(b) fog(x) = f(g (x)) = f(|5x – 2|) = ||5 x – 2|| = |5x-2|
(ii) f(x) = 8x³ and g(x) = ${ x }^{ 1/3 }$
(a) gof(x) = g(f(x)) = g(8x³) = ${ { (8x }^{ 3 }) }^{ 1/3 }$ = 2x
(b) fog (x) = f(g (x))=f( ${ x }^{ 1/3 }$ ) = 8.( ${ x }^{ 1/3 }$ )³ = 8x

Ex 1.3 Class 12 Maths Question 4.
If $f(x)=\frac { 4x+3 }{ 6x-4 } x\neq \frac { 2 }{ 3 }$ , show that fof (x) = x, for all $x\neq \frac { 2 }{ 3 }$ . What is the inverse of f?
Solution:
$f(x)=\frac { 4x+3 }{ 6x-4 } x\neq \frac { 2 }{ 3 }$
(a) fof (x) = f(f(x)) = $f\frac { 4x+3 }{ 6x-4 }$
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1

Ex 1.3 Class 12 Maths Question 5.
State with reason whether following functions have inverse
(i) f: {1,2,3,4}–>{10} with f = {(1,10), (2,10), (3,10), (4,10)}
(ii) g: {5,6,7,8}–>{1,2,3,4} with g = {(5,4), (6,3), (7,4), (8,2)}
(iii) h: {1,2,3,4,5}–>{7,9,11,13} with h = {(2,7), (3,9), (4,11), (5,13)}
Solution:
f: {1,2,3,4} –> {10} with f = {(1,10), (2,10), (3,10), (4,10)}
(i) f is not one-one since 1,2,3,4 have the same image 4.
=> f has no inverse.
(ii) g: {5,6,7,8} –> {1,2,3,4} with g = {(5,4), (6,3) , (7,4), (8,2)}
Here also 5 and 7 have the same image
∴ g is not one-one. Therefore g is not invertible.
(iii) f has an inverse

Ex 1.3 Class 12 Maths Question 6.
Show that f: [-1,1] –> R, given by f(x) = $\frac { x }{ (x+2) }$ is one-one. Find the inverse of the function f: [-1,1] –> Range f.
Hint – For y ∈ Range f, y = f (x) = $\frac { x }{ (x+2) }$ for some x in [- 1,1], i.e., x = $\frac { 2y }{ (1-y) }$
Solution:
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 2

Ex 1.3 Class 12 Maths Question 7.
Consider f: R –> R given by f (x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Solution:
f: R—>R given by f(x) = 4x + 3
f (x₁) = 4x₁ + 3, f (x₂) = 4x₂ + 3
If f(x₁) = f(x₂), then 4x₁ + 3 = 4x₂ + 3
or 4x₁ = 4x₂ or x₁ = x₂
f is one-one
Also let y = 4x + 3, or 4x = y – 3
∴ $x=\frac { y-3 }{ 4 }$
For each value of y ∈ R and belonging to co-domain of y has a pre-image in its domain.
∴ f is onto i.e. f is one-one and onto
f is invertible and f^-1 (y) = g (y) = $\frac { y-3 }{ 4 }$

Ex 1.3 Class 12 Maths Question 8.
Consider f: R₊ –> [4, ∞] given by f (x) = x² + 4. Show that f is invertible with the inverse f^-1 of f given by f^-1 (y) = √y-4 , where R₊ is the set of all non-negative real numbers.
Solution:
f(x₁) = x₁² + 4 and f(x₂) = x₂² + 4
f(x₁) = f(x₂) => x₁² + 4 = x₂² + 4
or x₁² = x₂² => x₁ = x₂ As x ∈ R
∴ x>0, x₁² = x₂² => x₁ = x₂ =>f is one-one
Let y = x² + 4 or x² = y – 4 or x = ±√y-4
x being > 0, -ve sign not to be taken
x = √y – 4
∴ f^-1 (y) = g(y) = √y-4 ,y ≥ 4
For every y ≥ 4, g (y) has real positive value.
∴ The inverse of f is f^-1 (y) = √y-4

Ex 1.3 Class 12 Maths Question 9.
Consider f: R₊ –> [- 5, ∞) given by f (x) = 9x² + 6x – 5. Show that f is invertible with
${ f }^{ -1 }(y)=\left( \frac { \left( \sqrt { y+6 } \right) -1 }{ 3 } \right)$
Solution:
Let y be an arbitrary element in range of f.
Let y = 9x² + 6x – 5 = 9x² + 6x + 1 – 6
=> y = (3x + 1)² – 6
=> y + 6 = (3x + 1)²
=> 3x + 1 = √y + 6
tiwari academy class 12 maths Chapter 1 Relations and Functions 9
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 9.1

Ex 1.3 Class 12 Maths Question 10.
Let f: X –> Y be an invertible function. Show that f has unique inverse.
Hint – suppose g₁ and g₂ are two inverses of f. Then for all y∈Y, fog₁(y)=I_y(y)=fog₂(y).Use one-one ness of f.
Solution:
If f is invertible gof (x) = I_x and fog (y) = I_y
∴ f is one-one and onto.
Let there be two inverse g₁ and g₂
fog₁ (y) = I_y, fog₂ (y) = I_y
I_y being unique for a given function f
=> g₁ (y) = g₂ (y)
f is one-one and onto
f has a unique inverse.

Ex 1.3 Class 12 Maths Question 11.
Consider f: {1,2,3} –> {a, b, c} given by f (1) = a, f (2)=b and f (3)=c. Find f^-1 and show that (f^-1)f^-1=f.
Solution:
f: {1,2, 3,} –> {a,b,c} so that f(1) = a, f(2) = b, f(3) = c
Now let X = {1,2,3}, Y = {a,b,c}
∴ f: X –> Y
∴ f^-1: Y –> X such that f^-1 (a)= 1, f^-1(b) = 2; f^-1(c) = 3
Inverse of this function may be written as
(f^-1)^-1 : X –> Y such that
(f^-1)^-1 (1) = a, (f^-1)^-1 (2) = b, (f^-1)^-1 (3) = c
We also have f: X –> Y such that
f(1) = a,f(2) = b,f(3) = c => (f^-1)^-1= f

Ex 1.3 Class 12 Maths Question 12.
Let f: X –> Y be an invertible function. Show that the inverse of f^-1 is f, i.e., (f^-1)^-1 = f.
Solution:
f: X —> Y is an invertible function
f is one-one and onto
=> g : Y –> X, where g is also one-one and onto such that
gof (x) = I_x and fog (y) = I_y => g = f^-1
Now f^-1 o (f^-1)^-1 = I
and fo[f^-1o (f^-1)^-1] =fol
or (fof^-1)^-1 o (f^-1)^-1 = f
=> Io (f^-1)^-1 = f
=> (f^-1)^-1 = f

Ex 1.3 Class 12 Maths Question 13.
If f: R –> R be given by f(x) = ${ \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }$ , then fof (x) is
(a) ${ x }^{ \frac { 1 }{ 3 } }$
(b) x³
(c) x
(d) (3 – x³)
Solution:
f: R-> R defined by f(x) = ${ \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }$
fof (x) = f[f(x)] = ${f{ \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }}$
= ${ \left[ 3-{ \left\{ { \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } } \right\} }^{ 3 } \right] }^{ \frac { 1 }{ 3 } }$
= ${ \left[ 3-{ \left\{ { \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } } \right\} } \right] }$
= ${ \left( { x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }$
= x

Ex 1.3 Class 12 Maths Question 14.
Let f: $R-\left\{ -\frac { 4 }{ 3 } \right\} \rightarrow R$ be a function defined as f (x) = $\frac { 4x }{ 3x+4 }$ . The inverse of f is the map g: Range f–> $R-\left\{ -\frac { 4 }{ 3 } \right\} \rightarrow R$ given by
(a) $g(y)=\frac { 3y }{ 3-4y }$
(b) $g(y)=\frac { 4y }{ 4-3y }$
(c) $g(y)=\frac { 4y }{ 3-4y }$
(d) $g(y)=\frac { 3y }{ 4-3y }$
Solution:
(b)

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions in Hindi Medium Ex 1.3

NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.3 Relations and functions all questions answers.

12 Maths Exercise 1.3 solutions free to all board in hindi.