# NCERT Solutions Class 12 maths Chapter-1(Relation And Functions)Exercise 1.3

**NCERT Solutions Class 12 Maths**from class 12th Students will get the answers of

**Chapter-1 (Relation And Functions) Exercise 1.3**This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.

We have given the answers of all the questions of

**NCERT Board Mathematics Textbook**in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

### Exercise 1.3

**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, 5), (5, 1)}. Write down gof.**

**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)}

**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

(f+g) oh(x) = (f+g) oh(x)

= (f+g) [h(x)]

= f[h(x)] + g[h(x)]

= foh(x) + goh(x)

(f+g) oh = foh + goh

(f * g) oh(x) = (f * g) oh(x)

= (f * g) [h(x)]

= f[h(x)] * g[h(x)]

= foh(x) * goh (x)

(f * g) oh = (foh) * (goh)

**Question 3. **

**Find gof and fog, if**

**(i) f(x) = |x| and g(x) = |5x – 2|**

**(ii) f(x) = 8x**^{3 }**and g(x) = x**^{1/3}

**Solution: **

**(i) **We have,

f(x) = |x| and g (x) = | 5x – 2 |

gof(x) = g(f(x)) = g(|x|)

=> gof(x) = | 5 |x|-2 |

fog(x) = f(g(x)) = f(|5x-2|)

=> fog(x) = || 5x-2|| = | 5x -2 |

**(ii) **We have,

f(x) = 8x^{3} and g(x) = x^{1/3}

gof(x) = g(f(x)) = g(8x^{3})

=> gof(x) = (8x^{3})^{1/3} = 2x

fog(x) = f(g(x)) = f(x^{1/3})

=> fog(x) = 8(x^{1/3})^{3} = 8x

**Question 4. **

**If f(x) = , show that fof(x) = x for all . What is the inverse of f ?**

**Solution: **

Given that,

Now,

fof(x) = f(f(x)) =

=

On simplifying by taking LCM = (6x-4)

fof(x)=

=> fof (x) *= **= x ** *

=> fof(x) = I_{A }(x) for all

=> fof(x) = I_{A} such that A = – which is the domain of f

=> f^{-1} = f

Hence, proved.

**Question 5. **

**State with reason whether the following functions have inverse. Find the inverse, if it exists.**

**(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 : {2, 3, 4, 5} -> {7, 9, 11, 13}**

**with h : {(2, 7), (3, 9), (4, 11), (5, 13)}**

**Solution: **

**(i)** We have f(1) = f(2) = f(3) = f(4) = 10 which means that f is many-one

and not one-one, therefore inverse of f does not exist.

**(ii)** Here g(5) = g(7) =4 i.e. g is many-one, so inverse of g does not exist.

**(iii) **Since range of h = {7, 9, 11, 13} = co-domain, therefore h is onto,

Also, each element of domain has a unique image in h, therefore h is one-one.

Now, since h is both one-one and onto,thus inverse of h exists.

h^{-1 }= {(7, 2), (9, 3), (11, 4), (13, 5)}

**Question 6. **

**Show that f : [-1, 1] -> R, given by f(x) **= **is one-one. Find the inverse of the function f : {-1,1} -> Range f.**

**Solution: **

Let x, y_{ } [-1, 1]

f(x) =

f(y) =

Now,

Let f(x) = f(y)

=> x(x + 2) = y(x + 2)

=> x y + 2x = x y + 2y

=> 2x = 2y

=> x = y

=> f is one-one

Also,

X = [-1, 1] and,

Y = { } = range of f.

=> f is onto

Since f is one-one and onto, therefore inverse of f exists.

Let y = f(x) => x =f^{-1}(y)

=> y * *=

=> x y + 2y = x

=> 2y = x(1 – y)

=> x =

Therefore, f : Y-> X is defined by f(y)= .

**Question 7. **

**Consider f : R -> R is given by f(x) = 4x + 3 . Show that f is invertible. Find the inverse of f.**

**Solution: **

It is given that,

f(x) = 4x + 3 where f : R -> R

Let,

f(x) = f(y)

=> 4x + 3 = 4y + 3

=> 4x = 4y

=> x = y

=> f is one-one function

Also,

Let y = 4x + 3 where y R

=> x =

Since for any . there exists such that

f(x) = = 4 +3 = y

=> f is onto

Since f is both one-one and onto, therefore f^{-1} exists

=> f^{-1}(y) =

**Question 8. **

**Consider f : R**_{+}** -> [4, **** ) given by f(x) = x**^{2}** + 4. Show that f is invertible with the inverse f**^{-1}** of f given by f**^{-1}**(y) **= **, where R**_{+}** is the set of all non-negative real numbers.**

**Solution: **

Let f(x) = f(y)

=> x + 4 = y + 4

=> x^{2} = y^{2}

=> x = y [ x,y R_{+} ]

=> f is one-one

Let y = x^{2} + 4 where y

=> x^{2} = y – 4 4 [ y

=> x

Therefore, for any y , there exists x = =

=> f is onto

Since, f is both one-one and onto, f^{-1} exists for every ,

=> f^{-1}(y) =

**Question 9. **

**Consider R**_{+ }**-> [ -5, ****) given by f (x) = 9x**^{2}** + 6x -5. Show that f is invertible with f**^{-1}** (y) **=

**Solution: **

Let f(x) = f(y)

=> 9x^{2} + 6x -5 = 9y^{2} + 6y – 5

=> 9x^{2} + 6x = 9y^{2} + 6y

=> 9(x^{2} – y^{2}) + 6 (x – y) = 0

=> (x – y) [9 (x + y) + 6] = 0

=> x – y =0

=> x = y

=> f is one-one

Now, let y = 9x^{2} + 6x – 5

=> 9x^{2} + 6x – 5 (x + y) = 0

=> x =

=> f(x) =

On simplifying, we have f (x) = y

=> f is onto

Since f is both one-one and onto. f^{-1} exists

f^{-1}(y) =

**Question 10. **

**Let f : X -> Y be an invertible function. Show that f has unique inverse. **

**Solution: **

We have,

f : X -> Y is an invertible function

Let g and h be two distinct inverses of f.

Then, for all y Y,

fog (y) = I (y) = foh (y)

=> f g (y)) = f(h (y))

=> g(y) = h(y) [f is one-one]

=> g = h [g is one-one]

which contradicts our supposition.

Hence, f has a unique inverse.

**Question 11. **

**Consider f : {1, 2, 3} -> {a, b, c} given by f (1) = a, f (2) = b, f (3) = c. Find f and show that (f**^{-1}**)f**^{-1}** = f.**

**Solution: **

Given that,

f(1) = a, f(2) = b, f(3) = c

We have,

f = {(1, a), (2, b), (c, 3)}

which shows that f is both one-one and onto and thus f is invertible.

Therefore,

f^{-1} = {(a, 1), (b, 2), (c, 3)}

Also,

(f^{-1})^{-1} = {(1, a), (2,b), (3, c)}

=> (f^{-1})^{-1} = f

Hence proved.

**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: **

Since, f is an invertible function,

=> f is both one-one and onto

Also,

Let g : Y -> X , where g is a one-one and onto function such that

gof (x) = I_{x} and fog (y) = I_{y} => g = f^{-1}

=> f^{-1} o (f^{-1})^{-1} = I

=> f o [f^{-1} o (f^{-1})^{-1}] = f o I

=> (f o f^{-1}) o (f^{-1})^{-1} = f

=> I o (f^{-1})^{-1} = f

Hence, (f^{-1})^{-1} = f

**Question 13. If f : R -> R given by f (x) = (3 – x**^{3}**)**^{1/3 }**,then fof (x) is :**

**(A) x**^{1/3 }**(B) x**^{3}** (C) x. (D) (3 – x**^{3}**)**

**Solution: **

**Answer: (C)**

We have,

f(x) = (3 – x^{3})^{1/3} where f : R -> R

Now,

fof(x) = f(f(x))

=> fof(x) = f((3 – x^{3})^{1/3})

=> fof(x) = [3 – ((3 – x^{3})^{1/3} )^{3}]^{1/3}

=> fof(x) = [3 – (3 – x^{3})]^{1/3}

=> fof(x) = (x^{3})^{1/3}

=> fof(x) = x

Hence, option C is correct.

**Question 14.**

Let f : R -{ -> R be a function defined as f(x) = . The

Let f : R -{ -> R be a function defined as f(x) = . The

** inverse of f is the map g : Range f -> R – **{ } given by

**(A) g(y) ** = ** **** (B) g(y) ****= **

**(C) g(y) = **** **** **** (D) g(y) ****= **

**Solution: **

**Answer: (B)**

Let y = f(x)

=> y =

=> 3xy + 4y = 4x

=> x( 4 – 3y) = 4y

=> x =

f^{-1}(y) = g (y) =

#### Chapter-1 (Relation And Functions)

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