# NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 1: Relations and Functions (Part IV)

Jun 19, 2017 10:00 IST

NCERT Solutions for CBSE Class 12 Maths, Chapter 1: Relations & Functions are available in this article. Here, you will get solutions to the questions of exercise 1.3 from question number 1 to question number 7. Most of the questions given in this exercise are related to composition of functions and invertible function. These questions are important CBSE Class 12 Maths board exam.

NCERT Solutions for CBSE Class 12th Maths, Chapter 1: Relations and Functions (Exercise 1.3) are given below

Question1: 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 gof.

Solution 1:

The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as

= {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}.

gof (1) = g [f (1)] = g (2) = 3

gof (3) = g [f (3)] = g (5) = 1

gof (4) = g [f (4)] = g (1) = 3

Thus, 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 2:

To prove: (f + g) o h = foh + goh

Proof:

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

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

= f [{h(x)} + g{h(a)}]

= (foh) (x) + (goh) (x)

= {foh + goh} x

Hence, (f + g)oh = foh + goh

To prove: (f.g)oh = (foh) (goh)

Proof:

[(f.g)oh] (x) = (f.g) [h(x)] = f [h(x)].g [h(x)] = (foh) (x). (goh) (x) = [(foh).(goh)] (x)

Hence, (f.g)oh = (foh) (goh)

NCERT Exemplar Class 12 Mathematics – Chapter 1 Relations and Functions

Question 3: Find gof and fog, if

(a) f (x) = | x | and g(x) = | 5x – 2 |

(b) f (x) = 8x3 and g(x) = x1/3.

Solution 3:

(a) f (x) = |x| & g (x) = |5x ‒ 2|

(gof ) (x) = g [f (x)] = g (|x|) = |5 |x| ‒ 2|

(fog) (x) = f [g (x)] = f (|5x ‒ 2|) = ||5x ‒ 2|| = |5x ‒ 2|.

(b) f (x) = 8 x3 & g (x) = x1/3

(gof) (x) = g [f (x)] = g (8x3)1/3 = 2x

(fog) (x) = f [g (x)] = f (x1/3) = 8(x1/3)3 = 8x.

Question 4: If f (x) = (4x + 3)/ (6x ‒ 4), x ≠ 2/3, show that fof (x) = x, for all x ≠ 2/3. What is the inverse of f?

Solution 4:

f (x) = (4x + 3)/(6x ‒ 4), x ≠ 2/3

(fof) (x) = f [f (x)]

Hence, the given function f is invertible and the inverse of f is f itself.

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

Solution 5:

(ii)

(iii)

Question 6: Show that f : [–1, 1] → R, given by f (x) = x/(x + 2) is one-one. Find the inverse of the function of the function f : [–1, 1] → Range f.

[Hint: For y ∈ Range f, y = f (x) = 2/(x + 2), for some x in [–1, 1], i.e., x = 2y/(1 ‒ y)]

Solution 6:

Question 7: Consider f : R → R given by f (x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Solution 7:

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