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

NCERT Solutions for CBSE Class 12 Maths - Chapter 1: Relations and Functions are available here. From this article, you will find solutions of exercise 1.3 from question number 8 to question number 14. Solutions of other exercises are available in further parts. All the questions given in NCERT textbook are very important for CBSE Class 12 Maths board exam. 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 8 to question number 14. These questions are related to composition of functions and invertible function.

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

Question8: Solution 8: NCERT Exemplar Class 12 Mathematics – Chapter 1 Relations and Functions

Question 9: Solution 9: NCERT Solutions for CBSE Class 12 Physics: All Chapters

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) = IY(y) = fog2(y). Use one-one ness of f).

Solution 10:

Let Y is an invertible function

Suppose g1 & g2 are two inverses of f

Now, for all y ϵ Y,

fog1 (y) = fog2 (y)

f[g1(y)] = f[g2(y)]

g1(y) = g2(y) {f is one-one}

g1 = g2 (g is one-one)

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)–1 = f.

Solution 11:

f: {1, 2, 3} → {abc}

f(1) = a

f(2) = b

f(3) = c

Let g: {abc} → {1, 2, 3}

g(a) = 1

g(b) = 2

g(c) = 3

Now,

(fog) (a) = f [g(a)] = f (1) = a

(fog) (b) = f [g(b)] = f(2) = b

(fog) (c) = f [g(c)] = f (3) = c

(gof) (1) = g [f(1)] = g(a) =1

(gof) (2) = g [f(2)] = g(b) =2

(gof) (3) = g [f(3)] = g(c) =3

Thus, gof = fog

gof  = IX

fog = IY

Where X = {1, 2, 3} and Y= {abc}

Thus, f−1 = g

Hence, f−1: {abc} → {1, 2, 3}

f−1(a) = 1

f−1(b) = 2

f-1(c) = 3

Let us calculate the inverse of f−1

h: {1, 2, 3} → {abc}

h(1) = ah(2) = bh(3) = c

Then,

(goh) (1) = g[h(1)] = g(a) = 1

(goh) (2) = g[h(2)] = g(b) = 2

(goh) (3) = g[h(3)] = g(c) = 3

(hog)(a) = h [g(a)]= h (1) = a

(hog)(b) = h [g(b)]= h (2) = b

(hog)(c) = h [g(c)]= h (3) = c

goh = IX

hog = IY

Now,

g−1 = h

(f−1)−1 = h

We can observe that

h = f

Hence, (f‒1)‒1 = f.

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 12: Question 13: If f : R → R be given by f (x) = (3 − x3 )1/3 , then fof (x) is

(A) x3

(B) x3

(C) x

(D) (3 – x3).

Solution 13: Question 14: Solution 14: Download NCERT Solutions for Class 12 Maths: Chapter 1 Relations and Functions in PDF format