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

Jun 20, 2017 14:00 IST

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

    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:

    NCERT Solutions for CBSE Class 12th Maths, Chapter 1: Relations and Functions, Exercise 1.3, Question 8

    Solution 8:

    NCERT Solutions for CBSE Class 12th Maths, Chapter 1: Relations and Functions, Exercise 1.3, Solution 8

    NCERT Exemplar Class 12 Mathematics – Chapter 1 Relations and Functions

    Question 9:

    NCERT Solutions for CBSE Class 12th Maths, Chapter 1: Relations and Functions, Exercise 1.3, Question 9

    Solution 9:

    NCERT Solutions for CBSE Class 12th Maths, Chapter 1: Relations and Functions, Exercise 1.3, 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:

    NCERT Solutions for CBSE Class 12th Maths, Chapter 1: Relations and Functions, Exercise 1.3, 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:

    NCERT Solutions for CBSE Class 12th Maths, Chapter 1: Relations and Functions, Exercise 1.3, Solution 13

    Question 14:

    NCERT Solutions for CBSE Class 12th Maths, Chapter 1: Relations and Functions, Exercise 1.3, Solution 14

    Solution 14:

    NCERT Solutions for CBSE Class 12th Maths, Chapter 1: Relations and Functions, Exercise 1.3, Solution 14

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

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