NCERT Class 12 Maths Solutions of Relations and Functions – Chapter 1 Ex 3

Sol. The functions f : {1, 3, 4} →{1, 2, 5} and g : {1, 2, 5} →{1, 3} are defined as f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}

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

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

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

gof = {(1, 3), (3, 1), (4, 3)}

Sol. To prove (f + g)oh = foh + goh

Consider

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

= f(h(x)) + g(h(x)) = (foh)(x) + (goh)(x) = {(foh) +(goh)} (x)

((f + g)oh)(x) = {(foh) + (goh)} (x) ∀ x ϵ R

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

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

Consider LHS = ((f.g)oh)(x)

=(f.g)(h(x)) = f(h(x)) . g(h(x)) = (foh)(x). (goh)(x)

= {(foh).(goh)}(x)

((f.g)oh)(x) = {(foh).(goh)}(x) ∀ x ϵ R

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

Sol. (i) f(x) = |x| and g(x) = |5x-2|

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

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

(ii) f(x) = 8x³ and g(x) = x^1/3

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

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

Sol. (i) Function f : {1, 2, 3, 4} → {10} defined as f = {(1, 10), (2, 10), (3, 10), (4, 10)}

From the given definition of f, we can see that f is a many one.

Function as f(1) = f(2) = f(3) = f(4) = 10. Therefore, f is not one-one. Hence, function f does not have an inverse.

(ii) Function g : {5, 6, 7, 8}→{1, 2, 3, 4} defined as g = {(5, 4), (6, 3), (7, 4), (8, 2)}

From the given definition of g, it is seen that g is a many one.

Function as g(5) = g(7) = 4. Therefore, g is not one-one. Hence, function g does not have an inverse.

(iii) Function h : {2, 3, 4, 5}→{7, 9, 11, 13} defined as

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

It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h. Therefore, function h is one-one.

Also, h is onto, since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5} have distinct images under h. Thus, h is a one-one and onto function. Hence, h has an inverse.

Sol. Function f : [-1, 1] → R is given f(x) =x/(x+2)

Let f(x) = f(y)

→x/(x+2)=y/(y+2) → xy + 2x = xy + 2y

→ 2x = 2y → x = y

Therefore, f is a one-one function.

Let y = x/(x+2) → x = xy + 2y → x = 2y/1-y

So, for every y except 1 in the range there exists x in the domain such that

f(x) = y. Hence, function f is onto.

Therefore, f : [-1, 1] → Range f is one-one and onto and therefore, the inverse of the function f : [-1, 1] →Range f exists.

Let y be an arbitrary element of range f.

Since, f : [-1, 1] → Range f is onto, we have y = f(x) for some x ϵ[-1, 1]

→ y =x/(x+2) → xy + 2y = x

→ x(1-y) = 2y → x =2y/1-y, y ≠ 1

Now, let us define g : Range f → [-1, 1] as g(y) =2y/1-y, y ≠ 1

Therefore, gof = fog = I_R,   Therefore, f^-1 = g

Therefore, f^-1(y) = 2y/1-y, y ≠ 1

Sol. Here, f : R→R is given by f(x) = 4x+3

Let x, y ϵ R, such that

f(x) = f(y) → 4x+3 = 4y+ 3

→ 4x = 4y → x = y

Therefore, f is a one-one function

Let y = 4x + 3

→ There exist, x =(y-3/4) ϵ R, ∀ y ϵ R

Therefore, for any y ϵ R, there exist x =y-3/4 ϵ R, such that

f(x) = f(y-3/4) = 4(y-3/4) + 3 = y

Therefore, f is onto function.

Thus, f is one-one and onto and therefore, f^-1 exists.

Let us define g : R →R by g(x) =(y-3/4)

Now, (gof)(x) = g(f(x)) = g(4x + 3) =(4x+3)-3/4) = x

(fog)(y) = f(g(y)) = f(y-3/4) = 4(y-3/4)+ 3 = y – 3 + 3 = y

Therefore, gof = fog = I_R

Hence, f is invertible and the inverse of f is given by f^-1(y) = g(y) =(y-3/4)

Sol. Let f : X → Y be an invertible function

Also, suppose f has two inverses (say g₁ and g₂)

Then, for all y ϵ Y, we have

→ fog₁(y) = I_y(y) = fog₂(y) → f(g₁(y)) = f(g₂(y))

→ g₁(y) = g₂(y)                                   [f is invertible → f is one-one]

→ g₁ = g₂                                            [g is one-one]

Hence, f has a unique inverse.

Sol. Here, function f :{1, 2, 3}→{a, b, c} is given by

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

If we define g : {a, b, c}→{1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have

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

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

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

and (gof)(1) = g(f(1)) = f(a) = 1

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

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

Therefore, gof = I_X and fog = I_Y, where X = {1, 2, 3} and Y = {a, b, c}

Thus, the inverse of f exists and f^-1 = g

Therefore, f^-1 : {a, b, c}→{1, 2, 3} is given by,

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

Let us now find the inverse of f^-1  i.e., find the inverse of g.

If we define h : {1, 2, 3}→{a, b, c} as

h(1) = a, h(2) = b, h(3) = c, then we have

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

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

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

and (hog)(a) = h(g(a)) = h(1) = a

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

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

Therefore, goh = I_X and hog = I_Y, where X = {1, 2, 3} and Y = {a, b, c}

Thus, the inverse of g exists and g^-1 = h → (f^-1)^-1 = h

It can be noted that h = f. Hence, (f^-1)^-1 = f.

Sol. Let f : X→Y be an invertible function

Then, there exists a function g : Y →X such that gof = I_X and fog = I_Y

Here, f^-1 = g

Now, gof = I_X and fog = I_Y

Therefore, f^-1 of = I_X and fof^-1 = I_Y

Hence, f^-1 : Y→X is invertible and f is the inverse of f^-1

i.e., (f^-1)^-1 = f

Sol. Here, function f : R→R is given as f(x) = (3 – x³)^1/3

fof(x) = f(f(x)) = f((3 – x³)^1/3) = [3 – ((3 – x³)^1/3)³]^1/3

= [3 – ((3 – x³)]^1/3 = (x³)^1/3 = x

fof(x) = x. Thus, the correct answer is option (c)