This page contains the

**CBSE mathematics class 12 chapter Relations and Functions Exercise 1.3 Solutions**. You can find the questions/answers/solutions for the**chapter 1/Exercise 1.3**of**CBSE class 12 mathematics**in this page. So is the case if you are looking for**CBSE class 12 Maths**related topic**Relations and Functions**. This page contains Exercise 1.3 solutions. If you’re looking for summary of the chapter or other exercise solutions, they’re available at●

●

●

●

●

Exercise 1.3

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, 3), (5, 1)}. Write down gof.

It is given that f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3}

So, gof : {1, 3, 4} → {1, 3}

The function f is given as

f(1) = 2

f(3) = 5

f(4) = 1

This can be diagramatically represented as follows:

The function g is given as

g(1) = 3

g(2) = 3

g(5) = 1

This can be diagramatically represented as follows:

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

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

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

This can be diagramatically represented as follows:

∴ gof = {(1,3), (3, 1), (4, 3)}

2. Let f, g and h be functions from R to R. Show that

(f + g)oh = foh + goh

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

(f + g)oh = foh + goh

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

We have

{((f + g)oh)(x)}

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

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

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

{∴ (f + g)oh = foh + goh}

Similarly,

((f.g)oh)(x)

{= (f.g)(h(x))}

{= f(h(x)).g(h(x))}

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

{∴ (f.g)oh = (foh).(goh)}

3. Find gof and fog, if

i.

{f(x) = |x|} and {g(x) = |5x - 2|}

ii.

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

To find gof and fog when {f(x) = |x|} and {g(x) = |5x - 2|}

We have,

gof(x)

{= g(f(x))}

{= g(|x|)} {(∵ f(x) = |x|)}

{= |5|x| - 2|} {(∵ g(x) = |5x - 2|)}

{∴ gof(x) = |5|x| - 2|}

Similarly,

fog(x)

{= f(g(x))}

{= f(|5x - 2|)} {(∵ g(x) = |5x - 2|)}

{= ||5x - 2||} {(∵ f(x) = |x|)}

{= |5x - 2|}

To find gof and fog when {f(x) = 8x^3} and {g(x) = x^{\frac{1}{3}}}

We have,

gof(x)

{= g(f(x))}

{= g(8x^3)} {(∵ f(x) = 8x^3)}

{= \left(8x^3\right)^\frac{1}{3}} {(∵ g(x) = x^\frac{1}{3})}

{= \left(\left(2x\right)^3\right)^\frac{1}{3}}

{= \left(2x\right)^{3\times\frac{1}{3}}} {(∵ \left(a^m\right)^n = a^{mn}})

{= 2x}

{∴ gof(x) = 2x}

Similarly,

fog(x)

{= f(g(x))}

{= f(x^\frac{1}{3})} {(∵ g(x) = x^\frac{1}{3})}

{= 8\left(x^\frac{1}{3}\right)^3} {(∵ f(x) = 8x^3)}

{= 8x^{\frac{1}{3}\times3}} {(∵ \left(a^m\right)^n = a^{mn})}

{= 8x}

{∴ fog(x) = 8x}

4. If {f(x) = \dfrac{(4x + 3)}{(6x - 4)}}, x ≠ \dfrac{2}{3}, show that {fof = x}, for all x ≠ \dfrac{2}{3}. What is the inverse of f?

We have,

fof(x)

{= f(f(x))}

{= f\left(\dfrac{4x - 3}{6x - 4}\right)} {\left(∵ f(x) = \dfrac{4x - 3}{6x - 4}\right)}

{= \dfrac{4\left(\dfrac{4x - 3}{6x - 4}\right) - 3}{6\left(\dfrac{4x - 3}{6x - 4}\right) - 4}} {\left(∵ f(x) = \dfrac{4x - 3}{6x - 4}\right)}

{= \dfrac{\dfrac{4(4x - 3) - 3(6x - 4)}{6x - 4}}{\dfrac{6(4x - 3) - 4(6x - 4)}{6x - 4}}}

{= \dfrac{16x - 12 - 18x + 12}{24x - 18 - 24x + 16}} (After cancelling/cutting {6x - 4} in both numerator and denominator)

{= \dfrac{-2x}{-2}}

= x

{∴ fof(x) = x}

{∵ fof(x) = x}, the inverse of f is f itself.

∴ We can say that {f^{-1} = f}

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)}

As we know, a function f will have an inverse only if the function is both one-one and onto.

Let’s now check one by one whether the given functions f, g and h are both one-one and onto.

To check whether f is one-one

In the function f, we have

{f(1) = 10}

{f(2) = 10}

{f(3) = 10} and

{f(4) = 10}

This can be diagramatically represented as follows:

As different elements in the domain have the same image in the co-domain, f is not one-one

∴ f does not have an inverse (as f is not one-one, the check for f being an onto is not required)

To check whether g is one-one

In the function g, we have

{g(5) = 4}

{g(6) = 3}

{g(7) = 4}

{g(8) = 2}

This can be diagramatically represented as follows:

As two different elements 5 and 7 in the domain have the same image 4 in the co-domain, g is not one-one.

∴ g does not have an inverse (as g is not one-one, the check for g being an onto is not required)

To check whether h is one-one

In the function h, we have

{h(2) = 7}

{h(3) = 9}

{h(4) = 11}

{h(5) = 13}

This can be diagramatically represented as follows:

Clearly different elements in the domain have the different images in the co-domain.

∴ h is one-one

To check whether h is onto

As every element in the co-domain {7, 9, 11, 13} has an inverse-image in the domain, h is also onto.

∴ h is both one-one and onto.

∴ h will have an inverse and the inverse of h will be

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

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

(Hint: For y ∈ Range f, {y = f(x) = \dfrac{x}{x + 2}}, for some x in [-1, 1], i.e., {x = \dfrac{2}{2 - y}})

The function f is given by

{f(x) = \dfrac{x}{2 + x}}

This can also be written as

{f(x) = \dfrac{x + 2 - 2}{x + 2} = 1 - \dfrac{2}{x + 2}}

To Check whether f is one-one

Consider the two elements in the domain x_1, x_2 ∈ [-1, 1] such that

{f(x_1) = f(x_2)}

{⇒ 1 - \dfrac{2}{x_1 + 2} = 1 - \dfrac{2}{x_2 + 2}}

{⇒ -\dfrac{2}{x_1 + 2} = -\dfrac{2}{x_2 + 2}} (After cutting/cancelling 1 on both sides)

{⇒ \dfrac{1}{x_1 + 2} = \dfrac{1}{x_2 + 2}} (After cutting/cancelling 2 on both sides)

{⇒ x_1 + 2 = x_2 + 2} (After inversing both sides)

{⇒ x_1 = x_2} (After cutting/cancelling 2 on both sides)

∴ f is one-one

To Check whether f is onto

We have {y = f(x) = 1 - \dfrac{2}{x + 2}}

{⇒ \dfrac{2}{x + 2} = 1 - y}

{⇒ \dfrac{x + 2}{2} = \dfrac{1}{1 - y}}

{⇒ x + 2 = \dfrac{2}{1 - y}}

{⇒ x = \dfrac{2}{y - 1} - 2}

{⇒ x = \dfrac{2 - 2(1 - y)}{1 - y}}

{⇒ x = \dfrac{2y}{1 - y}}

As the above equation is valid for all values of y except 1 (∵ when y = 1, 1 – y = 0 and division by 0 is not defined.

Also, as y is defined as {y = \dfrac{x}{x + 2}}, y can never be 1 {(∵ x} ≠ {x + 2)}

⇒ Range of f never includes 1.

Thus for every y ∈ Range of f, there exists an inverse image in the domain.

∴ f is onto.

As f is both one-one and onto f is invertible and is given as {f^{-1}(y) = \dfrac{2y}{1 - y}}

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

Method 1: To show that f is invertible by showing that f is both one-one and onto.

For a function f to be invertible, it should be both one-one and onto.

To check whether f is one-one

The function f is given as {f(x) = 4x + 3}

Consider two elements x_1, x_2 ∈ R such that

{f(x_1) = f(x_2)}

{⇒ 4x_1 + 3 = 4x_2 + 3}

{⇒ 4x_1 = 4x_2} (After cutting/cancelling 3 on both sides)

{⇒ x_1 = x_2} (After cutting/cancelling 4 on both sides)

∴ f is one-one

To check whether f is onto

The function f is given as {y = f(x) = 4x + 3}

{⇒ y = 4x + 3}

{⇒ y - 3 = 4x}

{⇒ 4x = y - 3}

{⇒ x = \dfrac{y - 3}{4}}

The above expression is valid for all real values of y.

In otherwords, every element y ∈ R will be an image of an element x in the domain R

∴ f is onto

As f is both one-one and onto, f is invertible and is given as

{f^{-1}(x) = \dfrac{y - 3}{4}}

Method 2: To show that f is invertible by showing that fog = gof = IR by defining the inverse function g

Let y be an arbitrary element in the range of f.

⇒ For some x ∈ R, {y = 4x + 3}

{⇒ y - 3 = 4x}

{⇒ 4x = y - 3}

{⇒ x = \dfrac{y - 3}{4}}

Let’s now define g : R → R given as {g(y) = \dfrac{y - 3}{4}}

Now,

{gof(x) = g(f(x))}

{= g(4x + 3)}

{= \dfrac{(4x + 3) - 3}{4}}

{= \dfrac{4x}{4}}

{= x}

Also,

{fog(y) = f(g(y))}

{= f\left(\dfrac{y - 3}{4}\right)}

{= 4\times\left(\dfrac{y - 3}{4}\right) + 3}

{= y - 3 + 3} (After cancelling/cutting 4 in the numerator and denominator)

{= y}

Hence gof = IR and also fog = IR

⇒ f is invertible and is given as {f^{-1} = g}

8. Consider f : R+ → [4, ∞) given by {f(x) = x² + 4}. Show that f is invertible with the inverse {f^{-1}} of f given by {f^{-1}(y) = \sqrt{y - 4}}, where R+ is the set of all non-negative real numbers.

Method 1: To show that f is invertible by showing that f is both one-one and onto.

For a function f to be invertible, it should be both one-one and onto.

To check whether f is one-one

The function f is given as {f(x) = x^2 + 4}

Consider two elements x_1, x_2 ∈ R+ in the domain such that

{f(x_1) = f(x_2)}

{⇒ x_1^2 + 4 = x_2^2 + 4}

{⇒ x_1^2 = x_2^2} (After cutting/cancelling 4 on both sides)

{⇒ x_1 = \pm x_2}

As it is given that the domain R+ is the set of non-nagative real numbers, we can ignore the negative values of x

∴ f is one-one

To check whether f is onto

The function f is given as {y = f(x) = x^2 + 4}

{⇒ x^2 + 4 = y}

{⇒ x^2 = y - 4}

{⇒ x = \sqrt{y - 4}}

The above equation is valid for all {y - 4 ≥ 0} (∵ square root of negative numbers is not defined in real numbers)

or {y \geq 4}

It is given that the co-domain of f is [4, ∞)

Every y ∈ [4, ∞) in the co-domain will be an image of an element x ∈ R+ in the domain.

∴ f is onto.

As f is both one-one and onto, f is invertible and the inverse of f is given as

{f^{-1}(y) = \sqrt{y - 4}}

Method 2: To show that f is invertible by showing that {fof^{-1} = f^{-1}of} = IR+ by using the inverse function f^{-1}

Let y be an arbitrary element in the range of f.

For some x ∈ R+, {y = x^2 + 4}

{⇒ y - 4 = x^2}

{⇒ x^2 = y - 4}

{⇒ x = \sqrt{y - 4}}

Let’s now define g : [4, ∞) → R+ given as {g(y) = \sqrt{y - 4}}

Now

{gof(x) = g(f(x))}

{= g(x² + 4)}

{= \sqrt{x^2 + 4 - 4}} {(∵ g(y) = \sqrt{y - 4})}

{= \sqrt{x^2}}

{= x}

Also,

{fog(y) = f(g(y))}

{= f(\sqrt{y - 4})}

{= \left({\sqrt{y - 4}}\right)^2 + 4} {(∵ f(x) = x²)}

{= y - 4 + 4}

{= y}

Hence gof = IR+ and also fog = IR+

⇒ f is invertible and is given as {f^{-1} = g}

9. Consider f : R+ → [-5, ∞) given by {f(x) = 9x^2 + 6x - 5}. Show that f is invertible with {f^{-1}(y) = \left(\dfrac{\left(\sqrt{y + 6}\right) - 1}{3}\right)}.

Method 1: To show that f is invertible by showing that f is both one-one and onto.

For a function f to be invertible, it should be both one-one and onto.

To check whether f is one-one

The function f is given as {f(x) = 9x^2 + 6x - 5}

This function can be written as {y = f(x) = 9x² + 6x + -5}

Consider two elements x_1, x_2 ∈ R+, in the domain R+ such that

{f(x_1) = f(x_2)}

{⇒ 9x_1^2 + 6x_1 - 5 = 9x_2^2 + 6x_2 - 5}

{⇒ 9x_1^2 + 6x_1 = 9x_2^2 + 6x_2} (After cutting/cancelling -5 on both sides)

{⇒ 9x_1^2 - 9x_2^2 + 6x_1 - 6x_2 = 0}

{⇒ 9(x_1^2 - x_2^2) + 6(x_1 - x_2) = 0}

{⇒ 9(x_1 + x_2)(x_1 - x_2) + 6(x_1 - x_2) = 0} {(∵ a^2 - b^2 = (a + b)(a - b))}

{⇒ 3(x_1 - x_2)[3(x_1 + x_2 + 2)] = 0}

{x_1 - x_2 = 0} (∵ x ∈ R+ and hence {3x_1 + 3x_2 + 2} being sum of 2 positive real numbers and 2 can never be 0)

{⇒ x_1 = x_2}

∴ f is one-one

To check whether f is onto:

The function f is given as {y = f(x) = 9x^2 + 6x - 5}

{⇒ 9x^2 + 6x - 5 = y}

{⇒ 9x^2 + 6x + 1 - 6 = y} (so that we can use the identity {a^2 + 2ab + b^2 = (a + b)^2})

{⇒ (3x)^2 + 2×(3x)×1 + 1^2 - 6 = y}

{⇒ (3x + 1)² - 6 = y} {(∵ a^2 + 2ab + b^2 = (a + b)^2)}

{⇒ (3x + 1)^2 = y + 6}

{⇒ (3x + 1) = \sqrt{y + 6}}

{⇒ 3x = \left(\sqrt{y + 6}\right) - 1}

⇒ x = \dfrac{\left(\sqrt{y + 6}\right) - 1}{3}

The above equation is valid when {y + 6 \geq 0}

{⇒ y ≥ -6}

As the co-domain of f is given as [-5, ∞), every element in y ∈ [-5, ∞) will have an inverse image in the domain.

∴ f is onto.

As f is both one-one and onto, f is invertible and the inverse of f is given as

{f^{-1}(y)\left(\dfrac{\left(\sqrt{y + 6}\right) - 1}{3}\right)}

Method 2: To show that f is invertible by using the inverse function {f{-1}}

Let y be an arbitrary element in the range of f.

For some x ∈ R+, {y = 9x^2 + 6x - 5}

{⇒ y = (3x)^2 + 2×(3x)×1 + 1^2 - 6}

{⇒ y + 6 = (3x + 1)^2}

{⇒ (3x + 1)^2 = y + 6}

{⇒ 3x + 1 = \sqrt{y + 6}}

{⇒ 3x = \sqrt{y + 6} - 1}

{⇒ x = \dfrac{\sqrt{y + 6} - 1}{3}}

Let’s now define g : [-6, ∞) → R+, given as

{g(y) = \dfrac{\sqrt{y + 6} - 1}{3}}

Now

{gof(x) = g(f(x))}

{= g(9x^2 + 6x - 5)} {(∵ f(x) = 9x² + 6x - 5)}

{= \dfrac{\sqrt{9x² + 6x - 5 + 6} - 1}{3}} {(∵ g(y) = \dfrac{\sqrt{y + 6} - 1}{3})}

{= \dfrac{\sqrt{9x² + 6x + 1} - 1}{3}}

{= \dfrac{\sqrt{(3x)² + 2 × 3x × 1 + 1²} - 1}{3}}

{= \dfrac{\sqrt{(3x + 1)²} - 1}{3}}

{= \dfrac{3x + 1 - 1}{3}}

{= \dfrac{3x}{3}}

{= x}

Also,

{fog(y) = f(g(y))}

{= f\left(\dfrac{\sqrt{y + 6} - 1}{3}\right)} {(∵ g(y) = \dfrac{\sqrt{y + 6} - 1}{3}}

{= 9\left(\dfrac{\sqrt{y + 6} - 1}{3}\right)^2 + 6\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) - 5} {(∵ f(x) = 9x^2 + 6x - 5)}

{= {9\left(\dfrac{\left(\sqrt{y + 6}\right)^2 + 2\times\sqrt{y + 6}\times1 + 1^2}{3^2}\right)} + 6\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) - 5}

{= {9\left(\dfrac{y + 6 + 2\sqrt{y + 6} + 1}{9}\right)} + 6\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) - 5}

{= y + 7 - 2\sqrt{y + 6} + 6\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) - 5} (After cutting/cancelling 9 in the numerator and denominator)

{= \dfrac{3y + 21 - 6\sqrt{y + 6} + 6\sqrt{y + 6} - 6 - 15}{3}}

{= \dfrac{3y}{3}}

{= y}

Hence gof = IR+ and fog = I[-5, ∞)

∴ f is convertible and is given as {f^{-1} = g}

10. Let f : X → Y be an invertible function. Show that f has unique inverse.

(Hint: suppose g_1 and g_2 are two inverses of f. Then for all {y ∈ Y}, fog_1(y) = IY{(y) = fog_2(y)}. Use one-one ness of f).

As f is defined as f : X → Y and also as it is given that f is invertible.

⇒ f is both one-one and onto.

As f is one-one, if there exists any two elements x_1, x_2 ∈ X such that

{f(x_1) = f(x_2)} then {x_1 = x_2}

Let’s now assume that the invertible function f has two different inverses g_1 and g_2 defined as g_1 : Y → X and g_2 : Y → X

{⇒ fog_1(y) = I_Y} and {fog_2(y) = I_Y}

{⇒ fog_1(y) = fog_2(y)}

{⇒ f(g_1(y)) = f(g_2(y))}

{⇒ g_1(y) = g_2(y)} {(∵ f(x_1) = f(x_2) ⇒ x_1 = x_2)}

{⇒ g_1 = g_2}

∴ f has unique inverse.

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}

It is given that

{f(1) = a} {⇒ f^{-1}(a) = 1}

{f(2) = b} {⇒ f^{-1}(b) = 2}

{f(3) = c} {⇒ f^{-1}(c) = 3}

This can be diagramatically represented as follows:

∴ f^{-1} is defined as {f^{-1} : \{a, b, c\}} → {1, 2, 3} and {f^{-1} = \{(a, 1), (b, 2), (c, 3)\}}

As {f^{-1}(a) = 1 ⇒ (f^{-1})^{-1}(1) = a}

As {f^{-1}(b) = 2 ⇒ (f^{-1})^{-1}(2) = b}

As {f^{-1}(c) = 3 ⇒ (f^{-1})^{-1}(3) = c}

And this can be diagramatically represented as follows:

∴ {(f^{-1})^{-1}} is defined as {(f^{-1})^{-1} = \{1, 2, 3\} → \{a, b, c\}} and {(f^{-1})^{-1} = \{(1, a), (2, b), (3, c)\}}

∴ From the above, it is clear that {(f^{-1})^{-1} = f}

12. Let f : X → Y be an invertible function. Show that the inverse of f^{-1} is f, i.e., {(f^{-1})^{-1} = f}.

The inverse of f is f^{-1}

It is given that f is invertible

⇒ f is both one-one and onto.

Consider an arbitrary elemnet x ∈ X, such that {f(x) = y}

{⇒ f^{-1}(y) = x}

{⇒ (f^{-1})^{-1}(x) = y}

But we know that {f(x) = y}

So {(f^{-1})^{-1}(x) = f(x)}

As x is an arbitrary element in X, the above statement is true for all x ∈ X

{∴ (f^{-1})^{-1} = f}

13. If f : R → R be given by {f(x) = \left(3 - x^3\right)^{\frac{1}{3}}}, then fof(x) is

A.

x^\frac{1}{3}

B.

x^3

C.

x

D.

{(3 - x^3)}

{fof(x) = f(f(x))}

{= f\left(\left(3 - x^3\right)^{\frac{1}{3}}\right)}

{= \left(3 - \left(\left(3 - x^3\right)^\frac{1}{3}\right)^3\right)^\frac{1}{3}}

{= \left(3 - \left(3 - x^3\right)^{3\times\frac{1}{3}}\right)^\frac{1}{3}}

{= \left(3 - \left(3 - x^3\right)^1\right)^\frac{1}{3}}

{= \left(3 - \left(3 - x^3\right)\right)^\frac{1}{3}}

{= \left(3 - 3 + x^3\right)^\frac{1}{3}}

{= \left(x^3\right)^\frac{1}{3}}

= x

∴ C is the correct answer.

14. Let f : R – \Bigl\{\dfrac{4}{3}\Bigr\} → R be a function defined as {f(x) = \dfrac{4x}{3x + 4}}. The inverse of f is the map g : Range f → R {- \Bigl\{\dfrac{4}{3}\Bigr\}} given by

A.

{g(y) = \dfrac{3y}{3 - 4y}}

B.

{g(y) = \dfrac{4y}{4 - 3y}}

C.

{g(y) = \dfrac{4y}{3 - 4y}}

D.

{g(y) = \dfrac{3y}{4 - 3y}}

The given function is {y = f(x)}

{⇒ f^{-1}(y) = x}

{⇒ g(y) = x} (∵ The inverse of f is given as g)

Also, it is given that {y = f(x) = \dfrac{4x}{3x + 4}}

{⇒ \dfrac{4x}{3x + 4} = y}

{⇒ 4x = (3x + 4)y}

{⇒ 4x = 3xy + 4y}

{⇒ 4x - 3xy = 4y}

{⇒ x(4 - 3y) = 4y}

{⇒ x = \dfrac{4y}{4 - 3y}}

{⇒ g(y) = \dfrac{4y}{4 - 3y}}

∴ B is the correct answer