Exercise 1.3 Solutions

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Relations and Functions

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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)

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

{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

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.

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}

{\Rightarrow y = 4x + 3}

{\Rightarrow y - 3 = 4x}

{\Rightarrow 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

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}

{\Rightarrow y - 3 = 4x}

{\Rightarrow 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 \geq 0}

(∵ square root of negative numbers is not defined in real numbers)

{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.

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}

{\Rightarrow (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}

{\Rightarrow y \geq -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.

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 \in Y}

fog_1(y)

= IY

{(y) = fog_2(y)}

. Use one-one ness of

f

f

is defined as

f

: X → Y and also as it is given that

f

is invertible.

⇒

f

is both one-one and onto.

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)\}}

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

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

{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 \to Y

be an invertible function. Show that the inverse of

f^{-1}

f

, i.e.,

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

The inverse of

f

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}

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

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)

x^\frac{1}{3}

x^3

x

{(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

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

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

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

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

The given function is

{y = f(x)}

{⇒ f^{-1}(y) = x}

{\Rightarrow 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}

{\Rightarrow 4x = (3x + 4)y}

{\Rightarrow 4x = 3xy + 4y}

{\Rightarrow 4x - 3xy = 4y}

{\Rightarrow x(4 - 3y) = 4y}

{⇒ x = \dfrac{4y}{4 - 3y}}

{⇒ g(y) = \dfrac{4y}{4 - 3y}}

∴ B is the correct answer