# Exercise 1.3 Solutions

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)
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 yR 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_2R+ 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 xR+ 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 xR+, {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_2R+, 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} (∵ xR+ 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 xR+, {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_2X 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 xX, 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 xX
{∴ (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
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}}