RD Sharma Class 12 Exercise 2.2 Functions Solutions Maths

RD Sharma Class 12 Exercise 2.2 Function Solutions Maths - Download PDF Free Online

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The RD Sharma Solutions for Class 12 Mathematics help the students to understand and work out the problems efficiently. It is essential to possess the right guide while working on homework or assignments, especially on complex topics like the RD Sharma Class 12th Exercise 2.2. This single compilation is adequate for the CBSE students who are preparing for their board exams.

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RD Sharma Class 12 Solutions Chapter 2 Functions - Other Exercise

Functions Exercise 2.2 Question 1 (i) .

Answer :

g o f (x) = 4 x^{2} + 12 x + 14 and f o g = 2 x^{2} + 13

Hint :

g o f

means

f (x)

is in

g (x)

function

f o g

means

g (x)

function is in

f (x)

function
Given :

f : R \to R and g : R \to R

f o g : R \to R and g o f : R \to R

f (x) = 2 x + 3, g (x) = x^{2} + 5

Solution :
Now first we find

g o f

g o f (x) = g (2 x + 3)

= {(2 x + 3)}^{2} + 5

g o f (x) = 4 x^{2} + 12 x + 9 + 5

∵ [{(a + b)}^{2} = a^{2} + b^{2} + 2 a b]

g o f (x) = 4 x^{2} + 12 x + 14

Now we find

f o g

f o g (x) = f (g (x))

f (x^{2} + 5) = 2 (x^{2} + 5) + 3

f o g (x) = 2 x^{2} + 13

Hence,

f o g (x) = 2 x^{2} + 13 and g o f (x) = 4 x^{2} + 12 x + 14

Then we find

f o g

f o g (x) = f (g (x))

= f (x^{3})

f o g (x) = 2 x^{3} + x^{6}

Hence,

f o g = 2 x^{3} + x^{6}

and

g o f = {(2 x + x^{2})}^{3}

Functions Exercise 2.2 Question 1 (ii) .

Answer :

f o g = 2 x^{3} + x^{6} and g o f = {(2 x + x^{2})}^{3}

Hint :

g o f

means

f (x)

function is in

g (x)

function

f o g

means

g (x)

function is in

f (x)

funcion
Given :

f (x) = 2 x + x^{2}

g (x) = x^{3}

f : R \to R and g : R \to R

f o g : R \to R and g o f : R \to R

Solution :
First, we find

g o f

g o f (x) = g (f (x))

= g (2 x + x^{2})

g o f (x) = {(2 x + x^{2})}^{3}

Then we find

f o g

f o g (x) = f (g (x))

= f (x^{3})

f o g (x) = 2 x^{3} + x^{6}

Hence,

f o g = 2 x^{3} + x^{6}

and

g o f = {(2 x + x^{2})}^{3}

Functions Exercise 2.2 Question 1 (iii) .

Answer :

g o f = 3 {(x^{2} + 8)}^{3} + 1 and f o g = 9 x^{6} + 6 x^{3} + 9

Hint :

g o f

means

f (x)

function is in

g (x)

function

f o g

means

g (x)

function is in

f (x)

function
Given :

f (x) = x^{2} + 8

g (x) = 3 x^{3} + 1

f : R \to R and g : R \to R

f o g : R \to R and g o f : R \to R

Solution :
First, we find

g o f

Thus,

g o f (x) = g [f (x)]

g o f (x) = g [x^{2} + 8]

g o f (x) = 3 {[x^{2} + 8]}^{3} + 1

Similarly,

f o g (x) = f [g (x)]

f o g (x) = f (3 x^{3} + 1)

f o g (x) = {[3 x^{3} + 1]}^{2} + 8

f o g (x) = {[9 x^{6} + 1 + 6 x^{3}]}^{2} + 8

∵ [{(a + b)}^{2} = a^{2} + b^{2} + 2 a b]

f o g (x) = 9 x^{6} + 1 + 6 x^{3} + 9

Hence,

f o g = 9 x^{6} + 1 + 6 x^{3} + 9 and g o f = 3 {(x^{2} + 8)}^{3} + 1

Functions Exercise 2.2 Question 1 (iv) .

Answer :

f o g = | x | and g o f = | x |

Hint :

g o f

means

f (x)

function is in

g (x)

function

f o g

means

g (x)

function is in

f (x)

function
Given :

f (x) = x

g (x) = | x |

Solution :
Since

f : R \to R and g : R \to R

f o g : R \to R and g o f : R \to R

Now,

g o f (x) = g (f (x))

= g (x)

g o f (x) = | x |

Similarly ,

f o f (x) = f (g (x))

= f | x |

f o g (x) = | x |

Hence,

f o g = | x | and g o f = | x |

Functions Exercise 2.2 Question 1 (v) .

Answer :

f o g = 9 x^{2} - 18 x + 5

f o g = 9 x^{2} - 18 x + 5 and g o f = 3 x^{2} + 6 x - 13

Hint :

g o f

means

f (x)

function is in

g (x)

function

f o g

means

g (x)

function is in

f (x)

function
Given :

f (x) = x^{2} + 2 x - 3

g (x) = 3 x - 4

Solution :
Since

f : R \to R and g : R \to R

f o g : R \to R and g o f : R \to R

Now,

g o f (x) = g (f (x)) = g (x)

g o f (x) = g (f (x)) = g (x^{2} + 2 x - 3)

g o f (x) = 3 (x^{2} + 2 x - 3) - 4

g o f (x) = 3 x^{2} + 6 x - 9 - 4

g o f (x) = 3 x^{2} + 6 x - 13

f o g (x) = f (g (x)) = f (x)

f o g (x) = f (g (x)) = f (3 x - 4)

f o g (x) = {(3 x - 4)}^{2} + 2 (3 x - 4) - 3

= 9 x^{2} + 16 - 24 x + 6 x - 8 - 3

∵ [{(a - b)}^{2} = a^{2} + b^{2} - 2 a b]

f o g (x) = 9 x^{2} - 18 x + 5

Hence,

f o g = 9 x^{2} - 18 x + 5 and g o f = 3 x^{2} + 6 x - 13

Functions Exercise 2.2 Question 1 (vi).
Hint :

g o f

means

f (x)

function is in

g (x)

function

f o g

means

g (x)

function is in

f (x)

function
Given :

f (x) = 8 x^{3}

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

Solution :
Since

f : R \to R and g : R \to R

f o g : R \to R and g o f : R \to R

Now,

g o f (x) = g (f (x)) = g (8 x^{3})

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

= {(8)}^{\frac{1}{3}} = {(2^{3})}^{\frac{1}{3}}

∵ [as we know, {(a^{m})}^{\frac{1}{n}} = {(a)}^{\frac{m}{n}}]

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

g o f (x) = 2 x

Similarly,

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

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

f o g (x) = 8 x

∵ [(a^{m})^{\frac{1}{n}} = (a)^{\frac{m}{n}}]

Hence,

f o g = 8 x and g o f = 2 x

Functions Exercise 2.2 Question 2

Answer :

f o g = {(1, 1), (3, 1), (4, 3), (5, 3)}

g o f = {(3, 3), (9, 3), (12, 9)}

Hint : The set which contains all the elements of all the ordered pairs of relation R is known as the domain of the relation. The set which contains all the second element, is known as the range of the relation.
Given :

f = {(3, 1), (9, 3), (12, 4)}

$g = {(- 1, - 2), (- 2, - 4), (- 3, - 6), (4, 8)}$
Prove : $g o f$ and $f o g$ are both defined
Solution : $f : {3, 9, 12} \to {1, 3, 4} and g : {1, 3, 4, 5} \to {3, 9}$
Co-domain of f is a subset of the domain g
So, $g o f$ exist and $g o f : {3, 9, 12} \to {3, 9}$
$(g o f) (3) = g (f (3)) = g (1) = 3$
$(g o f) (9) = g (f (9)) = g (3) = 3$
$(g o f) (12) = g (f (12)) = g (4) = 9$
$g o f = {(3, 3), (9, 3), (12, 9)}$
Co-domain of g is subset of the domain of f.
So, $f o g$ exist and $f o g : {1, 3, 4, 5} \to {3, 9, 12}$
$(f o g) (1) = f (g (1)) = f (3) = 1$
$(f o g) (3) = f (g (3)) = f (3) = 1$
$(f o g) (4) = f (g (4)) = f (9) = 3$
$(f o g) (5) = f (g (5)) = f (9) = 3$
$f o g = {(1, 1), (3, 1), (4, 3), (5, 3)}$
Hence proved, $g o f$ and $f o g$ are both defined.
$f o g = {(1, 1), (3, 1), (4, 3), (5, 3)}$ and
$g o f = {(3, 3), (9, 3), (12, 9)}$

Functions Exercise 2.2 Question 3.

Answer :

g o f = {(1, 2), (4, - 4), (9, - 6), (16, 8)}

Hint : The set which contains all the elements of all the ordered pairs of relation R is known as the domain of the relation. The set which contains all the second element is known as the range of the relation.
Given :

f = {(1, - 1), (4, - 2), (9, - 3), (16, 4)}

g = {(- 1, - 2), (- 2, - 4), (- 3, - 6), (4, 8)}

Prove :

g o f

is defined while

f o g

is not defined.
Solution
Now, Domain of

f = {1, 4, 9, 16}

Range of

f = {- 1, - 2, - 3, 4}

Domain of

g = {- 1, - 2, - 3, 4}

Range of

g = {- 2, - 4, - 6, 8}

Clearly, Range of f = domain of g

∴ g o f is defind .

But Range of

g \neq

domain of f

∴ f o g is not defind .

Now,

g o f (1) = g (- 1) = - 2

g o f (4) = g (- 2) = - 4

g o f (9) = g (- 3) = - 6

g o f (16) = g (4) = 8

g o f = {(1, - 2), (4, - 4), (9, - 6), (16, 8)}

Hence proved,

g o f

is defined but

f o g

is mot defined.

g o f = {(1, - 2), (4, - 4), (9, - 6), (16, 8)}

Functions Exercise 2.2 Question 4.

Answer :

g o f = {(a, a), (b, b), (c, c)} and

f o g = {(u, u), (v, v), (w, w)}

Hint : For any function to be a bijective, the given function should be one-one and onto.\
Given :

A = {a, b, c}

B = {u, v, w}

Solution :
f and g be two functions from A to B and from B to A ;

A \to B and g : B \to A

f = {(a, v), (b, u), (c, w)}

f = {(a, v), (b, u), (c, w)}

g = {(u, b), (v, a), (w, c)}

For both f and g, Different elements of the domain have different images.

∴

f and f are onto
Again, for each element in co-domain of f and g, there in a preimage in domain

∴

f and f are onto
Thus, f and g are bijectives.
Now,

g o f = {(a, a), (b, b), (c, c)} and

f o g = {(u, u), (v, v), (w, w)}

Hence prove, f and g both are bijections

g o f = {(a, a), (b, b), (c, c)} and

f o g = {(u, u), (v, v), (w, w)}

Functions Exercise 2.2 Question 5.

Answer :

f o g (2) = 633

g o f (1) = 2188

Hint :

g o f

means

f (x)

function is in

g (x)

function

f o g

means

g (x)

function is in

f (x)

function
Given :

f : R \to R; f (x) = x^{2} + 8

g : R \to R; g (x) = 3 x^{3} + 1

Solution :
First, we find

f o g

f o g (x) = f (g (x)) = f (3 x^{3} + 1)

f o g (x) = (3 x^{3} + 1)^{2} + 8

∵ [(a + b)^{2} = a^{2} + b^{2} + 2 a b]

f o g (2) = (3 \times (2)^{3} + 1)^{2} + 8

= {(3 \times 8 + 1)}^{2} + 8

= (25)^{2} + 8

= 625 + 8

f o g (2) = 633

Similarly,

g o f (x) = g (f (x)) = g (x^{2} + 8) = 3 (x^{2} + 8)^{3} + 1

g o f (1) = 3 (1 + 8)^{3} + 1 \Rightarrow 3 (9)^{3} + 1

= 3 \times 729 + 1

= 2188

Hence,

f o g (2) = 633 and g o f (1) = 2188

Function Exercise 2.2 Question 6.

Answer :

f o g = x and g o f = x

Hint :

g o f

means

f (x)

function is in

g (x)

function

f o g

means

g (x)

function is in

f (x)

function
Given :

f : R^{+} \to R^{+}

defined by

f (x) = x^{2}

g : R^{+} \to R^{+}

defined by

g (x) = \sqrt{x}

Solution :

f o g (x) = f (g (x)) = f (\sqrt{x})

f o g (x) = (\sqrt{x})^{2} = x

Similarly,

g o f (x) = g (f (x)) = g (x^{2})

g o f (x) = (\sqrt{x})^{2} = x

Thus

f o g (x) = g o f (x)

x = x

Hence proved functions are equal.

f o g = x and g o f = x

Function Exercise 2.2 Question 7.

Answer :

f o g \neq g o f

Hint :

g o f

means

f (x)

function is in

g (x)

function

f o g

means

g (x)

function is in

f (x)

function
Given :

f : R \to R

defined by

f (x) = x^{2}

g : R \to R

defined by

g (x) = x + 1

Solution :

f o g (x) = f (g (x)) = f (x + 1)

f o g (x) = (x + 1)^{2}

∴ [(a + b)^{2} = a^{2} + b^{2} + 2 a b]

f o g (x) = x^{2} + 1 + 2 x

......(i)
Similarly,

g o f (x) = g (f (x)) = g (x^{2})

g o f (x) = x^{2} + 1

.....(ii)
Now we have to make a comparison between eq. (i) and (ii)

f o g (x) \neq g o f (x)

x^{2} + 1 + 2 x \neq x^{2} + 1

Hence proved

f o g \neq g o f

Function Exercise 2.2 Question 8.

Answer :

f o g = g o f = I_{R}

Hint :

g o f

means

f (x)

function is in

g (x)

function

f o g

means

g (x)

function is in

f (x)

function
Given :

f : R \to R

and

g : R \to R

are defined by

f (x) = x + 1 g (x) = x - 1

Solution :

\begin{aligned} f \circ g (x) = f (g (x)) = f (x - 1) \\ f \circ g (x) = x - 1 + 1 \\ f \circ g (x) = x = I_{R} \end{aligned}

.....(i)
Similarly,

\begin{aligned} g \circ f (x) = g (f (x)) = g (x + 1) \\ g \circ f (x) = x + 1 - 1 \\ g \circ f (x) = x = I_{R} \end{aligned}

..... (ii)
From equation (i) and (ii)

f o g = g o f = I_{R}

Hence proved :

f o g = g o f = I_{R}

Function Exercise 2.2 Question 9.

Answer :

h o (g o f) = (h o g) o f

Hint :

g o f

means

f (x)

function is in

g (x)

function

f o g

means

g (x)

function is in

f (x)

function
For associative property

h o (g o f) = (h o g) o f

Given :

\begin{aligned} f : N \to Z_{0}, : Z_{0} \to Q, R : Q \to R \\ f (x) = 2 x \\ g (x) = \frac{1}{x} \\ h (x) = e^{x} \end{aligned}

Solution :

\begin{aligned} g \circ f : N \to Q and h \circ g : Z_{0} \to R \\ h \circ (g \circ f) : N \to R and (h \circ g) \circ f : N \to R \end{aligned}

So, both have same domains

[since f (x) = 2 x]

\begin{matrix} (g \circ f) (x) = g [f (x)] \\ = g (2 x) \\ = \frac{1}{2 x} \\ (h \circ g) (x) = h [g (x)] \\ = h (\frac{1}{x}) \Rightarrow e^{\frac{1}{x}} \end{matrix}

.....(ii)
Now,

\begin{matrix} [h \circ (g \circ f)] (x) = h [(g \circ f) (x)] \\ = h (\frac{1}{2 x}) \end{matrix}

....[From (i)]

= e^{\frac{1}{2 x}}

....[From (ii)]

[(h \circ g) \circ f (x)] = (h \circ g) [f (x)]

= (h \circ g) (2 x)

...[since

f (x) = 2 x

]

= e^{\frac{1}{2 x}}

....[From (ii)]

[h \circ (g \circ f)] (x) = [(h \circ g) \circ f (x)], \in N

So,

h \circ (g \circ f) = (h \circ g) \circ f

Hence, the associative property has benn verified.

Function Exercise 2.2 Question 10.

Answer :

h o (g o f) = (h o g) o f

Hint :

g o f

means

f (x)

function is in

g (x)

function

f o g

means

g (x)

function is in

f (x)

function
Given :

f : N \to N, g : N \to N, h : N \to R

defined as

\begin{array}{r} f (x) = 2 x \\ g (y) = 3 y + 4 \\ h (z) = \sin z \\ All x, y, z \in N \end{array}

Solution :

\begin{aligned} h \circ (g \circ f) (x) = & h (g \circ f (x)) = h (g (f (x))) \\ = h (g (2 x)) = h (3 (2 x) + 4) \\ = h (6 x + 4) \\ = \sin (6 x + 4) \forall x \in N \end{aligned}

Similarly,

\begin{aligned} ((h \circ g) \circ f) (x) & = (h \circ g) (f (x)) \\ = (h \circ g) (2 x) \\ = h (g (2 x)) = h (3 (2 x) + 4) \end{aligned}

= h (6 x + 4) = \sin (6 x + 4) \forall x ϵ N

This shows,

h o (g o f) = (h o g) o f

Functions Exercise 2.2 Question 11.

Answer :

\begin{aligned} f : N \to N by f (x) = x + 1 \\ g : N \to N by \\ g (x) = {x - 1 if x > 1, 1 if x = 1 \end{aligned}

Hint : Onto functions means every range has some preimage in the domain of function.
Given :

f : N \to N and g : N \to N

Solution :
Let

f : N \to N be f (x) = x + 1

and

g : N \to N be g (x) = {x - 1, x > 11, x = 1

We will first show that f is not onto
Checking f is not onto

\begin{aligned} f : N \to N be f (x) = x + 1 \\ y = f (x), where y \in N \\ y = x + 1; x = y - 1 \end{aligned}

for

y = 1; x = 1 - 1 = 0

But 0 is not a natural number

∴

f is not onto.
Finding

g o f :

\begin{aligned} f (x) = x + 1 \\ g (x) = {x - 1 x > 1 1 x = 1 \\ f (x) = x + 1; g (x) = 1 \end{aligned}

Since

g (x) = 1, g (f (x)) = 1

So,

g o f = 1

For

x > 1

Since,

\begin{aligned} g (x) = x - 1, g (f (x)) = f (x) - 1 \\ g \circ f = (x + 1) - 1 \\ g \circ f = x \end{aligned}

So,

g \circ f = {x, x > 11, x = 1}

Let

g \circ f = y, where y \in N

So,

y = {x, x > 11, x = 1}

Here, y is a natural number. As

y = x

So, x is also a natural number.
Hence,

g o f

is onto.

Functions Exercise 2.2 Question 12.

Answer :

f (x) = x and g (x) = | x |

where

f : N \to Z and g : Z \to Z

Hint : An injective function is a function that maps distinct elements of its domain to the distinct elements of its co-domain.
Given :

f : N \to Z and g : Z \to Z

Solution :
Let

f (x) = x and g (x) = | x |

g (x) = | x | = {x x \geq 0 - x x < 0

Checking

g (x)

injective (one-one)

g (1) = | 1 | = 1

g (- 1) = | - 1 | = 1

Since, different elements 1, -1 have the same image 1

∴

g is not injective (one -one).
Checking for injective (one-one)

\begin{aligned} f : N \to Z & g : Z \to Z \\ f (x) = x and g (x) = | x | \\ g \circ f (x) = g (f (x)) = | f (x) | \\ = | x | = {x, x \geq 0 - x, x < 0} \\ Here g \circ f (x) : N \to Z \end{aligned}

So, x is always a natural number.
Here

| x |

will always be a natural number.
So,

g o f (x)

has a unique image

∴

g o f (x)

is injective.
Hence,

g o f

is injective but g is not injective

Functions Exercise 2.2 Question 13.

Answer :

g o f

is a one -one function.
Hint : one-one functions means every domain has a distinct image. If one-one is given for any function

f (x) as if F (x_{1}) = F (x_{2}) then x_{1} = x_{2} where x_{1}, x_{2} \in domains of f (x)

Given :

f : A \to B and g : B \to C

one-one function
Solution :

\begin{aligned} Let, x, y \in A such that \\ g \circ f (x) = g \circ f (y) \\ g (f (x)) = g (f (y)) \end{aligned}

f (x) = f (y)

[

∵

g is one - one ]

x = y

[

∵

f is one - one]

g o f

is a one - one function

Functions Exercise 2.2 Question 14.

Answer :

g o f

is an onto function
Hint : Onto function means every range has some pre-image in the domain of function.
Given :

f : A \to B and g : B \to C

are onto function
Prove :

g o f : A \to C

is an onto function
Solution :
Let

y ϵ C

then

\begin{aligned} g \circ f (x) = y \\ g (f (x)) = y \end{aligned}

....(i)
Since, g is onto, for each element in C, then exists a pre-image in B.

g (x) = y

.....(ii)
From (i) and (ii)

f (x) = x

Since f is onto, for each element in B, there exist a pre image in A

f (x) = x

....(iii)
From (ii) and (iii) we can conclude that for each

y ϵ C

, there exist a preimage in A
Such that

g o f (x) = y

g o f

is onto.

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The 2nd chapter in RD Sharma Class 12th Exercise 2.2 is regarding Functions in mathematics. This chapter focuses on the various kinds of functions, compositions of functions, and inverse of functions. Again, this is a specific topic to be grasped by students compared to the other chapters in the book. The first exercise, 2.1, is a bit essential in the mathematical functions. The second exercise, 2.2, dives deeply into the concepts of fog and gof, injective, surjective, and bijective.

The Class 12 RD Sharma Chapter 2 Exercise 2.2 Solution will lend the students a helping hand. Making the students understand the concept of functions is essential as the upcoming chapters are based on it. For the convenience of the students, the solutions for RD Sharma Class 12th Exercise 2.2 are given in the same order as present in the textbooks.

RD Sharma Class 12th Exercise 2.2 is divided into two levels, Level 1 and Level 2, according to the questions' complexity and difficulty. Level 1 has 21 questions, and level 2 has a couple of questions. Even though numerous examples are provided in the book, solutions elaborately make the students understand the concept more easily and effectively. Therefore, every answer provided in the RD Sharma Class 12 Solutions Chapter 2 Ex 2.2 is given in a more manageable format for the welfare of the students.

With the help of this best solutions book, the students would find the Function chapter easier than before. The RD Sharma Class 12 Solutions Function Ex 2.2 makes the students connect more effortlessly with the concepts than expected. Moreover, once they start using it to prepare their homework and assignments, facing tests and exams in this chapter would become easier based on the NCERT syllabus. Hence, if you are a CBSE board class 12 student, it is better to possess the RD Sharma Class 12th Exercise 2.2 Solutions as there are many chances of questions being asked from this book on your board exam.

Chapter-wise RD Sharma Class 12 Solutions

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Frequently Asked Questions (FAQs)

1. Are the solutions provided in RD Sharma Solutions book updated?

Every solution provided in the RD Sharma Solutions books is updated frequently according to the change in the syllabus.

2. Which RD Sharma solutions book is highly recommended to learn Chapter 2 of Class 12 Mathematics?

The RD Sharma Class 12 Chapter 2 part is enough for the students to prepare the second chapter easily. They can download it from websites like Career360 for free of cost.

3. Is RD Sharma Solutions available according to particular exercises in Class 12 mathematics?

The students can refer according to the specific exercises they require. For instance, the RD Sharma Class 12 Chapter 2 Exercise 2.2 part is enough to work out the sums in exercise 2.2.

4. Are the solutions provided in the RD Sharma books prepared from the exam point of view?

The solutions provided in the RD Sharma book are prepared by experts and are constructed from the exam point of view.

5. Is the RD Sharma Solutions book enough for the CBSE board students to prepare for their public exams?

The RD Sharma Solutions books are more than enough for the class 12 students to prepare for their board exams. In addition, these books provide a deeper insight into the concepts to the students hence making it easier for them to learn.