# RD SHARMA Solutions for Class 12-science Maths Chapter 2 - Functions

## Chapter 2 - Functions Exercise MCQ

Let be defined by Then,

a.

b.

c. fof(x) = -x

d.

Given: be defined by

For inverse, we first need to check whether f is invertible.

Let such that f(x) = f(y)

So, f is one-one.

Let y = f(x)

Therefore, for every there exist y ∈ R such that f(x) = y.

Thus, f is bijective and so invertible.

Taking y = f(x), we get

Hence,

Let f: R → R be defined by Then, f is

a. one-one

b. onto

c. bijective

d. not defined

Given: f: R → R be defined by

As x ∈ R, so it can take the value 0 as well.

At x = 0, f(x) is not defined.

Thus, f is not defined.

Let
f: R → R be defined
by f(x) = 3x^{2} - 5 and g: R → R by Then, (gof)(x) is

a.

b.

c.

d.

f: R → R is given by
f(x) = 3x^{2} - 5

g: R → R is given by

Which of the following functions from Z to Z are bijections?

a. f(x) = x^{2}

b. f(x) = x + 2

c. f(x) = 2x + 1

d. f(x) = x^{2} + 1

a. If we take f: Z → Z defined by
f(x) = x^{2}

Clearly, it is not one-one as considering f(x) = f(y) will give us

x = ±y

Therefore, it is not a bijection.

b. Consider f: Z → Z defined by f(x) = x + 2

Let x, y ∈ Z such that f(x) = f(y)

∴ x + 2 = y + 2

∴ x = y

So, f is one-one.

For x in the domain Z, f(x) will also take integer values.

So, the range of f is Z.

Therefore, f is onto.

Thus, f is a bijection.

c. Consider f: Z → Z defined by f(x) = x + 2

Let x, y ∈ Z such that f(x) = f(y)

∴ x = y

So, f is one-one.

For x in the domain Z, f(x) = 2x + 1 will take odd integer values.

So, the range of f is not same as its co-domain.

Therefore, f is not onto.

Thus, f is not a bijection.

d. If we take f: Z → Z defined by
f(x) = x^{2} + 1

Clearly, it is not one-one as considering f(x) = f(y) will give us

x = ± y

Therefore, it is not a bijection.

Hence, option (b) is correct.

Let f: A → B and g: B → C be the bijective functions. Then

a.

b.

c.

d.

It is given that f: A → B and g: B → C are two bijective functions.

Consider,

Thus,

Let f: N → R be the function defined by and g: Q → R be another function defined by g(x) = x + 2. Then (g o f)(3/2) is

a. 1

b. 2

b.

c. none of these

It is given that f: N → R as and g: Q → R as g(x) = x + 2.

Now,

Hence,

Let A = {x ε R : - 1 ≤ x ≤ 1} = B and C = {x ε R : x ≥ 0} and let S = {(x, y) ε A × B : x^{2} + y^{2} = 1} and s_{0} = {(x, y) ε A × C : x^{2} + y^{2} = 1}. Then

(a) S defines a function from A to B

(b) S_{0} defines a function from A To C

(c) S_{0} defines a function from A to B

(d) S defines a function from A to C

(a) injective

(b) surjective

(c) bijective

(d) none of these

If f : A → B given by 3^{f(x)} + 2^{-x} = 4 is a bijection, then

(a) A = {x ε R : - 1 < x < ∞], B = {x ε R ; 2 < x < 4]

(b) A = {x ε R : - 3 < x < ∞}, B = {x ε R ; 0 < x < 4}

(c) A = {x ε R : - 2 < x ε R : 0 < x < 4}

(d) none of these

The function f : R → R defined by f(x) = 2^{x} + 2^{|x|} is

(a) one-one and onto

(b) many-one and onto

(c) one-one and into

(d) many-one and into

(a) f is one-one but not onto

(b) f is onto but not one-one

(c) f is both one-one and into

(d) none of these

The function f : A → B defined by f(x) = -x^{2} + 6x - 8 is a bijection, if

(a) A = ( -∞, 3] and B = ( -∞, 1]

(b) A = [-3, ∞) and B = (-∞, 1]

(c) A = ( -∞, 3] and B = [1, ∞)

(d) A = [3, ∞) and B = [1, ∞)

Let A = {x ε R : - 1 ≤ x ≤ 1} = B. then, the mapping f : A → B given by f(x) = x|x| is

(a) injective but not surjective

(b) surjective but not injective

(c) bijective

(d) none of these

Let f : R → R be given by f(x) = [x]^{2} + [x + 1] - 3, where [x] denotes the greatest integer less than or equal to x. Then, f(x) is

(a) many-one and onto

(b) Many-one and into

(c) one-one and into

(d) one-one and onto

let m be the set of all 2 × 2 matrices with entries from the set R of real numbers. Then the function f : M → R defined by f(A) = |A| for every A ε M, is

(a) one-one and onto

(b) neither one-one nor onto

(c) one-one but-not onto

(d) onto but not one-one

(a) one-one and onto

(b) one-one but not onto

(c) onto but not one-one

(d) neither one-one nor onto

The range of the function f (x) = ^{7-x}P_{x-3} is

(a) {1, 2, 3, 4, 5}

(b) {1, 2, 3, 4, 5, 6}

(c) {1, 2, 3, 4}

(d) {1, 2, 3}

A function f from the set of natural numbers to integers defined by

is

(a) neither one-one nor onto

(b) one-one but not onto

(c) onto but not one-one

(d) one-one and onto both

Let f be an injective map with domain {x, y, z} and range {1, 2, 3} such that exactly one of the following statements is correct and the remaining are false.

f(x) = 1, f(y) ≠ 1, f(z) ≠ 2.

The value of f^{-1} (1) is

(a) x

(b) y

(c) z

(d) none of these

Which of the following functions from Z to itself are bijections?

(a) f(x) = x^{3}

(b) f(x) = x + 2

(c) f(x) = 2x + 1

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

Which of the following from A = {x : -1 ≤ x ≤ 1} to itself are bijections?

Let A = {x : -1 ≤ x ≤ 1} and f; A → A such that f(x) = x|x|, then f is

(a) a bijection

(b) injective but not surjective

(c) surjective but not injective

(d) neither injective nor surjective

(a) R

(b) [0, 1]

(c) (0, 1]

(d) [0, 1)

if a function f : [2, ∞) → B defined by f(x) = x^{2} - 4x + 5 is a bijection, then B =

(a) R

(b) [1, ∞)

(c) [4, ∞)

(d) [5, ∞)

The function f : R → R defined by f(x) = (x - 1)(x - 2)(x - 3) is

(a) one-one but not onto

(b) onto but not one-one

(c) both one and onto

(d) neither one-one nor onto

(a) bijection

(b) injection but not a surjection

(c) surjection but not an injection

(d) neither an injection nor a surjection

(a) f is a bijection

(b) f is an injection only

(c) f is surjection on only

(d) f is neither an injection nor a surjection

(a) f is one-one onto

(b) f is one-one into

(c) f is many one onto

(d) f is many one into

(a) one-one but not onto

(b) one-one and onto

(c) onto but not one-one

(d) neither one-one nor onto

(a) one-one but not onto

(b) many-one but onto

(c) one-one and onto

(d) neither one-one nor onto

The function f ; R → R, f(x) = x^{2} is

(a) injective but not surjective

(b) surjective but not injective

(c) injective as well as surjective

(d) neither injective nor surjective

(a) neither one-one nor onto

(b) one-one but not onto

(c) onto but not one-one

(d) one-one and onto both

Which of the following functions from A = {x ε R : - 1 ≤ x ≤ 1} to itself are bijections?

(a) onto but not one-one

(b) one-one but not onto

(c) one-one and onto

(d) neither one-one nor onto

The function f ; R → R defined by f (x) = 6^{x} + 6^{|x|} is

(a) one-one and onto

(b) many one and onto

(c) one-one and into

(d) many one and into

Let f(x) = x^{2} and g(x) = 2^{x}. Then the solution set of the equation fog (x) = gof (x) is

(a) R

(b) {0}

(c) {0, 2}

(d) none of these

if f : R → R is given by f(x) = 3x - 5, then f^{-1}(x)

Let A = {x ε R : x ≥ 1}. The inverse of the function f ; A → A given by f(x) = 2^{x(x - 1)}, is

Let A = {x ε R : x ≤ 1} and f : A → 1 A be defined as f (x) = x(2 - x). Then, f^{-1} (x) is

ANSWER PENDING

If the function f: R→R be such that f(x) = x - [x], where [x] denotes the greatest integer less than or equal to x, then f^{-1}(x) is

ANSWER PENDING

(a) x

(b) 1

(c) f(x)

(d) g(x)

The distinct linear functions which map [-1, 1] onto [0, 2] are

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

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

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

(d) none of these

Let f: [2, ∞) → X be defined by f(x) = 4x - x^{2}. Then, f is invertible, if X=

(a) [2, ∞)

(b) ( -∞, 2]

(c) (-∞, 4]

(d) [4, ∞)

ANSWER PENDING

(a) 2x - 3

(b) 2x + 3

(c) 2x^{2} + 3x + 1

(d) 2x^{2} - 3x - 1

Let f(x) = x^{3} be a function with domain {0, 1, 2, 3}. Then domain of f^{-1} is

(a) {3, 2, 1, 0}

(b) {0, -1, -2, -3}

(c) {0, 1, 8, 27}

(d) {0, -1, -8, -27}

Let f : R → R be given by f(x) = x^{2} - 3. Then domain of f^{-1} is given by:

Let A = {1, 2, ..., n} and B = {a, b}. Then the number of subjections from A into B is

If the set A contains 5 elements and the set B 6 elements, then the number of one-one and onto mappings from A to B is

(a) 720

(b) 120

(c) 0

(d) none of these

If the set A contains 7 elements and the set B contains 10 elements, then the number one-one functions from A to B is

## Chapter 2 - Functions Exercise Ex. 2.1

Give an example of function which is one-one but not onto.

Give an example of a function which is not one-one but onto.

Given an example of a function which is neither one-one nor onto.

Classify the following functions as injection, surjection or bijection:

f : R → R, defined by f(x) = sin^{2}x + cos^{2}x

We have *f* : R → R given by *f*(x) = e^{x}

let x, y ∊ R, such that

*f*(x) = *f*(y)

⇒ e^{x} = e^{y}

⇒ e^{x-y} = 1 = e°

⇒ x - y = 0

⇒ x = y

∴ *f* is one-one

clearly range of *f* = (0, ∞) ≠ R

∴ *f* is not onto

when co-domain is replaced by i.e, (0, ∞) then *f* becomes an onto function.

If A = { 1, 2, 3}, show that a ono-one function f : A → A must be onto.

Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the co-domain {1, 2, 3} under f.

Hence, f has to be onto.

If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.

Suppose f is not one-one.

Then, there exists two elements, say 1 and 2 in the domain whose image in the co-domain is same.

Also, the image of 3 under f can be only one element.

Therefore, the range set can have at most two elements of the co-domain {1, 2, 3}

i.e f is not an onto function, a contradiction.

Hence, f must be one-one.

Classify the following function as injection, surjection or bijection:

f: R → R, defined by

Given: f: R → R, defined by

Injective: Let x, y ∈ R such that f(x) = f(y)

This does not imply x = y

Therefore, f is not one-one.

Surjective: Let y ∈ R be arbitrary, then

f(x) = y

This does not give any value for x.

Therefore, f is not onto.

Thus, it is not bijective.

Let A = [-1, 1]. Then, discuss whether the following function from A to itself are one-one, onto or bijective.

Given: f: A → A, defined by

Injective: Let x, y ∈ A such that f(x) = f(y)

Therefore, f is one-one.

Surjective: For f(x), x ∈ [-1, 1]

Therefore, range is the subset of codomain.

Thus, f is not onto.

Hence, f is one-one but not onto.

Let A = [-1, 1]. Then, discuss whether the following function from A to itself are one-one, onto or bijective.

Given: g: A → A, defined by

Injective: Let x, y ∈ A such that g(x) = g(y)

Therefore, f is not one-one.

Surjective: For f(x), x ∈ [-1, 1]

Therefore, range is the subset of codomain.

Thus, f is not onto.

Hence, f is neither one-one nor onto.

Let A = [-1, 1]. Then, discuss whether the following function from A to itself are one-one, onto or bijective.

Given: h: A → A, defined by

Injective: Let x, y ∈ A such that h(x) = h(y)

Therefore, f is not one-one.

Surjective: For f(x) = x^{2},
x ∈ [-1, 1]

∴ x^{2}∈ [0, 1]

So, the range is subset of co-domain.

Thus, f is not onto.

Hence, f is neither one-one nor onto.

Is the following set of ordered pairs function? If so, examine whether the mapping is injective or surjective.

{(x, y): x is a person, y is the mother of x}

Given set of ordered pair is {(x, y): x is a person, y is the mother of x}

Now biologically, each person 'x' has only one mother.

So, the above set of ordered pairs is a function.

Also, more than one person may have same mother.

So the function is many-one and surjective.

Is the following set of ordered pairs function? If so, examine whether the mapping is injective or surjective.

{(a, b): 'a' is a person, 'b' is an ancestor of 'a'}

Given set of ordered pair is {(a, b): 'a' is a person, 'b' is an ancestor of 'a'}

Clearly, a person can have more than one ancestors.

So, the above set of ordered pairs is not a function.

Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:

a. an injective map from A to B

b. a mapping from A to B which is not injective

c. a mapping from A to B

Given: A = {2, 3, 4}, B = {2, 5, 6, 7}

a. Consider f: A → B defined by f(x) = x + 3

Clearly, f is injective because each element of A has distinct output in B.

b. Consider g: A → B defined by g(x) = 3

Here, g is a constant function.

Therefore, each element in A has the same output in B.

Thus, g is not injective.

c. Consider h: A → B such that h = {(2, 2), (3, 5), (4, 7)}

Clearly, h is a mapping from A to B.

## Chapter 2 - Functions Exercise Ex. 2.2

Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and

h(z) = sin z for all x, y, z ϵ N.

show tht ho (gof) = (hog) of.

Given Examples of two functions f : N → N and g ; N → N such that gof is onto but f is not onto.

If then find fof(x).

Given:

Hence, fof(x) = x.

## Chapter 2 - Functions Exercise Ex. 2VSAQ

If f: [5, 6] → [2, 3] and g: [2, 3] → [5, 6] are given by f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}, find f o g.

We have f: [5, 6] → [2, 3] and g: [2, 3] → [5, 6] are given by:

f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}

Now, (f o g)(2) = f(g(2)) = f(5) = 2

And, (f o g)(3) = f(g(3)) = f(6) = 3

Therefore, fog = {(2, 2), (3, 3)}.

Let
f: R → R be the
function defined by f(x) = 4x - 3 for all x ∈ R. Then write
f^{-1}.

Let x, y ∈ R such that f(x) = f(y)

∴ 4x - 3 = 4y - 3

∴ x = y

So, f is one-one.

For x ∈ R, 4x - 3 ∈ R

Therefore, range and co-domain are same.

So, f is onto.

Thus, f is bijective and so it is invertible.

Let y = f(x)

∴ y = 4x - 3

Hence,

Which one of the following relations on A = {1, 2, 3} is a function?

f = {(1, 3), (2, 3), (3, 2)}, g = {(1, 2), (1, 3), (3, 1)}

Consider, f = {(1, 3), (2, 3), (3, 2)}

In f, every element in the domain A has unique image.

So, f is a function.

It is given that g = {(1, 2), (1, 3), (3, 1)}

For g, the element 1 in the domain A has two different images.

So, g is not a function.

Write the domain of the real function f defined by

Given:

The function f is defined only when

Hence, the domain is [-5, 5].

Let
A = {a, b, c, d} and f: A → A be given by f = {(a, b), (b,
d), (c, a), (d, c)}, write f^{-1}.

Given: A = {a, b, c, d} and f: A → A is given by f = {(a, b), (b, d), (c, a), (d, c)}

Here, f^{-1}
will be a function from A to A and given by

f^{-1} =
{(b, a), (d, b), (a, c), (c, d)}

Let
f, g: R → R be given by f(x) = 2x + 1 and
g(x) = x^{2} - 2 for all x ∈ R, respectively. Then, find gof.

Given: f, g: R → R

Therefore, gof: R → R will be

gof(x) = g(f(x))

= g(2x + 1)

= (2x + 1)^{2}
- 2

= 4x^{2} + 4x - 1

Hence, gof(x) = 4x^{2} + 4x - 1.

If the mapping f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3}, given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, write fog.

Given: f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3}

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

fog(2) = f(g(2)) = f(3) = 5

fog(5) = f(g(5)) = f(1) = 2

Therefore, fog: {1, 2, 5} → {5, 2}

Hence, fog = {(1, 5), (2, 5), (5, 2)}.

If a function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by g(x) = αx + β, find the values of α and β.

Given: g(x) = αx + β and g = {(1, 1), (2, 3), (3, 5), (4, 7)}

Now, g(1) = 1

∴ α + β = 1 … (i)

And, g(2) = 3

∴ 2α + β = 3 … (ii)

Solving (i) and (ii), we get

α = 2 and β = -1

If
f(x) = 4 - (x - 7)^{3}, write

Given: f(x) = 4 -
(x - 7)^{3}

Let y = 4 - (x -
7)^{3}

Hence,

Figure (a)

Figure (b)

Figure (a)

Figure (b)

If A = {1, 2, 3} B = {a, b}, write total number of functions from A to B.

A = {1, 2, 3} B = {a, b}

The total number of functions is 8

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1,4), (2,5), (3,6)} be a function from A to B.

State the whether f is one - one or onto

It is given that

A = {1, 2, 3}, B = {4, 5, 6, 7} and f = {(1,4), (2,5), (3,6)}

The function f is one-one from A to B

## Chapter 2 - Functions Exercise Ex. 2.3

If f, g: R → R be two functions defined as f(x) = |x| + x and g(x) = |x| - x for all x ∈ R. Then, find fog and gof. Hence, find fog(-3), fog(5) and gof(-2).

Given: f(x) = |x| + x and g(x) = |x| - x for all x ∈ R

∴ fog(-3) = -4(-3) = 12, fog(5) = 0 and

gof(-2) = 0

Let f be a real function given by

Find each of the following

fof

Let f be a real function given by

Find each of the following

fofof

Let f be a real function given by

Find each of the following

(f0f0f) (38)

Let f be a real function given by

Find each of the following

f^{2}^{}Also, show that fof ≠ f^{2 }.

## Chapter 2 - Functions Exercise Ex. 2.4

Let
f: N → N be a
function defined as f(x) = 9x^{2} + 6x - 5. Show that f: N → S, where S is
the range of f, is invertible. Find the inverse of f and hence find f^{-1}(43)
and f^{-1}(163).

Given: f(x) = 9x^{2}
+ 6x - 5

Let x, y ∈ N such that f(x) = f(y)

∴ 9x^{2}
+ 6x - 5 = 9y^{2} + 6y - 5

∴ 9(x^{2}
- y^{2}) + 6(x - y) = 0

∴ (x - y){9(x + y) + 6} = 0

As x, y ∈ N, 9(x + y) + 6 can't be zero.

∴ x - y = 0

∴ x = y

So, f is one-one.

Clearly, f is onto as range and co-domain are same.

Therefore, f is bijective.

Thus, f is invertible.

Let y = f(x) = 9x^{2}
+ 6x - 5

Let
be a function
defined as Show that is one-one
and onto. Hence, find f^{-1}.

Given:

Let x, y ∈ such that f(x) = f(y)

So, f is one-one.

Clearly, f is onto as the co-domain of is same as its range.

Therefore, f is bijective.

Thus, f is invertible.

Let y = f(x)

Let
A = R - {2} and B = R - {1}. If f: A → B is a function defined by show that f
is one-one and onto. Find f^{-1}.

Given:

Let x, y ∈ A such that f(x) = f(y)

So, f is one-one.

Let y ∈ B

So, for every x ∈ A, there is a y ∈ B such that f(x) = y.

Therefore, f is onto.

Thus, f is bijective and so it is invertible.

Hence,

Show
that the function f: N → N defined by f(x) = x^{2}
+ x + 1 is one-one but onto. Find the inverse of f:
N → S, where S is
range of f.

Given: f: N → N defined by
f(x) = x^{2} + x + 1

Let x, y ∈ N such that f(x) = f(y)

∴ x^{2}
+ x + 1 = y^{2} + y + 1

∴ x^{2}
- y^{2} + x - y = 0

∴ (x - y)(x + y + 1) = 0

As x, y ∈ N, so (x^{2}
+ x + 1) can't be zero.

∴ x - y = 0

∴ x = y

So, f is one-one.

Since x ∈ N, x^{2}
+ x + 1 > 3 as the minimum value x can take is 1.

Therefore, range of f = (3, ∞)

As the co-domain and range does not match, f is not onto.

But, f: N → (3, ∞) is onto as the range matches with co-domain.

Thus, f: N → S is a bijective function and so it is invertible.

Let y = f(x)

∴ y = x^{2}
+ x + 1

∴ x^{2}
+ x + 1 - y = 0

Hence,

Consider
the bijective function f: R^{+}→ (7, ∞) given by f(x)
= 16x^{2} + 24x + 7, where R^{+} is the set of positive real
numbers. Find the inverse function of f.

Given: f: R^{+}→ (7, ∞) given by f(x)
= 16x^{2} + 24x + 7

As f is a bijective function, so it is invertible.

Let y = f(x)

∴ y = 16x^{2}
+ 24x + 7

∴ 16x^{2}
+ 24x + 7 - y = 0

Hence,

Thus, h is a bijection and is invertible.

Consider and defined as .

Show that are invertible. Find and show that