Request a call back

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

## Functions Exercise MCQ

### Solution 55

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,

### Solution 56

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.

### Solution 57

f: R R is given by f(x) = 3x2 - 5

g: R R is given by

### Solution 58

a. If we take f: Z Z defined by f(x) = x2

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) = x2 + 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.

### Solution 59

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

Consider,

Thus,

### Solution 60

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

Now,

Hence,

## Functions Exercise Ex. 2.1

### Solution 12

We have f : R → R given by f(x) = ex

let x, y  R, such that

f(x) = f(y)

⇒ ex = ey

⇒ ex-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.

### Solution 14

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.

### Solution 15

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.

### Solution 5(xvii)

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.

### Solution 8(i)

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.

### Solution 8(ii)

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.

### Solution 8(iii)

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) = x2, x [-1, 1]

x2 [0, 1]

So, the range is subset of co-domain.

Thus, f is not onto.

Hence, f is neither one-one nor onto.

### Solution 9(i)

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.

### Solution 9(ii)

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.

### Solution 21

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.

## Functions Exercise Ex. 2.2

### Solution 13

Given:

Hence, fof(x) = x.

## Functions Exercise Ex. 2VSAQ

### Solution 37

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)}.

### Solution 38

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,

### Solution 39

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.

### Solution 40

Given:

The function f is defined only when

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

### Solution 41

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)}

### Solution 42

Given: f, g: R R

Therefore, gof: R R will be

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

= g(2x + 1)

= (2x + 1)2 - 2

= 4x2 + 4x - 1

Hence, gof(x) = 4x2 + 4x - 1.

### Solution 43

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)}.

### Solution 44

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

### Solution 45

Given: f(x) = 4 - (x - 7)3

Let y = 4 - (x - 7)3

Hence,

### Solution 3

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

The total number of functions is 8

### Solution 36

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

## Functions Exercise Ex. 2.3

### Solution 13

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

## Functions Exercise Ex. 2.4

### Solution 15

Given: f(x) = 9x2 + 6x - 5

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

9x2 + 6x - 5 = 9y2 + 6y - 5

9(x2 - y2) + 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) = 9x2 + 6x - 5

### Solution 16

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)

### Solution 17

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,

### Solution 18

Given: f: N N defined by f(x) = x2 + x + 1

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

x2 + x + 1 = y2 + y + 1

x2 - y2 + x - y = 0

(x - y)(x + y + 1) = 0

As x, y N, so (x2 + x + 1) can't be zero.

x - y = 0

x = y

So, f is one-one.

Since x N, x2 + 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 = x2 + x + 1

x2 + x + 1 - y = 0

Hence,

### Solution 19

Given: f: R+ (7, ) given by f(x) = 16x2 + 24x + 7

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

Let y = f(x)

y = 16x2 + 24x + 7

16x2 + 24x + 7 - y = 0

Hence,

### Solution 1

Thus, h is a bijection and is invertible.