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

Page / Exercise

## Chapter 2 - Functions Exercise MCQ

Question 55

Let  be defined by  Then,

a.

b.

c. fof(x) = -x

d.

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,

Question 56

Let f: R R be defined by  Then, f is

a. one-one

b. onto

c. bijective

d. not defined

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.

Question 57

Let f: R R be defined by f(x) = 3x2 - 5 and g: R R by  Then, (gof)(x) is

a.

b.

c.

d.

Solution 57

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

g: R R is given by

Question 58

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

a. f(x) = x2

b. f(x) = x + 2

c. f(x) = 2x + 1

d. f(x) = x2 + 1

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.

Question 59

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

a.

b.

c.

d.

Solution 59

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

Consider,

Thus,

Question 60

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

Solution 60

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

Now,

Hence,

Question 1

Let A = {x ε R : - 1 ≤ x ≤ 1} = B and C = {x ε R : x ≥ 0} and let S = {(x, y) ε A × B : x2 + y2 = 1} and s0 = {(x, y) ε A × C : x2 + y2 = 1}. Then

(a) S defines a function from A to B

(b) S0 defines a function from A To C

(c) S0 defines a function from A to B

(d) S defines a function from A to C

Solution 1

Question 2

(a)  injective

(b) surjective

(c) bijective

(d) none of these

Solution 2

Question 3

If f : A → B given by 3f(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

Solution 3

Question 4

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

(a) one-one and onto

(b) many-one and onto

(c) one-one and into

(d) many-one and into

Solution 4

Question 5

(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

Solution 5

Question 6

The function f : A → B defined by f(x) = -x2 + 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, ∞)

Solution 6

Question 7

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

Solution 7

Question 8

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

Solution 8

Question 9

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

Solution 9

Question 10

(a) one-one and onto

(b) one-one but not onto

(c) onto but not one-one

(d) neither one-one nor onto

Solution 10

Question 11

The range of the function f (x) = 7-xPx-3 is

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

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

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

(d) {1, 2, 3}

Solution 11

Question 12

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

Solution 12

Question 13

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

Solution 13

Question 14

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

(a) f(x) = x3

(b)  f(x) = x + 2

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

(d) f(x) = x2 + x

Solution 14

Question 15

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

Solution 15

Question 16

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

Solution 16

Question 17

(a) R

(b) [0, 1]

(c) (0, 1]

(d) [0, 1)

Solution 17

Question 18

if a function f : [2, ∞) → B defined by f(x) = x2 - 4x + 5 is a bijection, then B =

(a) R

(b) [1, ∞)

(c) [4, ∞)

(d) [5, ∞)

Solution 18

Question 19

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

Solution 19

Question 20

(a) bijection

(b) injection but not a surjection

(c) surjection but not an injection

(d) neither an injection nor a surjection

Solution 20

Question 21

(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

Solution 21

Question 22

(a) f is one-one onto

(b) f is one-one into

(c) f is many one onto

(d) f is many one into

Solution 22

Question 23

(a) one-one but not onto

(b) one-one and onto

(c) onto but not one-one

(d) neither one-one nor onto

Solution 23

Question 24

(a) one-one but not onto

(b) many-one but onto

(c) one-one and onto

(d) neither one-one nor onto

Solution 24

Question 25

The function f ; R → R, f(x) = x2 is

(a) injective but not surjective

(b) surjective but not injective

(c) injective as well as surjective

(d) neither injective nor surjective

Solution 25

Question 26

(a) neither one-one nor onto

(b) one-one but not onto

(c) onto but not one-one

(d) one-one and onto both

Solution 26

Question 27

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

Solution 27

Question 28

(a) onto but not one-one

(b) one-one but not onto

(c) one-one and onto

(d) neither one-one nor onto

Solution 28

Question 29

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

(a) one-one and onto

(b) many one and onto

(c) one-one and into

(d) many one and into

Solution 29

Question 30

Let f(x) = x2 and g(x) = 2x. Then the solution set of the equation fog (x) = gof (x) is

(a) R

(b) {0}

(c) {0, 2}

(d) none of these

Solution 30

Question 31

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

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

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

Solution 34

Question 35

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

Solution 35

Question 36

Solution 36

Question 37

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

Solution 37

Question 38

Solution 38

Question 39

(a) x

(b) 1

(c) f(x)

(d) g(x)

Solution 39

Question 40

Solution 40

Question 41

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

Solution 41

Question 42

Let f: [2, ∞) → X be defined by f(x) = 4x - x2. Then, f is invertible, if X=

(a) [2, ∞)

(b) ( -∞, 2]

(c) (-∞, 4]

(d) [4, ∞)

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

(a) 2x - 3

(b) 2x + 3

(c) 2x2 + 3x + 1

(d) 2x2 - 3x - 1

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Let f(x) = x3 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}

Solution 48

Question 49

Let f : R → R be given by f(x) = x2 - 3. Then domain of f-1 is given by:

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

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

Solution 52

Question 53

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

Solution 53

Question 54

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

Solution 54

## Chapter 2 - Functions Exercise Ex. 2.1

Question 1(i)

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

Solution 1(i)

Question 1(ii)

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

Solution 1(ii)

Question 1(iii)

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

Solution 1(iii)

Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5(i)
Solution 5(i)
Question 5(ii)
Solution 5(ii)
Question 5(iii)
Solution 5(iii)
Question 5(iv)
Solution 5(iv)
Question 5(v)
Solution 5(v)
Question 5(vi)
Solution 5(vi)
Question 5(vii)
Solution 5(vii)
Question 5(viii)

Solution 5(viii)

Question 5(ix)
Solution 5(ix)
Question 5(x)

Solution 5(x)

Question 5(xi)

Classify the following functions as injection, surjection or bijection:

f : R → R, defined by f(x) = sin2x + cos2x

Solution 5(xi)

Question 5(xii)
Solution 5(xii)
Question 5(xiii)

Solution 5(xiii)

Question 5(xiv)

Solution 5(xiv)

Question 5(xv)

Solution 5(xv)
Question 5(xvi)
Solution 5(xvi)
Question 6
Solution 6
Question 7
Solution 7

Question 10
Solution 10
Question 11
Solution 11
Question 12

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.

Question 13
Solution 13
Question 14

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

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.

Question 15

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

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.

Question 16
Solution 16
Question 17
Solution 17
Question 18
Solution 18
Question 19
Solution 19
Question 20
Solution 20
Question 22

Solution 22

Question 23

Solution 23

Question 5(xvii)

Classify the following function as injection, surjection or bijection:

f: R R, defined by

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.

Question 8(i)

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

Question 8(ii)

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

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.

Question 8(iii)

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

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.

Question 9(i)

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}

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.

Question 9(ii)

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'}

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.

Question 21

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

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.

## Chapter 2 - Functions Exercise Ex. 2.2

Question 1(i)
Solution 1(i)
Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)
Solution 1(iv)
Question 1(v)
Solution 1(v)
Question 1(vi)
Solution 1(vi)
Question 2
Solution 2
Question 3
Solution 3

Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10

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.

Solution 10

Question 11

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

Solution 11

Question 12
Solution 12
Question 14
Solution 14
Question 13

If  then find fof(x).

Solution 13

Given:

Hence, fof(x) = x.

## Chapter 2 - Functions Exercise Ex. 2VSAQ

Question 37

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.

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)}.

Question 38

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

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,

Question 39

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)}

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.

Question 40

Write the domain of the real function f defined by

Solution 40

Given:

The function f is defined only when

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

Question 41

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.

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)}

Question 42

Let f, g: R R be given by f(x) = 2x + 1 and g(x) = x2 - 2 for all x R, respectively. Then, find gof.

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.

Question 43

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.

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)}.

Question 44

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.

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

Question 45

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

Solution 45

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

Let y = 4 - (x - 7)3

Hence,

Question 1

Figure (a)

Figure (b)

Solution 1

Question 2

Figure (a)

Figure (b)

Solution 2

Question 3

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

Solution 3

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

The total number of functions is 8

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

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

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

## Chapter 2 - Functions Exercise Ex. 2.3

Question 13

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).

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

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 2
Solution 2

Question 3
Solution 3
Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11(i)

Let f be a real function given by

Find each of the following

fof

Solution 11(i)

Question 11(ii)

Let f be a real function given by

Find each of the following

fofof

Solution 11(ii)

Question 11(iii)

Let f be a real function given by

Find each of the following

(f0f0f) (38)

Solution 11(iii)

Question 11(iv)

Let f be a real function given by

Find each of the following

f2Also, show that fof ≠ f2 .

Solution 11(iv)

Question 12

Solution 12

## Chapter 2 - Functions Exercise Ex. 2.4

Question 15

Let f: N N be a function defined as f(x) = 9x2 + 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).

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

Question 16

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

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)

Question 17

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.

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,

Question 18

Show that the function f: N N defined by f(x) = x2 + x + 1 is one-one but onto. Find the inverse of f: N S, where S is range of f.

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,

Question 19

Consider the bijective function f: R+ (7, ) given by f(x) = 16x2 + 24x + 7, where R+ is the set of positive real numbers. Find the inverse function of f.

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,

Question 1

Solution 1

Thus, h is a bijection and is invertible.

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3

Consider  and  defined as .

Show that  are invertible. Find  and show that

Solution 3