NCERT Solutions Class 12 Maths (Relation And Functions) Miscellaneous Exercise

NCERT Solutions Class 12 Maths (Relation And Functions) Miscellaneous Exercis

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-1 (Relation And Functions) Miscellaneous Exercise This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Set 1

Question 1. 

Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g o f = f o g = 1_R.

Solution:

As, it is mentioned here

f : R → R be defined as f(x) = 10x + 7

To, prove the function one-one

Let’s take f(x) = f(y)

10x + 7 = 10y + 7

x = y

Hence f is one-one.

To, prove the function onto

y ∈ R, y = 10x+7

So, it means for y ∈ R, there exists 

 

Hence f is onto.

As, f is one-one and onto. This f is invertible function.

Let’s say g : R → R be defined as 

Hence, g : R → R such that g o f = f o g = 1R.

g : R → R is defined as 

Question 2. 

Let f : W → W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

Solution:

The function f is defined as

As, we know f is invertible, if and only if f is one-one and onto.

ONE-ONE

For the pair of number, we will deal with three cases:

Case 1: When both numbers p and q are odd numbers.

f(p) = p-1

f(q) = q-1

f(p) = f(q)

p-1 = q-1

p – q = 0

Case 2: When both numbers p and q are even numbers.

f(p) = p+1

f(q) = q+1

f(p) = f(q)

p+1 = q+1

p – q = 0

Case 3: When p is odd and q is even

f(p) = p-1

f(q) = q+1

f(p) = f(q)

p-1 = q+1

p – q = 2

Subtracting an odd number and even always gives a odd number, not even. Hence, the case 3 result is impossible.

So, the function f is one-one, for case 1 and case 2 only.

ONTO

Case 1: When p is odd number

f(p) = p-1

y = p-1

p = y+1

Hence, when p is odd y is even.

Case 2: When p is even number

f(p) = p+1

y = p+1

p = y-1

Hence, when p is even y is odd.

So, it means for y ∈ W, there exists p = y+1 and y-1 for odd and even value of p respectively.

Hence f is onto.

As, f is one-one and onto. This f is an invertible function.

Let’s say g : W → W be defined as

f = g

Hence, The inverse of f is f itself

Question 3. 

If f : R → R is defined by f(x) = x²– 3x + 2, find f (f(x)).

Solution:

f(x) = x²– 3x + 2

f(f(x)) = f(x²– 3x + 2)

= (x²– 3x + 2)2 – 3(x²– 3x + 2) + 2

= x⁴ + 9x² + 4 -6x³ – 12x + 4x² – 3x² + 9x – 6 + 2

f(f(x)) = x⁴ – 6x³ + 10x² – 3x

Question 4. 

Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by f(x)= ,x ∈ R is one one and onto function.

Solution:

As, it is mentioned here

f : R → {x ∈ R : – 1 < x < 1} defined by , x ∈ R

As, we know f is invertible, if and only if f is one-one and onto.

ONE-ONE

For the pair of number, we will deal with three cases:

Case 1: When both numbers p and p are positive numbers.

The function f is defined as

Case 1: When both numbers p and q are positive numbers.

f(p) = f(q)

p(1+q) = q(1+p)

p = q

Case 2: When number p and q are negative numbers.

f(p) = f(q)

p(1-q) = q(1-p)

p = q

Case 3: When p is positive and q is negative

f(p) = f(q)

p(1-q) = q(1+p)

p + q = 2pq

Here, RHS will be negative and LHS will be positive. Hence, the case 3 result is impossible.

So, the function f is one-one, for case 1 and case 2.

ONTO

Case 1: When p>0.

Case 2: When p <0

Hence, p is defined for all the values of y, p∈ R

Hence f is onto.

As, f is one-one and onto. This f is an invertible function.

Question 5. 

Show that the function f : R → R given by f(x) = x³ is injective.

Solution:

As, it is mentioned here

f : R → R defined by f(x) = x³, x ∈ R

To prove f is injective (or one-one).

ONE-ONE

The function f is defined as

f(x) = x³

f(y) = y³

f(x) = f(y)

x³ = y³

x = y

The function f is one-one, so f is injective.

Question 6. 

Give examples of two functions f : N → Z and g : Z → Z such that g o f is injective but g is not injective.

(Hint : Consider f(x) = x and g (x) = | x |).

Solution:

Two functions, f : N → Z and g : Z → Z

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

Let’s check, whether g is injective or not

g(5) = |5| = 5

g(-5) = |-5| = 5

As, we can see here that

Taking two integers, 5  and -5

g(5) = g(-5)

but, 5 ≠ -5

So, g is not an injective function.

Now, g o f: N → Z is defined as

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

Now, as x,y∈ N

g(x) = |x|

g(y) = |y|

g(x) = g(y)

|x| = |y|

x = y (both x and y are positive)

Hence, g o f is an injective.

Question 7. 

Give examples of two functions f : N → N and g : N → N such that g o f is onto but f is not onto.

(Hint : Consider f(x) = x + 1 and 

Solution:

Two functions, f : N → N and g : N → N

Taking f(x) = x+1 and 

 As, f(x) = x+1

y = x+1

x = y-1

But, when y=1, x = 0. Which doesn’t satiny this relation f : N → N.

Hence. f is not an onto function.

Now, g o f: N → N is defined as

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

When x+1=1, we have

g(x+1) = 1 (1∈ N)

And, when x+1>1, we have

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

y = x, which also satisfies x,y∈ N

Hence, g o f is onto.

Question 8. 

Given a non empty set X, consider P(X) which is the set of all subsets of X.

Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.

Solution:

Given, A and B are the subsets of P(x), A⊂ B

To check the equivalence relation on P(X), we have to check

• ·         Reflexive

As, we know that every set is the subset of itself.

Hence, A⊂ A and B⊂ B

ARA and BRB is reflexive for all A,B∈ P(X)

• ·         Symmetric

As, it is given that A⊂ B. But it doesn’t make sure that B⊂ A.

To be symmetric it has to be A = B

ARB is not symmetric.

• ·         Transitive

When A⊂ B and B⊂ C

Then of course, A⊂ C

Hence, R is transitive.

So, as R is not symmetric. 

R is not an equivalence relation on P(X).

Question 9. 

Given a non-empty set X, consider the binary operation ∗ : P(X) × P(X) → P(X) given by A ∗ B = A ∩ B ∀ A, B in P(X), where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation ∗.

Solution:

Given, P(X) × P(X) → P(X) is defined as A*B = A∩B ∀ A, B ∈ P(X)

This implies, A⊂  X and B ⊂  X

So, A∩X = A and B∩X = B ∀ A, B ∈ P(X)

⇒ A*X = A and B*X = B

Hence, X is the identity element for intersection of binary operator.

Question 10. 

Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.

Solution:

Onto function from the set {1,2,3,…..,n} to itself is just same as the permutations of n.

1×2×3×4×…….×n

Which is n!.

Set 2

Question 11: 

Let S = {a, b, c} and T = {1, 2, 3}. Find F^–1 of the following functions F from S to T, if it exists.

(i) F = {(a, 3), (b, 2), (c, 1)} 

Solution:

As, F = {(a, 3), (b, 2), (c, 1)} and S = {a,b,c} and T={1,2,3}

F: S→T is defined as

F(a) = 3, F(b) = 2 and F(c) = 1

F is one-one and onto.

Taking F^-1, so F^-1: T→S

a = F^-1(3), b = F^-1(2) and c = F^-1(1)

F^-1 = {(3,a),(2,b),(1,c)}

(ii) F = {(a, 2), (b, 1), (c, 1)}

Solution:

As, F = {(a, 2), (b, 1), (c, 1)}

F: S→T is defined as

F(a) = 2, F(b) = 1 and F(c) = 1

Here, F(b) = F(c) but b ≠ c

Hence, F is not one-one.

So, F is not invertible and F^-1 doesn’t exists.

Question 12. 

Consider the binary operations ∗ : R × R → R and o : R × R → R defined as a ∗b = |a – b| and a o b = a, ∀ a, b ∈ R. Show that ∗ is commutative but not associative, o is associative but not commutative. Further, show that ∀ a, b, c ∈ R, a ∗ (b o c) = (a ∗ b) o (a ∗ c). [If it is so, we say that the operation ∗ distributes over the operation o]. Does o distribute over ∗? Justify your answer.

Solution:

Binary operations ∗ : R × R → R defined as a ∗b = |a – b|

a*b = |a-b|

b*a = |b-a| = |-(a-b)| = |a-b|

a*b = b*a

Hence, ∗ is commutative.

Now, let’s take a=1, b=2 and c=3 for better understanding

a*(b*c) = a*|b-c| = |a-|b-c|| = |1-|2-3|| = 0

(a*b)*c = |a-b|*c = ||a-b|-c| = ||1-2|-3| = 2

a*(b*c) ≠ (a*b)*c

Hence, ∗ is not associative.

Binary operations o : R × R → R defined as a o b = a, ∀ a, b ∈ R

a o b = a

b o a = b

a o b ≠ b o a

Hence, o is not commutative.

a o (b o c) = a o b = a

(a o b) o c = a o c = a

a o (b o c) ≠ (a o b) o c

Hence, o is associative.

Let’s check for a ∗ (b o c) = (a ∗ b) o (a ∗ c) a, b, c ∈ R

a ∗ (b o c) = a * b = |a-b|

(a ∗ b) o (a ∗ c) = |a-b| o |a-c| = |a-b|

Hence, a ∗ (b o c) = (a ∗ b) o (a ∗ c)

Now, let’s check for a o (b * c) = (a o b) * (a o c)

a o (b * c) = a

(a o b) * (a o c) = a * a = |a-a| = 0

Hence, a o (b * c) ≠ (a o b) * (a o c)

o does not distribute over ∗

Question 13. 

Given a non-empty set X, let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A^–1 = A. 

(Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ).

Solution:

Set X, such that P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X)

φ*A = (φ-A) U (A-φ) = φ U A = A

A*φ = (A-φ) U (φ-A) = A U φ = A

Hence, φ is the identity element for the operation * on P(X)

A*A = (A-A) U (A-A) = φ U φ = φ

⇒ A = A^-1

Hence, all the elements A of P(X) are invertible with A^–1 = A. 

Question 14. 

Define a binary operation ∗ on the set {0, 1, 2, 3, 4, 5} as 

 Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 – a being the inverse of a.

Solution:

Let the set x = {0, 1, 2, 3, 4, 5}

Let’s take i as identity element, where a*i = a = i*a ∀ a ∈ x

a*0 = a

0*a = a, when (a+0<6)

Hence, zero is the identity for this operation

An element a ∈ x is invertible if there exists b ∈ x such that a*b = b*a = 0

From above equations, we have

a = -b or b = 6-a

But, as x = {0, 1, 2, 3, 4, 5} and a,b∈ x. Then a≠-b

Hence, b = 6-a is the inverse of an element a∈ x

a≠0

a^-1 = 6-a

Question 15. 

Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be 

functions defined by f(x) = x2 – x, x ∈ A and  x ∈ A. Are f and g equal? Justify your answer. 

(Hint: One may note that two functions f : A → B and g : A → B such that f(a) = g (a) ∀ a ∈ A, are called equal functions).

Solution:

Given, f, g : A → B be functions defined by f(x) = x² – x, x ∈ A and g(x) 

  x ∈ A

At x = -1

f(0) = (-1)² – (-1) = 2

g(0) =  = 2

Here, f(-1) = g(-1) and 2=2

At x = 0

f(0) = 0² – 0 = 0

g(0) =  = 0

Here, f(0) = g(0) and 0=0

At x = 1

f(1) = 1² – 1 = 0

g(1) =  = 0

Here, f(1) = g(1) and 1=1

At x = 2

f(1) = 2² – 2 = 2

g(1) =  = 2

Here, f(2) = g(2) and 2=2

For, every c∈ A, f(c) = g(c)

Hence, f and g are equal functions.

Question 16. 

Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is

(A) 1 

(B) 2 

(C) 3 

(D) 4

Solution:

R = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1),(3,3)}

Reflexive : (1,1), (2,2), (3,3) ∈ R

Symmetric: (1,2), (2,1)∈ R and (1,3), (3,1) ∈ R

R is not Transitive because, (1,2), (1,3) ∈ R but (3,2) ∉R

So, if we will add (3,2) and (2,3) or both, then R will become transitive.

New, R = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}

Hence, A is the correct option.

Question 17. 

Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is

(A) 1 

(B) 2 

(C) 3 

(D) 4

Solution:

Smallest equivalence relations containing (1, 2):

R = {(1,1),(2,2),(1,2),(2,1),(3,3)}

or R = {(1,1),(2,2),(1,2),(2,1),(3,2),(2,3)(3,3)}

Hence, B is the correct option.

Question 18. 

Let f : R → R be the Signum Function defined as

and g : R → R be the Greatest Integer Function given by g (x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?

Solution:

Given, f : R → R and g : R → R

when x ∈ (0,1]

[x] = 1, when x=1

[x] = 0, when 0<x<1

Now, fog(x)=f(g(x)) = f([x])

 

And, Now gof(x) = g(f(x))

g(1) = [1] = 1

g(0) = [0] = 0

g(-1) = [-1] = -1

When x ∈ (0,1), fog = 0 and gof = 1. fog(1) ≠ gof(1)

Hence, fog and gof do not coincide in (0, 1].

Question 19. 

Number of binary operations on the set {a, b} are

(A) 10 

(B) 16 

(C) 20 

(D ) 8

Solution:

Let A = {a,b}

A x A = {a,b} x {a,b}

R = {(a,a),(a,b),(b,a),(b,b)}

Number of elements are 4.

Hence, the number of binary operations on the set will be 2⁴ = 16

Hence, B is the correct option.