Class 12 NCERT Solutions- Mathematics Part I – Chapter 1 Relations And Functions -Miscellaneous Exercise on Chapter 1

Question 1. Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = [Tex]\frac{x}{1+|x|}   [/Tex], x ∈ R is one one and onto function.

Solution:

As, it is mentioned here

f : R → {x ∈ R : – 1 < x < 1} defined by [Tex]f(x) = \frac{x}{1+|x|}  [/Tex], 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.

[Tex]f(p) = \frac{p}{1+|p|}[/Tex]

[Tex]f(q) = \frac{q}{1+|q|}[/Tex]

f(p) = f(q)

[Tex]\frac{p}{1+|p|} = \frac{q}{1+|q|}[/Tex]

[Tex]\frac{p}{1+p} = \frac{q}{1+q}[/Tex]

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

p = q

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

[Tex]f(p) = \frac{p}{1+|p|}[/Tex]

[Tex]f(q) = \frac{q}{1+|q|}[/Tex]

f(p) = f(q)

[Tex]\frac{p}{1+|p|} = \frac{q}{1+|q|}[/Tex]

[Tex]\frac{p}{1-p} = \frac{q}{1-q}[/Tex]

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

p = q

Case 3: When p is positive and q is negative

[Tex]f(p) = \frac{p}{1+|p|}[/Tex]

[Tex]f(q) = \frac{q}{1+|q|}[/Tex]

f(p) = f(q)

[Tex]\frac{p}{1+|p|} = \frac{q}{1+|q|}[/Tex]

[Tex]\frac{p}{1+p} = \frac{q}{1-q}[/Tex]

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.

[Tex]f(p) = \frac{p}{1+|p|}[/Tex]

[Tex]y = \frac{p}{1+p}[/Tex]

[Tex]p = \frac{y}{1-y} (y≠1)[/Tex]

Case 2: When p <0

[Tex]f(p) = \frac{p}{1+|p|}[/Tex]

[Tex]y = \frac{p}{1-p}[/Tex]

[Tex]p = \frac{y}{1+y} (y≠-1)[/Tex]

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 2. 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 3. 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 4. 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!.

Question 5. 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 [Tex]g(x) = 2|x-\frac{1}{2}|-1    [/Tex] 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) = [Tex]2|x-\frac{1}{2}|-1  [/Tex]  x ∈ A

At x = -1

f(0) = (-1)² – (-1) = 2

g(0) = [Tex]2|-1-\frac{1}{2}|-1  [/Tex] = 2

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

At x = 0

f(0) = 0² – 0 = 0

g(0) = [Tex]2|0-\frac{1}{2}|-1  [/Tex] = 0

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

At x = 1

f(1) = 1² – 1 = 0

g(1) = [Tex]2|1-\frac{1}{2}|-1  [/Tex] = 0

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

At x = 2

f(1) = 2² – 2 = 2

g(1) = [Tex]2|2-\frac{1}{2}|-1  [/Tex] = 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 6. 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 7. 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.

Deleted Questions

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

[Tex]x = \frac{y-7}{10} ∈ R[/Tex]

So, it means for y ∈ R, there exists [Tex]x = \frac{y-7}{10}[/Tex]

[Tex]f(x) = f(\frac{y-7}{10}) = 10(\frac{y-7}{10})+7 = y – 7 + 7 = y[/Tex]

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 [Tex]g(y) = \frac{y-7}{10}[/Tex]

[Tex]g o f = g(f(x)) = g(10x+7) = \frac{(10x+7)-7}{10} = \frac{10x}{10} = x[/Tex]

[Tex]f o g = f(g(x)) = f(\frac{x-7}{10}) = 10(\frac{x-7}{10})+7 = x – 7 + 7 = x[/Tex]

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

g : R → R is defined as [Tex]g(y) = \frac{y-7}{10}[/Tex]

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

[Tex]f(n)= \begin{cases} n-1, \hspace{0.2cm}n \hspace{0.2cm}is\hspace{0.2cm} odd\\ n+1,\hspace{0.2cm}n \hspace{0.2cm}is\hspace{0.2cm} even \end{cases}[/Tex]

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 [Tex]g(y)= \begin{cases} y-1, \hspace{0.2cm}y \hspace{0.2cm}is\hspace{0.2cm} odd\\ y+1,\hspace{0.2cm}y \hspace{0.2cm}is\hspace{0.2cm} even \end{cases}[/Tex]

f = g

Hence, The inverse of f is f itself

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

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.

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 [Tex]g(x)= \begin{cases} x-1, \hspace{0.2cm}x>1\\ 1,\hspace{0.2cm}x=1 \end{cases}[/Tex]

Solution:

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

Taking f(x) = x+1 and [Tex]g(x)= \begin{cases} x-1, \hspace{0.2cm}x>1\\ 1,\hspace{0.2cm}x=1 \end{cases}[/Tex]

 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.

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.

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.

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 ∗

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. 

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

[Tex]a*b= \begin{cases} a+b, \hspace{0.2cm}a+b<6\\ a+b-6,\hspace{0.2cm}a+b\geq6 \end{cases}[/Tex]

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

[Tex]i.e.,\begin{cases} a+b=0=b+a, \hspace{0.2cm}a+b<6\\ a+b-6=0=b+a-6,\hspace{0.2cm}a+b\geq6 \end{cases}[/Tex]

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

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

[Tex]f(x)= \begin{cases} 1, \hspace{0.2cm}x<0\\ 0,\hspace{0.2cm}x=0\\ -1\hspace{0.2cm}x>0 \end{cases}[/Tex]

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

[Tex]g(x)= \begin{cases} 1, \hspace{0.2cm}x=1\\ 0,\hspace{0.2cm}0<x<1 \end{cases}[/Tex]

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

[Tex]f([x])= \begin{cases} f(1), \hspace{0.2cm}x=1\\ f(0),\hspace{0.2cm}0<x<1 \end{cases} =  [/Tex] [Tex]\begin{cases} 1, \hspace{0.2cm}x=1\\ 1,\hspace{0.2cm}0<x<1 \end{cases}[/Tex]

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].

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.