NCERT Solutions Class 12- 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) = , 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 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  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 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.