Practice AP Inter 2nd Year Maths Study Material Chapter 1 Relations and Functions MCQ to identify your strengths and weak areas.
AP Inter 2nd Year Maths Relations and Functions MCQ
I. Select the correct option from the given choices.
Question 1.
Let R he the relation in the set {1, 2, 3, 4} given by I
R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)). Choose the correct answer.
1) R is reflexive and symmetric but not transitive.
2) R Is reflexive and transitive but not symmetric.
3) R is symmetric and transitive but not reflexive.
4)R is an equivalence relation.
Solution:
2) R Is reflexive and transitive but not symmetric.
Let A= {1, 2, 3, 4} and R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
(i) (a, a) ∈ R for every a ∈A. So, R is reflexive.
(ii) (1, 2) ∈ R but (2, 1)∉ R. So, R is not symmetric.
(iii) (a, b), (b, c) ∈ R⇒ (a, c) ∈ R ∀ a, b, c ∈ A. So, R is transitive.
R is reflexive and transitive but not symmetric.
Question 2.
Let R be the relation in the set N given by R = {(a, b): a = b – 2. b > 6}.
Choose the correct answer.
1) (2, 4) ∈ R
2) (3, 8) ∈ R
3) (6, 8) ∈ R
4) (8, 7) ∈ R
Solution:
3) (6, 8) ∈ R
Given set is the set of Natural numbers N
Given that b > 6 ⇒ b = 7, 8, 9, … and a = b – 2,
Now, b = 7 ⇒ a = 7 – 2 = 5; (5, 7) ∈ R
b = 8 ⇒ a = 8 – 2 = 6; (6, 8) ∈ R
Question 3.
Let f : R → R be defined as f(x) = x4. Choose the correct answer.
1) f is one-one onto
2) f is many-one onto
3) f is one-one but not onto
4) f is neither one-one nor onto.
Solution:
4) f is neither one-one nor onto.
f: R → R and f(x) = x4
⇒ f(-1) = (-1)4 =1; f(1) = 14 = 1
∴ f is not – one – one
Also x4 ≥ 0 (Domain) ⇒ Range = [0, ∞) co-domain ∀ x ∈ R
⇒ f is not onto ∴ f is neither one-one nor onto
Question 4.
Let f : R → R be defined as f (x) = 3x. Choose the correct answer.
1) f is one-one onto
2) f is many-one onto
3) f is one-one but not onto
4) f is neither one-one nor onto.
Solution:
1) f is one-one onto
f: R → R and f(x) = 3x . This is a linear function of the form f(x) = ax + b
⇒ f is both one-one and onto
Question 5.
Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
1) 1
2) 2
3) 3
4) 4
Solution:
1) 1
The given set is A= {1, 2, 3}
The smallest relation containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is given by R={(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}
(i) {(1, 1), (2, 2), (3, 3)} ∈ R ⇒R is reflexive.
(ii) {(1, 2), (2, l)} ∈ R and {(1, 3)(3, 1)} ∈ R ⇒ R is symmetric.
(iii) {(3, 1), (1, 2)} ∈ R but (3, 2) ∉ R ⇒ R is not transitive.
Now, if we add only any of the two pairs (3, 2) and (2, 3) (or both) to relation R,
then relation R will become transitive.
Hence, the total number of desired relations is one.
Question 6.
Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
1) 1
2) 2
3) 3
4) 4
Solution:
2) 2
The given set is A-{1, 2, 3}
The smallest equivalence relation containing (1, 2) is given by:
R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Now, we are left with only four other pairs (2, 3), (3, 2), (1, 3) and (3, 1).
If we add any one pair say(2, 3) to R1, then for symmetry we must add (3, 2), and for transitivity we are required to add (1, 3) and (3, 1).
Hence, the only equivalence relation (bigger than R1) is the universal relation. This shows that the total number of equivalence relations containing (1.2) is two.
Question 7.
Set A has 3 elements and the set B has 4 elements. Then the number of injective mappings that can be defined from A to B :
1) 144
2) 12
3) 24
4) 64
Solution:
3) 24
n(A) = 3, n(B) = 4
no.of injective mappings from A to B = n(B)Pn(A) = 4P3 = 4 × 3 × 2 = 24
Question 8.
Let A = {3, 5}. Then the number of reflexive relations on set A is
1) 2
2) 4
3) 0
4) 8
Solution:
2) 4
A = {3, 5}
A × A ={3, 5} × {3, 5} == {(3, 3), (5, 5), (3, 5), (5, 3)}
Here, reflexive relations are
R1 = {(3, 3) (5, 5)}; R2 = {(3, 3) (5, 5) (3, 5)};
R3 = {(3, 3) (5, 5) (5, 3)}; R4 = {(3, 3) (5, 5) (3, 5) (5, 3)};
∴ Number of reflexive relations on set A is 4.
Question 9.
Let f : R → R be the function defined by f(x) = \(\frac{2 x-1}{2}\) and g: R → R be another function defined by g(x) = x + 2, then (gof) \(\left(\frac{3}{2}\right)\) is
1) 2
2) 3
3) \(\frac{7}{2}\)
4) \(\frac{5}{2}\)
Solution:
2) 3
g of \(\left(\frac{3}{2}\right)=g\left[f\left(\frac{3}{2}\right)\right]=g\left[\frac{2 \times \frac{3}{2}-1}{2}\right]=g\left[\frac{2}{2}\right]\)
= g(1) = 1 + 2 = 3
Question 10.
If f(x) = \(\) then f-1(x) is .
1) \(\frac{4}{2 x+3}\)
2) \(\frac{2 x-3}{4}\)
3) 2x + 3
4) \(\frac{4 x-3}{2}\)
Solution:
4) \(\frac{4 x-3}{2}\)
Put f-1(x) = y …………. (1) ⇒ x = f(y)
x = \(\frac{2 y+3}{4}\) ⇒ 4x = 2y + 3 ⇒ 2y = 4x – 3 ⇒ y = \(\frac{4 x-3}{2}\)
∴ f-1(x) = \(\frac{4 x-3}{2}\) (∵ from(1))