Attempting AP Inter 2nd Year Maths Study Material Chapter 1 Relations and Functions MCQ helps students become familiar with objective questions.
AP Inter 2nd Year Maths Relations and Functions MCQ
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.
Answer:
2) R is reflexive and transitive but not symmetric.
Explanation:
Given R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3) (3, 2)}
Reflexive : (a, a) ∈ R v a ∈ A
i.e.,(1, 1), (2, 2), (3, 3), (4, 4) ∈ R
⇒ R is reflexive.
Symmetric : If (a, b) ∈ R then (b, a) ∈ R (1, 2) ∈ R but (2, 1) ∉ R
⇒ R is not symmetric.
Transitive : If (a, b) & (b, c) ∉ R then (a, c) ∉ R
⇒ R is transitive.
Hence option (2) is correct.
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
Answer:
3) (6, 8) ∈ R
Explanation:
(6, 8) ∈ R, 8 > 6 & 6 = 8 – 2
Hence option (3) is correct.
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.
Answer:
4) f is neither one-one nor onto.
Explanation:
f is not one-one because f(1) = f(-1) = 1 two different elements – 1 & 1 have same image 1. f is not onto because negative numbers in the codomain R has no pre-image.
Hence option (4) is correct.
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.
Answer:
1) f is one-one onto
Explanation:
f is one-one because f(x1) = f(x2) ⇒ x1 = x2
f is onto because for any y ∈ R ∃ x = \(\frac{y}{3}\) ∈ R
Hence option (1) is correct.
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
Answer:
1) 1
Explanation:
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)} is reflexive and symmetric but not transitive because (2, 1) & (1, 3) ∈ R but (2, 3) ∉ R.
Thus there is only one such relation.
Hence option (1) is correct.
Question 6.
Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
1) 1
2) 2
3) 3
4) 4
Answer:
2) 2
Explanation:
The number of equivalence relations containing (1,2) is 2.
R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R2 = A × A
Hence option (2) is correct.
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
Answer:
3) 24
Explanation:
The number of injective mappings that can be defined from A to B is mPn = 4P3 = 4 × 3 × 2 = 24
Hence option (3) is correct.
Question 8.
Let A = {3, 5}. Then the number of reflexive relations on set A is
1) 2
2) 4
3) 0
4) 8
Answer:
2) 4
Explanation:
The number of reflexive relations on the set A is
2n2 – n = 222 – 2 = 4 (∵ n(A) = 2 = n)
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) \(\frac{5}{2}\)
Answer:
2) 3
Explanation:
Given f(x) = \(\frac{2 x-1}{2}\), g(x) = x + 2
(gof)\(\left(\frac{3}{2}\right)\) = g(f(\(\left(\frac{3}{2}\right)\)) = g\(\left(\frac{2\left(\frac{3}{2}\right)-1}{2}\right)\)
= 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) \(\frac{4 x-3}{2}\)
Answer:
4) \(\frac{4 x-3}{2}\)
Explanation:
Let f(x) = y
⇒ \(\frac{1}{2}\) = y ⇒ 2x + 3 = 4y
⇒ 2x = 4y – 3
⇒ x = \(\frac{4 y-3}{2}\) ⇒ f-1(y) = \(\frac{4 y-3}{2}\)
∴ f-1(x) = \(\frac{4 x-3}{2}\)