MTH202 Quiz 1 by Sahirullah

MTH202 QUIZ 01 SOLVED BY SAHIRULLAH

Q1. Vertical line test is used to determine that whether the graph of a relation is a function or not.

1. False

2. True ✔

Reason: The vertical line test checks if any vertical line cuts the graph more than once. If yes, it’s not a function.

Q2. In the matrix representation of an irreflexive relation all the entries in the main diagonal are ...

1. 2
2. 3

3. 0 ✔

4. 1

Reason: Irreflexive means no element relates to itself, so diagonal entries are 0.

Q3. Let A = {1,2,3,4} and R = {(1,1), (1,3), (2,4), (3,1), (4,2)}. Which statement about R is true?

1. R is not symmetric
2. R is transitive

3. R is symmetric ✔

4. R is reflexive

Reason: Every (a,b) in R has a corresponding (b,a). It’s not reflexive because (2,2), (3,3), (4,4) are missing.

Q4. Let A = {1,2,3,4} and R = {(1,3), (2,2), (2,4), (3,1), (4,2)}. Then R is:

1. Symmetric ✔

2. Antisymmetric

Reason: For every (a,b), (b,a) is also in R. Not antisymmetric because 1≠3 but both (1,3) and (3,1) exist. Discrete Mathematics - Complete Revision Guide

Q5. If x ≡ 17 (mod 5), which integer is a valid solution for x?

1. 12 ✔

2. -44
3. 8
4. 4

Reason: 17 mod 5 = 2, and 12 mod 5 = 2.

Q6. In matrix representation of a reflexive relation, the diagonal entries are always:

1. Transitive
2. Symmetric

3. 1 ✔ (Reflexive)

4. Null

Reason: Reflexive relations have (a,a) for all a, so diagonal = 1.

Q7. The function f(x) = 1/[(x+2)(x-2)] from Z to Z is not well defined because:

1. Function is defined at x=2 but not defined at x=-2
2. Function gives imaginary values for x < 0
3. Each input has two outputs

4. Function is not defined at x=-2 and x=2 ✔

Reason: Denominator becomes 0 at x = ±2, so f(x) is undefined there.

Q8. Let f = {(2, 6), (4, 2), (5, 1)} from X = {2, 4, 5} to Y = {1, 2, 4, 6}. The range of f is:

1. {1, 2, 4, 6}
2. {2, 4, 5}
3. {1, 4, 6}

4. {1, 2, 6} ✔

Reason: Range is the set of second elements: 6, 2, 1.

Q9. The union of R and S is R ∪ S = {(1,2), (1,3), (2,1), (2,2), (2,4), (3,3), (4,2), (4,3)}. Then R ∪ S is:

1. Not transitive ✔

2. Transitive

Reason: (1,2) and (2,4) are in R∪S but (1,4) is not, so it fails transitivity.