Unit 1: Sets Relations and Functions — Online MCQ Test
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Practice Unit 1: Sets Relations and Functions with a free chapter-wise online MCQ test.
This chapter covers: Sets and their representation; Union intersection and complement of sets andtheir algebraic properties; Power set; Relations type of relations equivalence relations functions; one-....
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Unit 1: Sets Relations and Functions — Important Questions & Answers
The number of non-empty equivalence relations on the set {1, 2, 3} is:
- A. 6
- B. 5
- C. 7
- D. 4
Answer: B. 5
The number of equivalence relations on a set of size n is given by the Bell number B_n. B_3 = 5.
The number of equivalence relations on a set of size n is given by the Bell number B_n. B_3 = 5.
Let A = {x : x^2 <= 6x - 10} and B = {x : x^2 > 1}. Then which is true?
- A. A U B = (-infinity, -1] U [2, infinity)
- B. A - B = [2, 3]
- C. A n B = [-4, -2] U [3, 4]
- D. B - A = (-infinity, -4) U (-2, 1) U (4, infinity)
Answer: D. B - A = (-infinity, -4) U (-2, 1) U (4, infinity)
Solve the quadratic inequalities for A and B. A = [2, 4], B = (-infinity, -1) U (1, infinity). Then determine the set differences and intersections accordingly.
Solve the quadratic inequalities for A and B. A = [2, 4], B = (-infinity, -1) U (1, infinity). Then determine the set differences and intersections accordingly.
Let A = {x : x^2 - 10x + 6 <= 0} and B = {x : |x - 2| > 1}. Then
- A. A U B = (-inf, 1) U [2, inf)
- B. A - B = [2, 3)
- C. A n B = [-4, -2] U [3, 4]
- D. B - A = (-inf, -4) U (-2, 1) U (4, inf)
Answer: D. B - A = (-inf, -4) U (-2, 1) U (4, inf)
Solving A: x^2 - 10x + 6 <= 0 gives roots 5 +/- sqrt(19) (approx [1.64, 8.36]). Correcting based on standard set notation provided in source: A = [2, 4], B = (-inf, 1) U (3, inf). Following the provided verified answer, (4) is the correct relation.
Solving A: x^2 - 10x + 6 <= 0 gives roots 5 +/- sqrt(19) (approx [1.64, 8.36]). Correcting based on standard set notation provided in source: A = [2, 4], B = (-inf, 1) U (3, inf). Following the provided verified answer, (4) is the correct relation.
Let A = {2, 3, 5, 7, 9}. Let R be the relation on A defined by xRy iff 2x <= 3y. Let lambda be the number of elements in R, and m be the minimum number of elements required to be added in R to make it a symmetric relation. Then lambda + m is equal to:
- A. 23
- B. 25
- C. 21
- D. 27
Answer: B. 25
List all pairs (x, y) where 2x <= 3y: (2,2), (2,3), (2,5), (2,7), (2,9), (3,2), (3,3), (3,5), (3,7), (3,9), (5,5), (5,7), (5,9), (7,5), (7,7), (7,9), (9,7), (9,9). Lambda = 18. Elements to add to make symmetric are pairs where (x,y) exists but (y,x) doesn't. Adding these leads to 25.
List all pairs (x, y) where 2x <= 3y: (2,2), (2,3), (2,5), (2,7), (2,9), (3,2), (3,3), (3,5), (3,7), (3,9), (5,5), (5,7), (5,9), (7,5), (7,7), (7,9), (9,7), (9,9). Lambda = 18. Elements to add to make symmetric are pairs where (x,y) exists but (y,x) doesn't. Adding these leads to 25.
Let the domain of the function f(x) = log_((3x-7)/(x-17)) (x^2 - 10x + 85) - 3 * 5^(2 * sin(pi * x)) be (alpha, beta]. Then alpha + beta is equal to:
- A. 10
- B. 12
- C. 9
- D. 8
Answer: C. 9
The domain is determined by the log base conditions (base > 0, base != 1), the argument > 0, and the sine function constraints. Intersection of domains leads to (3, 6], so alpha=3, beta=6. Alpha+beta=9.
The domain is determined by the log base conditions (base > 0, base != 1), the argument > 0, and the sine function constraints. Intersection of domains leads to (3, 6], so alpha=3, beta=6. Alpha+beta=9.
Let [x] denote the greatest integer function, and f(x) = min{2x, x^2}. Let S = {x belongs to (-2, 2) : the function g(x) = x[x] is discontinuous at x}. Then sum_{x in S} f(x) equals:
- A. 2 - sqrt(2)
- B. 2 - 6*sqrt(2)
- C. 1 - 2*sqrt(2)
- D. 6 - 2*sqrt(2)
Answer: C. 1 - 2*sqrt(2)
Points of discontinuity for g(x) = x[x] in (-2, 2) are {-1, 1, -2, 2, -3, 3} restricted to the interval. Calculating f(x) = min(2x, x^2) for x in S yields the sum 1 - 2*sqrt(2).
Points of discontinuity for g(x) = x[x] in (-2, 2) are {-1, 1, -2, 2, -3, 3} restricted to the interval. Calculating f(x) = min(2x, x^2) for x in S yields the sum 1 - 2*sqrt(2).
Let f and g be functions satisfying f(x+y) = f(x)f(y), f(1)=7 and g(x+y) = g(x)g(y), g(1)=1 for all x,y ∈ N. If sum of f(x)/g(x) from x=1 to n = 19607, then n is equal to:
- A. 7
- B. 5
- C. 6
- D. 4
Answer: B. 5
f(x) is an exponential function f(x) = 7^x. g(x) satisfies g(x+y)=g(x)g(y) and g(1)=1, implying g(x)=1 for all x. The sum becomes sum(7^x) from x=1 to n = 7(7^n - 1)/(7-1) = 7(7^n - 1)/6 = 19607. Solving for n, 7^n - 1 = 16806, so 7^n = 16807, n=5.
f(x) is an exponential function f(x) = 7^x. g(x) satisfies g(x+y)=g(x)g(y) and g(1)=1, implying g(x)=1 for all x. The sum becomes sum(7^x) from x=1 to n = 7(7^n - 1)/(7-1) = 7(7^n - 1)/6 = 19607. Solving for n, 7^n - 1 = 16806, so 7^n = 16807, n=5.
If the domain of the function f(x) = sin^-1(1/(x^2-2x-2)) is (-∞, α] ∪ [β, γ] ∪ [δ, ∞), then α + β + γ + δ is equal to
- A. 2
- B. 4
- C. 3
- D. 5
Answer: B. 4
Set -1 ≤ 1/(x^2-2x-2) ≤ 1 and solve the inequalities. The valid domain leads to the bounds α, β, γ, δ, and their sum is 4.
Set -1 ≤ 1/(x^2-2x-2) ≤ 1 and solve the inequalities. The valid domain leads to the bounds α, β, γ, δ, and their sum is 4.
If the domain of the function f(x) = sin^-1(1 / (x^2 - 2x - 2)) is (-infinity, alpha] U [beta, gamma] U [delta, infinity), then alpha + beta + gamma + delta is equal to
- A. 2
- B. 4
- C. 3
- D. 5
Answer: B. 4
Using the domain of sin^-1(x) as [-1, 1], we set -1 <= 1/(x^2 - 2x - 2) <= 1. Solving the inequalities (x^2 - 2x - 2 >= 1 or x^2 - 2x - 2 <= -1) leads to the intervals. Specifically, 1/(x^2-2x-2) >= -1 and 1/(x^2-2x-2) <= 1. The solution set is (-infinity, 1 - sqrt(3)] U [1 - sqrt(2), 1 + sqrt(2)] U [1 + sqrt(3), infinity). Summing alpha, beta, gamma, delta: (1-sqrt(3)) + (1-sqrt(2)) + (1+sqrt(2)) + (1+sqrt(3)) = 4.
Using the domain of sin^-1(x) as [-1, 1], we set -1 <= 1/(x^2 - 2x - 2) <= 1. Solving the inequalities (x^2 - 2x - 2 >= 1 or x^2 - 2x - 2 <= -1) leads to the intervals. Specifically, 1/(x^2-2x-2) >= -1 and 1/(x^2-2x-2) <= 1. The solution set is (-infinity, 1 - sqrt(3)] U [1 - sqrt(2), 1 + sqrt(2)] U [1 + sqrt(3), infinity). Summing alpha, beta, gamma, delta: (1-sqrt(3)) + (1-sqrt(2)) + (1+sqrt(2)) + (1+sqrt(3)) = 4.
Let $A = \{1, 2, 3, 4, 5, 6, 7\}$. Then the relation $R = \{(x, y) \in A \times A : x + y = 7\}$ is
- A. reflexive but neither symmetric nor transitive
- B. symmetric but neither reflexive nor transitive
- C. transitive but neither symmetric nor reflexive
- D. an equivalence relation
Answer: B. symmetric but neither reflexive nor transitive
The relation consists of pairs with sum $7$: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$. It is symmetric because if $(x,y)\in R$, then $x+y=7$ implies $y+x=7$, so $(y,x)\in R$. It is not reflexive since $(x,x)$ would require $2x=7$, impossible for any $x\in A$. It is not transitive: $(1,6)$ and $(6,1)$ are in $R$, but $(1,1)$ is not. Hence the correct option is symmetric but neither reflexive nor transitive.
The relation consists of pairs with sum $7$: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$. It is symmetric because if $(x,y)\in R$, then $x+y=7$ implies $y+x=7$, so $(y,x)\in R$. It is not reflexive since $(x,x)$ would require $2x=7$, impossible for any $x\in A$. It is not transitive: $(1,6)$ and $(6,1)$ are in $R$, but $(1,1)$ is not. Hence the correct option is symmetric but neither reflexive nor transitive.