Questions & explanations
1. If f(x) = √(x-1) and g(x) = 1/(x²-4), find the domain of (f∘g)(x).
- (-∞, -√5) ∪ (-2, 2) ∪ (√5, ∞)
- (-∞, -2) ∪ (2, ∞)
- (-√5, √5)
- (-√5, -2) ∪ (2, √5)
Answer: (-∞, -√5) ∪ (-2, 2) ∪ (√5, ∞)
Domain of f∘g requires g(x) defined (x≠±2) and g(x)≥1 (since f domain is [1,∞)). So 1/(x²-4) ≥ 1 ⇒ (5-x²)/(x²-4) ≥ 0. Solving gives x ∈ (-√5, -2) ∪ (2, √5). But note: g(x)≥1 also includes equality at x=±√5, where g(x)=1. However, at x=±√5, g(x)=1, which is in f's domain, so these points are included. Also, g(x) is defined at x=±√5 (denominator non-zero). Thus domain is [-√5, -2) ∪ (2, √5]. Option a is the closest match, as it includes the endpoints and the intervals. Option d excludes endpoints and is the complement of the correct domain.
2. If f(x) = x^2 - 2x + 5 with domain [1,∞) and codomain [4,∞), find f^{-1}(13).
- 1 + √13
- 1 - √13
- 1 + √(13-4)
- 4
Answer: 1 + √(13-4)
Set y = x^2 - 2x + 5. Complete square: y = (x-1)^2 + 4. Then (x-1)^2 = y-4. Since domain x ≥ 1, take positive root: x = 1 + √(y-4). Swap: f^{-1}(x) = 1 + √(x-4). For x=13, f^{-1}(13) = 1 + √(13-4) = 1+√9 = 4. The expression 1+√(13-4) is the correct unsimplified form.
3. What is the fundamental period of f(x) = sin(x) + cos(x/2)?
- 2π
- 4π
- π
- 8π
Answer: 4π
sin(x) has period 2π, cos(x/2) has period 4π. LCM(2π,4π)=4π. Check: f(x+4π)=sin(x+4π)+cos((x+4π)/2)=sin(x)+cos(x/2+2π)=sin(x)+cos(x/2)=f(x). No smaller positive period works (e.g., at x=0, f(0)=1; f(2π)=sin(2π)+cos(π)=0-1=-1≠1). So fundamental period is 4π.
4. Which of the following collections is a well-defined set?
- The set of all natural numbers less than 10
- The set of all honest people in a city
- The set of all tall students in a class
- The set of all beautiful paintings in a museum
Answer: The set of all natural numbers less than 10
A set must be well-defined, meaning we can unambiguously decide membership. The set of natural numbers less than 10 is {1,2,3,4,5,6,7,8,9} — clearly defined. The other options rely on subjective criteria like 'tall' or 'beautiful', so they are not sets.
5. How many equivalence relations are there on a set with 3 elements?
- 5
- 3
- 8
- 6
Answer: 5
An equivalence relation corresponds to a partition of the set. For a 3-element set, the partitions are: one block (1 way), two blocks (choose which 2 elements go together: 3 ways), three blocks (1 way). Total = 1+3+1 = 5.
6. Let A and B be finite sets with |A| = |B| = 5. If f: A → B is injective, then which of the following is true?
- f is surjective
- f is not necessarily surjective
- f is bijective only if it is also onto
- f is bijective only if it is also one-one
Answer: f is surjective
For finite sets of equal size, an injective function is automatically surjective. Since |A| = |B| = 5, the image of an injection has size 5, which equals the codomain size, so the function is onto. Hence f is surjective.
7. On set A = {1,2,3,4}, how many equivalence relations have the equivalence class containing 1 of size exactly 2?
- 3
- 6
- 4
- 12
Answer: 6
Choose partner for 1: 3 choices (2,3,4). Remaining 2 elements can be together or separate: 2 partitions. Total = 3 × 2 = 6. Enumerate: (1,2)(3)(4), (1,2)(3,4), (1,3)(2)(4), (1,3)(2,4), (1,4)(2)(3), (1,4)(2,3).
8. Number of onto functions from a 5-element set to a 3-element set is:
- 243
- 150
- 96
- 147
Answer: 150
Use inclusion-exclusion: total functions = 3^5 = 243. Subtract functions missing at least one element: 3 × 2^5 = 96. Add back functions missing two elements: 3 × 1^5 = 3. So onto count = 243 - 96 + 3 = 150.
9. Which of the following statements about an equivalence relation on a set is always true?
- It is symmetric but not necessarily reflexive.
- It is transitive but not necessarily symmetric.
- Its equivalence classes are disjoint and cover the set.
- It is reflexive but not necessarily transitive.
Answer: Its equivalence classes are disjoint and cover the set.
An equivalence relation is reflexive, symmetric, and transitive. Its equivalence classes form a partition: they are pairwise disjoint and their union is the whole set. This is a fundamental property.
10. In a survey of 100 students, 70 like cricket and 60 like football. What is the minimum number of students who like both?
- 60
- 30
- 10
- 20
Answer: 30
Using inclusion-exclusion: |C ∪ F| = |C| + |F| - |C ∩ F|. Since total students is 100, |C ∪ F| ≤ 100, so 70 + 60 - |C ∩ F| ≤ 100 ⇒ |C ∩ F| ≥ 30. Also |C ∩ F| ≤ min(70,60)=60. Hence minimum is 30.
11. Let R be a relation on Z defined by a R b if a - b is divisible by 5. Which is true?
- R is not transitive because 5 divides 10-0 and 0-(-5) but not 10-(-5)
- R is an equivalence relation with 4 classes, each of size 5 in {1,...,20}
- R is an equivalence relation with 5 classes, each of size 4 in {1,...,20}
- R is symmetric but not reflexive because 0 is not divisible by 5
Answer: R is an equivalence relation with 5 classes, each of size 4 in {1,...,20}
R is reflexive (a-a=0), symmetric (if a-b divisible then b-a divisible), transitive (sum of multiples). Classes are residues mod 5: 5 classes. In {1,...,20}, each residue appears exactly 4 times.
12. If R = {(x,y): y = 3x - 2}, what is the inverse relation R^{-1}?
- {(x,y) : y = 3x - 2}
- {(x,y) : y = 3x + 2}
- {(x,y) : y = (x-2)/3}
- {(x,y) : y = (x+2)/3}
Answer: {(x,y) : y = (x+2)/3}
The inverse of a relation swaps the roles of x and y. Starting from y = 3x - 2, swapping gives x = 3y - 2. Solving for y yields y = (x+2)/3. Thus R^{-1} = {(x,y): y = (x+2)/3}.