If f(x) = 2x + 3 and (f∘g)(x) = 6x − 1, then g(x) is
- (f∘g)(x) = f(g(x)) = 2g(x) + 3.
- Set this equal to 6x − 1: 2g(x) + 3 = 6x − 1, so 2g(x) = 6x − 4.
- Divide by 2: g(x) = 3x − 2. Check: 2(3x − 2) + 3 = 6x − 1.
Answer: g(x) = 3x − 2
Sets and functions
NUST NET (Engineering) · Maths · Numbers, sets and functions. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.
The idea
Set questions are counting questions. A Venn diagram shows the union A ∪ B (in A or B or both), the intersection A ∩ B (in both) and the complement A′ (not in A). When two groups overlap, the overlap is counted twice unless you subtract it once.
A function gives each input exactly one output. Its domain is the set of allowed inputs and its range is the set of outputs it actually produces. To find a domain, look for what must be avoided: a zero denominator, or a negative number under a square root.
A composite f∘g means g acts first, then f. An inverse undoes the function and exists only when the function is one-one.
Rules to remember
- n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
- A set with n elements has 2n subsets, of which 2n − 1 are proper subsets
- De Morgan's laws: (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′
- Domain: denominator ≠ 0; expression under a square root ≥ 0 (strictly > 0 if the root is in a denominator)
- (f∘g)(x) = f(g(x)): work out g first, then put the result into f
- Inverse: write y = f(x), swap x and y, solve for y; domain of f−1 = range of f
- Even: f(−x) = f(x). Odd: f(−x) = −f(x), so an odd function defined at 0 has f(0) = 0
Common mistake
Working out f∘g in the wrong order. (f∘g)(x) means f(g(x)), so g is applied first; f(x) × g(x) and g(f(x)) are different things.
Worked example
Practice questions
Try each one, then open the answer.
1. A set A has 128 subsets. The number of elements in A is
- A64
- B7
- C8
- D6
Show answer
Answer: B. A set with n elements has 2n subsets, and 27 = 128, so n = 7. Halving 128 to get 64 confuses subsets with elements.
2. If f(x) = 3x − 1 and g(x) = x2 + 2, then (f∘g)(1) is
- A8
- B6
- C5
- D3
Show answer
Answer: A. (f∘g)(1) = f(g(1)) = f(3) = 3(3) − 1 = 8. Working out g(f(1)) = g(2) = 6 uses the wrong order.
3. U = {1, 2, 3, ..., 10}, A is the set of primes in U and B is the set of odd numbers in U. Then A ∩ B′ is
- A{3, 5, 7}
- B{1, 9}
- C{2}
- D{1, 2, 9}
Show answer
Answer: C. A = {2, 3, 5, 7} and B′ is the set of even numbers {2, 4, 6, 8, 10}. The only even prime is 2. {3, 5, 7} is A ∩ B and {1, 9} is B − A.
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