Unit 8
9 marks · NormalLet the function be defined as , for all real values of . In a probability experiment, the results are summarised in the table:
State the domain and range of .
| Score (x) | Frequency (f) |
|---|---|
| 1 | 6 |
| 2 | 4 |
| 3 | 5 |
| 4 | 2 |
| 5 | 3 |
A score is chosen at random. Work out the probability that the score is in the domain of and that .
A score is selected at random. Another score is selected at random without replacement. Find the probability that both scores are greater than 3. Give your answer as a fraction in its lowest terms.
The function is now restricted to the domain . For this domain, is one-to-one or many-to-one? Explain your answer.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) Domain: All real x; Range: All real values f(x) with minimum value at vertex (2, -1), so range is f(x) ≥ -1.
(b) Scores in table: 1, 2, 3, 4, 5 are all in domain of f. Compute f(x) for these. f(1) = 0, f(2) = -1, f(3) = 0, f(4) = 3, f(5) = 8. The only x with f(x) < 0 is x = 2 (f(2) = -1). Frequency of 2 = 4. Total frequency = 20. Probability = 4/20 = 1/5.
(c) Scores greater than 3: 4 (2), 5 (3); total 5/20. Probability both drawn (without replacement): (5/20)×(4/19) = 20/380 = 1/19.
(d) Function is one-to-one for x ≥ 3, since quadratic opens upwards and is increasing on x ≥ 2.
Common mistakes: Incorrectly finding function range; Assuming discrete domain; Forgetting to reduce probabilities
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · States correct domain: all real x or all scores in the table.
B2 (a) · 1 mark · States correct range: f(x) ≥ -1.
M1 (b) · 1 mark · Calculates which x values satisfy f(x) < 0.
A1 (b) · 1 mark · Correctly uses frequency: probability is 4/20 = 1/5.
B1 (b) · 1 mark · Answer as 1/5 or 0.2.
M1 (c) · 1 mark · Identifies both probabilities without replacement needed.
A1 (c) · 1 mark · States final probability as 1/19 in lowest terms.
B1 (d) · 1 mark · Recognises f is one-to-one for x ≥ 3, and justifies that quadratic is increasing.
B2 (d) · 1 mark · Explanation is clear: f strictly increasing for x ≥ 3, so one-to-one.
Worked solution
a) f(x) is defined for all real x, but with discrete data it's the scores 1–5. As a full function: domain is all real x; the minimum value for x in reals is at vertex x = 2, f(2) = -1, so range is f(x) ≥ -1.
b) Only x = 2 gives f(x) < 0, and frequency for x = 2 is 4. All table scores are possible, so probability = 4/20 = 0.2 (or 1/5).
c) Probability both scores > 3: scores 4 (2) and 5 (3), total 5. First member: 5/20 = 1/4; Second member: 4/19; so final probability = (5/20) * (4/19) = 20/380 = 1/19.
d) For x ≥ 3, f(x) is a quadratic with vertex at x = 2; for x ≥ 3, f(x) is strictly increasing, so for each x value there is only one f(x) value. Therefore, f is one-to-one on this domain.