The diagram shows the graph of the probability density function, f, of a random variable X which takes values between 0 and 4 only. Between these two values the graph is a straight line. Show that f(x) = kx for 0 ≤ x ≤ 4, where k is a constant to be determined.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [1] (½ × 4 × a = 1 Or let f(x) = kx and attempt ∫⁴₀ kx dx = 1.) For use of area = 1.
2. [1] [a = ½] f(x) = ⅛x (Or kx²/2 |⁴₀ = 1; 8k = 1; k = ⅛.) f(x) = ⅛x or k = ⅛.
3. [1] ∫⁴₀ x × (their f(x)) dx Attempt to integrate x × their f(x). Ignore limits accept in terms of k.
4. [1] x³/24 |⁴₀ Their integral and correct limits accept in terms of k.
5. [1] = 64/3 or 2.67 (3 sf) Note: Final answer of 64k/3 scores 2/3.
6. [1] (∫¹₀ g(w)dw = ½ i.e. (½ * w * (2/a - (2/a²)w))dw = ½, or integral from 1 to a. g(w) must be linear of form g(w) = mw (+c). Or area attempt: attempt to calculate heights using their linear equation (h₁=2/a and h₂= −2/a² + 2/a) and use in either area trapezium = 0.5, or area trapezium =area small triangle or area small triangle = 0.5 . Area trapezium =1/2 × 1 (2/a + −2/a² + 2/a) Area triangle = 1/2(a – 1)(–2/a² + 2/a)) For use of area = 1.
7. [1] a - 1/a = 1/√2 Or a²-4a + 2 = 0. Any correct equation in a, a not in denominator.
8. [1] a = 2 + √2 = 3.41