S2 June 2016 Q4

4. A continuous random variable \(X\) has cumulative distribution function \(\mathrm{F}(x)\) given by

\[\mathrm{F}(x) = \begin{cases} 0 & x \lt 2 \\ k(ax + bx^2 - x^3) & 2 \leqslant x \leqslant 3 \\ 1 & x \gt 3 \end{cases}\]

Given that the mode of \(X\) is \(\dfrac{8}{3}\)

(a) show that \(b = 8\) (6)

(b) find the value of \(k\). (4)

Mark scheme (a)

Mark (a) and (b) together – allow a missing \(k\) throughout

Scheme	Marks
\(\mathrm{f}(x) = ak + 2bkx - 3kx^2\)	M1
\(\left[\dfrac{\mathrm{df}(x)}{\mathrm{d}x} =\right] 2kb - 6kx\)	M1dA1
\(2kb - 6kx = 0\) \(k(2b - 6x) = 0\) \(2b - 6x = 0\)	M1d
\(2b - 6 \times \dfrac{8}{3} = 0\)	M1d
\(b = 8*\)	A1 cso
	(6)

M1: Attempting to differentiate F(\(x\)) at least one \(x^n \to x^{n-1}\)

M1d: Attempting to differentiate f(\(x\)) at least one \(x^n \to x^{n-1}\). Dependent on previous M mark being awarded.

A1: Condone missing \(\dfrac{\mathrm{df}(x)}{\mathrm{d}x}\)

M1d: Putting 2nd differential = 0 Dependent on previous Method mark being awarded

M1d: Subst \(x = \dfrac{8}{3}\). Allow with \(k\) in. Dependent on previous Method mark being awarded

A1: Answer given so must have been awarded all previous marks with no errors

Scheme	Marks
\(-3k\left(x^2 - \dfrac{2bx}{3} - \dfrac{a}{3}\right)\)	M1
\(-3k\left(\left(x - \dfrac{b}{3}\right)^2 - \left(\dfrac{b}{3}\right)^2 - \dfrac{a}{3}\right)\) or quoting \(\dfrac{-b}{2a}\)	M1d
\(-3k\left(x - \dfrac{b}{3}\right)^2 + \dfrac{b^2k}{3} + ak\)	A1
Max at \(x = \dfrac{b}{3}\)	M1d
\(\dfrac{b}{3} = \dfrac{8}{3}\)	M1d
\(b = 8*\)	A1 cso

M1: factorising by taking \(-3k\) out

M1: Attempting to complete the square dependent on previous M mark being awarded. \(\left(x - \dfrac{b}{3}\right)^2 \pm c\)

A1: Correct completed square form

M1d: Selecting their \(b/3\) Dependent on previous Method mark being awarded

M1: Putting their \(\dfrac{b}{3} = \dfrac{8}{3}\). Dependent on previous Method mark being awarded

A1: Answer given all steps must have shown all the required steps

SC if \(-b/2a\) quoted and not proved do not award the A marks. Max mark is M1M1A0M1M1A0

Mark scheme (b)

Scheme	Marks
\(\mathrm{F}(2) = 0\) eg \(\quad k(2a + 32 - 8) = 0\) Or \(\quad k(2a + 4b - 8) = 0\) oe	M1
\(a = -12\)	A1
\(\mathrm{F}(3) = 1\) eg \(\quad k(-36 + 72 - 27) = 1\) \(k(-36 + 9b - 27) = 1\) oe	M1
\(k = \dfrac{1}{9}\)	A1
	(4)
	(10 marks)

M1: Attempting to form an equation using F(2) = 0, or F(3) = 1 or F(3) – F(2) = 1. Need to subst in the \(x\) value and equate

A1: –12 - may be implied by \(k\) = 1/9. Do not award if the M1 is not given

M1: Forming an equation using two of F(2) = 0 or F(3) = 1 or F(3) – F(2) = 1

A1: Allow equivalent fractions or awrt 0.111

NB If you see \(k\) = 1/9 award full marks. You may award marks in part (b) for equations seen in (a)