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Date	May 2022	Marks available	6	Reference code	22M.1.AHL.TZ2.8
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Find	Question number	8	Adapted from	N/A

Question

A continuous random variable $X$ $X$ has the probability density function

$f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{2}{(b - a) (c - a)} (x - a), & a \leq x \leq c \frac{2}{(b - a) (b - c)} (b - x), & c < x \leq b 0, & otherwise \end{matrix}$ $f (x) = \{\begin{matrix} \frac{2}{(b - a) (c - a)} (x - a), & a \leq x \leq c \\ \frac{2}{(b - a) (b - c)} (b - x), & c < x \leq b \\ 0, & otherwise \end{matrix}$ .

The following diagram shows the graph of $y = f (x)$ $y = f (x)$ for $a \leq x \leq b$ $a \leq x \leq b$ .

Given that $c \geq \frac{a + b}{2}$ $c \geq \frac{a + b}{2}$ , find an expression for the median of $X$ $X$ in terms of $a, b$ $a, b$ and $c$ $c$ .

Markscheme

let $m$ $m$ be the median

EITHER

attempts to find the area of the required triangle M1

base is $(m - a)$ $(m - a)$ (A1)

and height is $\frac{2}{(b - a) (c - a)} (m - a)$ $\frac{2}{(b - a) (c - a)} (m - a)$

area $= \frac{1}{2} (m - a) \times \frac{2}{(b - a) (c - a)} (m - a) (= \frac{{(m - a)}^{2}}{(b - a) (c - a)})$ $= \frac{1}{2} (m - a) \times \frac{2}{(b - a) (c - a)} (m - a) (= \frac{{(m - a)}^{2}}{(b - a) (c - a)})$ A1

attempts to integrate the correct function M1

$m \int a \frac{2}{(b - a) (c - a)} (x - a) d x$ $\int_{a}^{m} \frac{2}{(b - a) (c - a)} (x - a) d x$

$= \frac{2}{(b - a) (c - a)} {[\frac{1}{2} {(x - a)}^{2}]}_{a}^{m}$ $= \frac{2}{(b - a) (c - a)} {[\frac{1}{2} {(x - a)}^{2}]}_{a}^{m}$ OR $\frac{2}{(b - a) (c - a)} {[\frac{x^{2}}{2} - a x]}_{a}^{m}$ $\frac{2}{(b - a) (c - a)} {[\frac{x^{2}}{2} - a x]}_{a}^{m}$ A1A1

Note: Award A1 for correct integration and A1 for correct limits.

THEN

sets up (their) $m \int a \frac{2}{(b - a) (c - a)} (x - a) d x$ $\int_{a}^{m} \frac{2}{(b - a) (c - a)} (x - a) d x$ or area $= \frac{1}{2}$ $= \frac{1}{2}$ M1

Note: Award M0A0A0M1A0A0 if candidates conclude that $m > c$ $m > c$ and set up their area or sum of integrals $= \frac{1}{2}$ $= \frac{1}{2}$ .

$\frac{{(m - a)}^{2}}{(b - a) (c - a)} = \frac{1}{2}$ $\frac{{(m - a)}^{2}}{(b - a) (c - a)} = \frac{1}{2}$

$m = a \pm \sqrt{\frac{(b - a) (c - a)}{2}}$ $m = a \pm \sqrt{\frac{(b - a) (c - a)}{2}}$ (A1)

as $m > a$ $m > a$ , rejects $m = a - \sqrt{\frac{(b - a) (c - a)}{2}}$ $m = a - \sqrt{\frac{(b - a) (c - a)}{2}}$

so $m = a + \sqrt{\frac{(b - a) (c - a)}{2}}$ $m = a + \sqrt{\frac{(b - a) (c - a)}{2}}$ A1

[6 marks]

Examiners report

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.7—Solutions of quadratic equations and inequalities, discriminant and nature of roots

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22M.2.AHL.TZ1.8a:
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22M.2.AHL.TZ1.8b:
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