21M.2.AHL.TZ1.7

pestleMathematicsAAHLPaper 221M· sl-5-10-indefinite-integration-reverse-chain-by-substitution, sl-2-4-key-features-of-graphs-intersections-using-technologysource ↗

A continuous random variable $X$ has the probability density function $f$ given by

$f (x) = \{\begin{cases} \frac{x}{\sqrt{{(x^{2} + k)}^{3}}} 0 \leq x \leq 4 \\ 0 otherwise \end{cases}$

where $k \in ℝ^{+}$ .

Show that $\sqrt{16 + k} - \sqrt{k} = \sqrt{k} \sqrt{16 + k}$ .

[5]

Find the value of $k$ .

[2]

Markscheme / solution

recognition of the need to integrate $\frac{x}{\sqrt{{(x^{2} + k)}^{3}}}$ (M1)

$\int \frac{x}{\sqrt{{(x^{2} + k)}^{3}}} d x (= 1)$

EITHER

$u = x^{2} + k \Rightarrow \frac{d u}{d x} = 2 x$ (or equivalent) (A1)

$\int \frac{x}{\sqrt{{(x^{2} + k)}^{3}}} d x = \frac{1}{2} \int u^{- \frac{3}{2}} d u$

$= - u^{- \frac{1}{2}} (+ c) (= - {(x^{2} + k)}^{- \frac{1}{2}} (+ c))$ A1

$\int \frac{x}{\sqrt{{(x^{2} + k)}^{3}}} d x = \frac{1}{2} \int \frac{2 x}{\sqrt{{(x^{2} + k)}^{3}}} d x$ (A1)

$= - {(x^{2} + k)}^{- \frac{1}{2}} (+ c)$ A1

THEN

attempt to use correct limits for their integrand and set equal to $1$ M1

${[- u^{- \frac{1}{2}}]}_{k}^{16 + k} = 1$ OR ${[- {(x^{2} + k)}^{- \frac{1}{2}}]}_{0}^{4} = 1$

$- {(16 + k)}^{- \frac{1}{2}} + k^{- \frac{1}{2}} = 1 (\Rightarrow \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{16 + k}} = 1)$ A1

$\sqrt{16 + k} - \sqrt{k} = \sqrt{k} \sqrt{16 + k}$ AG

[5 marks]

attempt to solve $\sqrt{16 + k} - \sqrt{k} = \sqrt{k} \sqrt{16 + k}$ (M1)

$k = 0.645038 \dots$

$= 0.645$ A1

[2 marks]

Examiners’ report

[N/A]