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Date	November 2019	Marks available	5	Reference code	19N.1.AHL.TZ0.H_7
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Find and Hence	Question number	H_7	Adapted from	N/A

Question

Write $2x - {x^2}$ in the form $a{\left( {x - h} \right)^2} + k$ , where $a{\text{, }}h{\text{, }}k \in \mathbb{R}$ .

[2]

Hence, find the value of $\int_{\frac{1}{2}}^{\frac{3}{2}} {\frac{1}{{\sqrt {2x - {x^2}} }}} {\text{d}}x$ .

[5]

Markscheme

attempt to complete the square or multiplication and equating coefficients (M1)

$2x - {x^2} = - {\left( {x - 1} \right)^2} + 1$ A1

$a = - 1$ , $h = 1$ , $k = 1$

[2 marks]

use of their identity from part (a) $\left( {\int_{\frac{1}{2}}^{\frac{3}{2}} {\frac{1}{{\sqrt {1 - {{\left( {x - 1} \right)}^2}} }}} {\text{d}}x} \right)$ (M1)

$= \left[ {{\text{arc}}\,{\text{sin}}\left( {x - 1} \right)} \right]_{\frac{1}{2}}^{\frac{3}{2}}$ or $\left[ {{\text{arc}}\,{\text{sin}}\left( u \right)} \right]_{ - \frac{1}{2}}^{\frac{1}{2}}$ A1

Note: Condone lack of, or incorrect limits up to this point.

$= {\text{arc}}\,{\text{sin}}\left( {\frac{1}{2}} \right) - {\text{arc}}\,{\text{sin}}\left( { - \frac{1}{2}} \right)$ (M1)

$= \frac{\pi }{6} - \left( { - \frac{\pi }{6}} \right)$ (A1)

$= \frac{\pi }{3}$ A1

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » AHL 5.15—Further derivatives and indefinite integration of these, partial fractions

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