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Date	November 2018	Marks available	7	Reference code	18N.2.AHL.TZ0.H_2
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	H_2	Adapted from	N/A

Question

A function $f$ satisfies the conditions $f\left( 0 \right) = - 4$ , $f\left( 1 \right) = 0$ and its second derivative is $f''\left( x \right) = 15\sqrt x + \frac{1}{{{{\left( {x + 1} \right)}^2}}}$ , $x$ ≥ 0.

Find $f\left( x \right)$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$f'\left( x \right) = \int {\left( {15\sqrt x + \frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right)} \,{\text{d}}x = 10{x^{\frac{3}{2}}} - \frac{1}{{x + 1}}\left( { + c} \right)$ (M1)A1A1

Note: A1 for first term, A1 for second term. Withhold one A1 if extra terms are seen.

$f\left( x \right) = \int {\left( {10{x^{\frac{3}{2}}} - \frac{1}{{x + 1}} + c} \right)} \,{\text{d}}x = 4{x^{\frac{5}{2}}} - {\text{ln}}\left( {x + 1} \right) + cx + d$ A1

Note: Allow FT from incorrect $f'\left( x \right)$ if it is of the form $f'\left( x \right) = A{x^{\frac{3}{2}}} + \frac{B}{{x + 1}} + c$ .

Accept ${\text{ln}}\left| {x + 1} \right|$ .

attempt to use at least one boundary condition in their $f\left( x \right)$ (M1)

$x = 0$ , $y = - 4$

⇒ $d = - 4$ A1

$x = 1$ , $y = 0$

⇒ $0 = 4 - {\text{ln}}\,2 + c - 4$

⇒ $c = {\text{ln}}\,2\left( { = 0.693} \right)$ A1

$f\left( x \right) = 4{x^{\frac{5}{2}}} - {\text{ln}}\left( {x + 1} \right) + x\,{\text{ln}}\,2 - 4$

[7 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.10—Indefinite integration, reverse chain, by substitution

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