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Date	May 2017	Marks available	6	Reference code	17M.1.SL.TZ2.S_5
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Find	Question number	S_5	Adapted from	N/A

Question

Let $f’(x) = \frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}$ . Given that $f(0) = 1$ , find $f(x)$ .

Markscheme

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

valid approach (M1)

eg $\,\,\,\,\,$ $\int {f'{\text{d}}x,{\text{ }}\int {\frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}{\text{d}}x} }$

correct integration by substitution/inspection A2

eg $\,\,\,\,\,$ $f(x) = - \frac{1}{4}{({x^3} + 1)^{ - 4}} + c,{\text{ }}\frac{{ - 1}}{{4{{({x^3} + 1)}^4}}}$

correct substitution into their integrated function (must include $c$ ) M1

eg $\,\,\,\,\,$ $1 = \frac{{ - 1}}{{4{{({0^3} + 1)}^4}}} + c,{\text{ }} - \frac{1}{4} + c = 1$

Note: Award M0 if candidates substitute into $f^{'}$ or $f^{″}$ .

$c = \frac{5}{4}$ (A1)

$f(x) = - \frac{1}{4}{({x^3} + 1)^{ - 4}} + \frac{5}{4}{\text{ }}\left( { = \frac{{ - 1}}{{4{{({x^3} + 1)}^4}}} + \frac{5}{4},{\text{ }}\frac{{5{{({x^3} + 1)}^4} - 1}}{{4{{({x^3} + 1)}^4}}}} \right)$ A1 N4

[6 marks]

Examiners report

[N/A]

Syllabus sections

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

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18N.1.SL.TZ0.S_6:
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17M.2.SL.TZ1.S_7a.ii:
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19N.2.SL.TZ0.S_10c:
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Find the total time taken for the two stages.
18N.2.AHL.TZ0.H_2:
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.

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19M.2.SL.TZ2.S_8d:
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22M.1.SL.TZ1.2b:
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22M.1.SL.TZ1.9b.ii:
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18M.1.SL.TZ1.S_5b:
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19M.2.SL.TZ2.S_8e.i:
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SPM.1.SL.TZ0.9d:
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18M.2.AHL.TZ1.H_5b:
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19M.1.SL.TZ2.S_10b:
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