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Date May 2017 Marks available 6 Reference code 17M.1.sl.TZ2.5
Level SL only Paper 1 Time zone TZ2
Command term Find Question number 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

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 6 - Calculus » 6.4 » Integration by inspection, or substitution of the form \(\mathop \int \nolimits f\left( {g\left( x \right)} \right)g'\left( x \right){\text{d}}x\) .

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