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]