| Date | None Specimen | Marks available | 3 | Reference code | SPNone.3ca.hl.TZ0.1 |
| Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
| Command term | Determine | Question number | 1 | Adapted from | N/A |
Question
The function f is defined on the domain \(\left] { - \frac{\pi }{2},\frac{\pi }{2}} \right[{\text{ by }}f(x) = \ln (1 + \sin x)\) .
Show that \(f''(x) = - \frac{1}{{(1 + \sin x)}}\) .
(i) Find the Maclaurin series for \(f(x)\) up to and including the term in \({x^4}\) .
(ii) Explain briefly why your result shows that f is neither an even function nor an odd function.
Determine the value of \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) - x}}{{{x^2}}}\).
Markscheme
\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\) A1
\(f''(x) = \frac{{ - \sin x(1 + \sin x) - \cos x\cos x}}{{{{(1 + \sin x)}^2}}}\) M1A1
\( = \frac{{ - \sin x - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{(1 + \sin x)}^2}}}\) A1
\( = - \frac{1}{{1 + \sin x}}\) AG
[4 marks]
(i) \(f'''(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\) A1
\({f^{(4)}}(x) = \frac{{ - \sin x{{(1 + \sin x)}^2} - 2(1 + \sin x){{\cos }^2}x}}{{{{(1 + \sin x)}^4}}}\) M1A1
\(f(0) = 0,{\text{ }}f'(0) = 1,{\text{ }}f''(0) = - 1\) M1
\(f'''(0) = 1,{\text{ }}{f^{(4)}}(0) = - 2\) A1
\(f(x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} - \frac{{{x^4}}}{{12}} + \ldots \) A1
(ii) the series contains even and odd powers of x R1
[7 marks]
\(\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) - x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + \ldots - x}}{{{x^2}}}\) M1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ - 1}}{2} + \frac{x}{6} + \ldots }}{1}\) (A1)
\( = - \frac{1}{2}\) A1
Note: Use of l’Hopital’s Rule is also acceptable.
[3 marks]
