Standard +0.3 This is a straightforward application of differentiation and Maclaurin series formula. Students need to differentiate ln(cos 3x) twice using chain rule, evaluate at x=0, then recognize that f(0)=0 and f'(0)=0 means the series starts with the x² term. The mechanics are routine for FP2 students who have practiced these techniques, though slightly above average difficulty due to the composite function and the small conceptual step of identifying which term is 'first'.
Given that \(f(x) = \ln(\cos 3x)\), find \(f'(0)\) and \(f''(0)\). Hence show that the first term in the Maclaurin series for \(f(x)\) is \(ax^2\), where the value of \(a\) is to be found. [4]
Use of standard cos and ln series can earn second M1 A1
[4]
If \(f''(0) = f'(0) = 0\) then M0
$f'(x) = \frac{-3\sin 3x}{\cos 3x} = -3\tan 3x \Rightarrow f'(0) = 0$ | M1 | For differentiating f(x) twice (y' as a function of a function)
$f''(x) = -9\sec^2 3x \Rightarrow f''(0) = -9$ | A1 | For correct f'(0) and f''(0) www (soi by correct expansion)
$\Rightarrow f(x) = -\frac{9}{2}x^2$ | M1 | For use of Maclaurin soi
| A1 | For correct series (condone $a = -\frac{9}{2}x^2$)
| [4] |
**ALT:** $\ln(\cos 3x) = \ln\left(1-\frac{1}{2}(3x)^2\right) = -\frac{9}{2}x^2$ | SC | Use of standard cos and ln series can earn second M1 A1
| [4] |
If $f''(0) = f'(0) = 0$ then M0
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Given that $f(x) = \ln(\cos 3x)$, find $f'(0)$ and $f''(0)$. Hence show that the first term in the Maclaurin series for $f(x)$ is $ax^2$, where the value of $a$ is to be found. [4]
\hfill \mbox{\textit{OCR FP2 2012 Q1 [4]}}