It is given that $\mathrm{f}(x)=\ln \left( 1+\sin 3x \right)$. Show that $\mathrm{f}''(x)=\frac{k}{1

Question

It is given that $\mathrm{f}(x)=\ln \left( 1+\sin 3x \right)$. Show that $\mathrm{f}''(x)=\frac{k}{1+\sin 3x}$, where $k$ is a constant to be found. [3] Hence find the first three non-zero terms of the Maclaurin expansion of $\mathrm{f}(x)$. [4]

Accepted Answer

(i) $k=-9$; (ii) $f(x)=3x-\frac92x^2+\frac92x^3+\cdots$

N2020 P1 Q3

N2020 P1 Q3

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