It is given that $\mathrm{f}(x)=\ln(a+a\sin x)$ for \(-\dfrac{1}{2}\pi

Question

It is given that $\mathrm{f}(x)=\ln(a+a\sin x)$ for $-\dfrac{1}{2}\pi < x < \dfrac{1}{2}\pi$, where $a$ is a positive constant. Show that $\mathrm{f}''(x)=\dfrac{k}{1+\sin x}$, where $k$ is a constant to be found.[3] By further differentiation of the result in 9(a), find the Maclaurin series for $\mathrm{f}(x)$, up to and including the term in $x^3$.[4] It is now given that $a=2$. Use the series found in 9(b) to find an approximate value of $\displaystyle\int_0^{1.2}\mathrm{f}(x)\,\mathrm{d}x$, giving your answer to 2 decimal places.[2] Use a calculator to evaluate $\displaystyle\int_0^{1.2}\mathrm{f}(x)\,\mathrm{d}x$, giving your answer to 2 decimal places.[1] Explain why your answers to 9(c)(i) and 9(c)(ii) are different.[1]

Accepted Answer

(a) $k=-1$ (b) $\mathrm{f}(x)\approx\ln a+x-\dfrac{x^2}{2}+\dfrac{x^3}{6}$ (c)(i) $1.35$ (ii) $1.32$ (iii) The series is an approximation for values of $x$ close to $0$, and the integration runs up to $x=1.2$, which is not close to $0$

N2026 P1 Q9

N2026 P1 Q9