It is given that $\mathrm{f}(x) = \ln(a + x)$, $x \in \mathbb{R}$, $x -a$, where&nbs

Question

It is given that $\mathrm{f}(x) = \ln(a + x)$, $x \in \mathbb{R}$, $x > -a$, where $a$ is a constant. Using the standard series from the List of Formulae (MF27), find the series expansion for $\mathrm{f}(x)$, up to and including the term in $x^3$.[2] It is given that $a=1$. Hence, or otherwise, show that the series expansion of $\sin[\mathrm{f}(x)]$, up to and including the term in $x^3$ is given by $x - \frac{1}{2}x^2 + \frac{1}{6}x^3$.[2] Deduce the Maclaurin series for $\cos[\mathrm{f}(x)]$ up to and including the term in $x^2$.[2] Find $\displaystyle \displaystyle\int_{1}^{3} (x - \frac{1}{2}x^2 + \frac{1}{6}x^3) \, \mathrm{d}x$. Without the use of a calculator or any further calculation, explain, whether this value is a good approximation to the value of $\displaystyle \displaystyle\int_{1}^{3} \sin[\mathrm{f}(x)] \, \mathrm{d}x$.[2]

Accepted Answer

(a) $\ln a+\dfrac xa-\dfrac{x^2}{2a^2}+\dfrac{x^3}{3a^3}+\cdots$ (b) $x-\dfrac12x^2+\dfrac16x^3+\cdots$ (c) $1-\dfrac12x^2+\cdots$ (d) $3$; not a good approximation

2025 VJC P2 Q1

2025 VJC P2 Q1

Video Solution:

Solution:

Solution locked