Expand $\displaystyle \frac1{\sqrt{1+3x}}$ in ascending powers of $x$, up to and including the t

Question

Expand $\displaystyle \frac1{\sqrt{1+3x}}$ in ascending powers of $x$, up to and including the term in $x^2$. State the range of values of $x$ for which the expansion is valid.[3] It is given that $y+xy=2+\sin x$ and $y=\mathrm{f}(x)$. Find the Maclaurin series for $\mathrm{f}(x)$, up to and including the term in $x^2$.[4] Use your results from parts (a) and (b) to find the Maclaurin series for $\displaystyle \frac{\mathrm{f}(x)}{\sqrt{1+3x}}$, up to and including the term in $x^2$. [2] Given that $\theta$ is sufficiently small for $\theta^5$ and higher powers of $\theta$ to be neglected, show that $\displaystyle \frac{\mathrm{f}(\sin^2\theta)}{\sqrt{1+3\sin^2\theta}}\approx p+q\theta^2+r\theta^4$, where $p$, $q$ and $r$ are constants to be determined.[3]

Accepted Answer

(a) $1-\frac32x+\frac{27}{8}x^2$, $|x|<\frac13$ (b) $2-x+x^2$ (c) $2-4x+\frac{37}{4}x^2$ (d) $p=2$, $q=-4$, $r=\frac{127}{12}$

2026 HCI P1 Q9

2026 HCI P1 Q9

Solution:

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