Given that $\ln \left( ky ight)={{ an }^{-1}}\left( kx ight)$, where $k$ is a non-zero constant

Question

Given that $\ln \left( ky ight)={{	an }^{-1}}\left( kx ight)$, where $k$ is a non-zero constant, show that $\left( 1+{{k}^{2}}{{x}^{2}} ight)\frac{\mathrm{d}y}{\mathrm{d}x}=ky$. By further differentiation of this result, find the Maclaurin’s series for $y$, in terms of $k$, up to and including the term in ${{x}^{3}}$. [5] Let $k=1$. By using a suitable substitution for $~x$, find, in terms of $\sqrt{3}$, an approximate value of ${{\mathrm{e}}^{\frac{\pi }{6}}}$. [4]

Accepted Answer

$e^{\pi/6} \approx \frac{7}{6} + \frac{17}{18\sqrt{3}}$

2013 HCI P2 Q2

2013 HCI P2 Q2

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