It is given that $y = \ln \sqrt{1 + \mathrm{e}^{3x}}$. Show that \(2\mathrm{e}^{2y} \fra

Question

It is given that $y = \ln \sqrt{1 + \mathrm{e}^{3x}}$. Show that $2\mathrm{e}^{2y} \frac{\mathrm{d}y}{\mathrm{d}x} = 3\mathrm{e}^{3x}$.[2] By further differentiation of the result in part (a), find the Maclaurin series for $y$, up to and including the term in $x^2$, giving the coefficients in exact form.[4] Deduce the Maclaurin series for $\ln \left( \frac{\sqrt{1 + \mathrm{e}^{-3x}}}{2} \right)$, up to and including the term in $x^2$.[2]

Accepted Answer

(a) $2\mathrm e^{2y}\dfrac{\mathrm dy}{\mathrm dx}=3\mathrm e^{3x}$ (b) $y=\dfrac12\ln2+\dfrac34x+\dfrac9{16}x^2+\cdots$ (c) $-\dfrac12\ln2-\dfrac34x+\dfrac9{16}x^2+\cdots$

2025 TJC P2 Q1

2025 TJC P2 Q1

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