It is given that the curve $y = \mathrm{f}(x)$ satisfies the equation \(\ln y = \frac{\pi}{4} - \t

Question

It is given that the curve $y = \mathrm{f}(x)$ satisfies the equation $\ln y = \frac{\pi}{4} - \tan^{-1}(e^x)$. Show that $(1 + e^{2x}) \frac{\mathrm{d}y}{\mathrm{d}x} + ye^x = 0$.[2] Find the Maclaurin series for $y$, up to and including the term in $x^2$.[5] Deduce the series expansion for $\frac{e^{\frac{\pi}{4} - \tan^{-1}(e^x)}}{\sqrt{1 - x}}$, up to and including the term in $x^2$.[4]

Accepted Answer

(a) Identity shown. (b) $y=1-\frac x2+\frac{x^2}{8}+\cdots$. (c) $1+\frac{x^2}{4}+\cdots$.

2025 CJC P1 Q7

2025 CJC P1 Q7

Solution:

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