Given that $y = \sqrt{\ln(x + \mathrm{e})}$ for $x 1 - \mathrm{e}$, show that \[2

Question

Given that $y = \sqrt{\ln(x + \mathrm{e})}$ for $x > 1 - \mathrm{e}$, show that $$2y\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{-y^2}.$$ Hence find the first three terms of the Maclaurin expansion of $y$. Give the coefficients as exact fractions in their simplest form.[4] Using standard series from the List of Formulae (MF27), expand $\ln(x + \mathrm{e})$ as far as the term in $x^2$. Hence, verify the correctness of your expansion in (a), showing your working clearly.[4] Use your expansion in part (a) to find an approximation to $\sqrt{\ln\left(\frac{1 + 10\mathrm{e}}{10}\right)}$, leaving your answer in terms of $\mathrm{e}$.[2]

Accepted Answer

(a) $1+\dfrac{x}{2\mathrm e}-\dfrac{3x^2}{8\mathrm e^2}+O(x^3)$ (b) $\ln(x+\mathrm e)=1+\dfrac x{\mathrm e}-\dfrac{x^2}{2\mathrm e^2}+O(x^3)$ (c) $1+\dfrac1{20\mathrm e}-\dfrac3{800\mathrm e^2}$

2024 NYJC Promo Q8

2024 NYJC Promo Q8

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