It is given that $y=\ln \left( 2-{{\mathrm{e}}^{-2x}} \right)$. Show that \(\frac{\mathr

Question

It is given that $y=\ln \left( 2-{{\mathrm{e}}^{-2x}} \right)$. Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=4{{\mathrm{e}}^{-y}}-2$.[2] Hence find the Maclaurin series for $y$, up to and including the term in ${{x}^{2}}$.[3] Using standard series from the List of Formulae (MF26), expand $\ln \left( 2-{{\mathrm{e}}^{-2x}} \right)$ as far as the term in ${{x}^{2}}$, and use this expansion as a check on the correctness of the series found in part (b). [4]

Accepted Answer

(b) $y=\left( 0 \right)+\left( 2 \right)x+\left( \frac{-8}{2!} \right){{x}^{2}}+\ldots =2x-4{{x}^{2}}+\ldots $ (c) $y=2x-4{{x}^{2}}+\ldots $ This is the same expression as found in part **(b)** and hence we can conclude that the expansion is correct.

2023 EJC Promo Q9

2023 EJC Promo Q9

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