2024 CJC Promo Q6

2024 CJC Promo Q6

7 marks

It is given that \(y = \sin\left(\mathrm{e}^x - 1\right)\).

  1. Show that \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{\mathrm{d}y}{\mathrm{d}x} - y\mathrm{e}^{2x}\).[3]
  2. Hence, find the Maclaurin series of \(y\), up to and including the term in \(x^2\).[2]
  3. Using the Maclaurin series found in part (b), determine the series expansion of \(\sqrt{1 + \sin\left(\mathrm{e}^x - 1\right)}\), up to and including the term in \(x^2\).[2]

Solution:

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Answer:(a) \(\frac{\mathrm d^2y}{\mathrm dx^2}=\frac{\mathrm dy}{\mathrm dx}-ye^{2x}\) (shown). (b) \(y=x+\frac12x^2+\cdots\). (c) \(1+\frac12x+\frac18x^2+\cdots\).

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