202212231507 Solution to 1990-CE-AMATH-II-1

Let $f(x)=\sqrt{x^2+k}\sin 2x$ , where $k$ is a constant. If $f'(0)=1$ , find the value of $k$ .

Roughwork.

$\begin{aligned} f(x) & = \sqrt{x^2+k}\sin 2x \\ f'(x) & = \sqrt{x^2+k}\cdot \frac{\mathrm{d}}{\mathrm{d}x}(\sin 2x) + \sin 2x\cdot \frac{\mathrm{d}}{\mathrm{d}x} \big( \sqrt{x^2+k}\big) \\ & = \sqrt{x^2+k}\cdot (2\cos 2x) + \sin 2x\cdot \bigg(\frac{x}{\sqrt{x^2+k}}\bigg)\\ f'(0) & = \sqrt{(0)^2+k}\cdot (2\cos 2(0)) + \sin 2(0)\cdot \bigg(\frac{(0)}{\sqrt{(0)^2+k}}\bigg)\\ 1 & =: 2\sqrt{k} \\ k & = \frac{1}{4} \\ \end{aligned}$

This problem is not to be attempted.

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