In the triangle $ABC$, $AC=BC=1$ and angle $ACB=\left( \frac{1}{2}\pi +\theta \right)$ radians

Question

In the triangle $ABC$, $AC=BC=1$ and angle $ACB=\left( \frac{1}{2}\pi +\theta \right)$ radians. Given that $\theta $ is sufficiently small angle, show that $AB\approx \sqrt{2}+\frac{\sqrt{2}}{2}\theta -\frac{\sqrt{2}}{8}{{\theta }^{2}}$[5] Let $\mathrm{f}(x)=\ln (2-x)$. Write down the first three non-zero terms of the Maclaurin series for $\mathrm{f}(x)$, where $-2\le x<2$.[3] Denote the answer to part (i) by $\mathrm{g}(x)$. Find the range of values of $x$ for which the percentage error between the value of $\mathrm{g}(x)$ and the value of $\mathrm{f}(x)$ is within $10\%$.[2]

Accepted Answer

(a) $AB\approx\sqrt2+\frac{\sqrt2}{2}\theta-\frac{\sqrt2}{8}\theta^2$ (b)(i) $f(x)\approx\ln2-\frac12x-\frac18x^2$ (b)(ii) $-1.73<x<0.737$ (3 s.f.)

2021 TJC P2 Q3

2021 TJC P2 Q3

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