UPSC NDA I 2018 Mathematics Solved Paper

Concept:The function $f(x) = \sin\left(x + \frac{\pi}{5}\right) + \cos\left(x + \frac{\pi}{5}\right)$ can be expressed as $f(x) = \sqrt{2}\,\sin\left(x + \frac{\pi}{5} + \frac{\pi}{4}\right) = \sqrt{2}\,\sin\left(x + \frac{9\pi}{20}\right)$.The maximum of $\sqrt{2}\,\sin\theta$ occurs when $\sin\theta = 1$.Explanation:We need the value of $x$ in $(0, \frac{\pi}{2})$ where $f(x)$ attains its maximum.Rewrite $f(x)$ using the identity $a\sin\theta+b\cos\theta = \sqrt{a^2+b^2}\,\sin(\theta+\phi)$ with $\phi = \frac{\pi}{4}$.Here $a=1$, $b=1$, so $f(x) = \sqrt{2}\,\sin\left(x + \frac{\pi}{5} + \frac{\pi}{4}\right) = \sqrt{2}\,\sin\left(x + \frac{9\pi}{20}\right)$.For maximum, $\sin\left(x + \frac{9\pi}{20}\right) = 1$.Thus $x + \frac{9\pi}{20} = \frac{\pi}{2} + 2n\pi$, but within $(0,\frac{\pi}{2})$ we take $n=0$: $x + \frac{9\pi}{20} = \frac{\pi}{2}$.Solve: $x = \frac{\pi}{2} - \frac{9\pi}{20} = \frac{10\pi - 9\pi}{20} = \frac{\pi}{20}$.Alternatively, differentiate: $f'(x) = \cos\left(x+\frac{\pi}{5}\right) - \sin\left(x+\frac{\pi}{5}\right) = 0$ gives $\tan\left(x+\frac{\pi}{5}\right)=1$, so $x+\frac{\pi}{5}=\frac{\pi}{4}$, yielding $x=\frac{\pi}{20}$.Second derivative negative confirms maximum.Answer:The maximum is attained at $x = \frac{\pi}{20}$, which corresponds to option A.