UPSC NDA II 2024 Mathematics Paper

Concept:To find the greatest value of a function on a closed interval, find critical points by setting the derivative to zero, then compare the function values at those points and at the endpoints.Explanation:Given $f(x)=\cos 2x + x$ on the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.Compute the derivative: $f'(x)=-2\sin 2x + 1$.Set $f'(x)=0$ to find critical points: $-2\sin 2x+1=0 \Rightarrow \sin 2x = \frac12$.The solutions for $2x$ in $[0,2\pi]$ are $2x=\frac{\pi}{6}$ or $2x=\frac{5\pi}{6}$.Thus $x=\frac{\pi}{12}$ or $x=\frac{5\pi}{12}$.Only $x=\frac{\pi}{12}$ lies in the given interval.Evaluate $f$ at the critical point and endpoints:$f\left(-\frac{\pi}{2}\right)=\cos(-\pi)-\frac{\pi}{2}=-1-\frac{\pi}{2}$.$f\left(\frac{\pi}{2}\right)=\cos\pi+\frac{\pi}{2}=-1+\frac{\pi}{2}$.$f\left(\frac{\pi}{12}\right)=\cos\frac{\pi}{6}+\frac{\pi}{12}=\frac{\sqrt3}{2}+\frac{\pi}{12}$.The value $\frac{\sqrt3}{2}+\frac{\pi}{12}$ is greater than the endpoint values.Therefore the greatest value of $f(x)$ is $\frac{\sqrt3}{2}+\frac{\pi}{12}$.Answer:Option B: $\frac{\sqrt3}{2}+\frac{\pi}{12}$.