UPSC NDA I 2021 Mathematics Solved Paper

Concept:Use the trigonometric identity $\sin 2\theta = 2\sin\theta\cos\theta$ to rewrite the product as a single sine function. The maximum value of sine is $1$, so find the maximum of the expression.Explanation:Let $f(x) = \sin 2x \cdot \cos 2x$.Recognise that $\sin 2x \cos 2x$ is half of $\sin 4x$ because $\sin 4x = 2\sin 2x \cos 2x$.Thus, $f(x) = \frac{1}{2} \sin 4x$.The range of $\sin 4x$ is $[-1, 1]$.Therefore, the maximum value of $f(x)$ occurs when $\sin 4x = 1$.The maximum value is $\frac{1}{2} \times 1 = \frac{1}{2}$.This value is attained for infinitely many $x$, for example at $4x = \frac{\pi}{2} + 2k\pi$, i.e., $x = \frac{\pi}{8} + \frac{k\pi}{2}$ for integer $k$.Answer:$\dfrac{1}{2}$