UPSC CDS II 2024 Math Paper

Concept:To find the maximum value of a trigonometric expression, treat $\sin\theta$ as a variable $x$ and maximize the quadratic function $8x - 4x^2$ where $0 \le x \le 1$.Explanation:Write the expression as $y = 8\sin\theta - 4\sin^2\theta$.Let $x = \sin\theta$, so $y = 8x - 4x^2$.Since $\sin\theta$ lies between $0$ and $1$, the maximum of $y$ occurs within $0 \le x \le 1$.Differentiate with respect to $x$: $\frac{dy}{dx} = 8 - 8x$.Set $\frac{dy}{dx} = 0$ to find critical point: $8 - 8x = 0 \Rightarrow x = 1$.Put $x = 1$ into $y$: $y = 8(1) - 4(1)^2 = 8 - 4 = 4$.Since $x=1$ is the upper bound of the domain and the quadratic opens downward, this gives the maximum value.Thus, the maximum value of $8\sin\theta - 4\sin^2\theta$ is $4$.Answer:4