UPSC NDA II 2023 Mathematics Solved Paper

Concept:The maximum value of a function $y = f(x)$ is found by setting its first derivative $f'(x)$ to zero and verifying that the second derivative $f''(x)$ is negative at the critical point.Explanation:Given $y = \sqrt{9 - 4x^2}$.Differentiate with respect to $x$: $\frac{dy}{dx} = \frac{1}{2\sqrt{9 - 4x^2}} \cdot (-8x) = \frac{-4x}{\sqrt{9 - 4x^2}}$.Set $\frac{dy}{dx} = 0$: $\frac{-4x}{\sqrt{9 - 4x^2}} = 0 \Rightarrow x = 0$.Now find the second derivative:$\frac{d^2y}{dx^2} = \frac{(-4)\sqrt{9 - 4x^2} - (-4x)\cdot \frac{-8x}{2\sqrt{9 - 4x^2}}}{(9 - 4x^2)}$.Simplify numerator: $-4\sqrt{9 - 4x^2} - \frac{16x^2}{\sqrt{9 - 4x^2}} = \frac{-4(9 - 4x^2) - 16x^2}{\sqrt{9 - 4x^2}} = \frac{-36 + 16x^2 - 16x^2}{\sqrt{9 - 4x^2}} = \frac{-36}{\sqrt{9 - 4x^2}}$.Thus $\frac{d^2y}{dx^2} = \frac{-36}{(9 - 4x^2)^{3/2}}$.At $x = 0$: $\frac{d^2y}{dx^2} = \frac{-36}{9^{3/2}} = \frac{-36}{27} = -\frac{4}{3} < 0$.Since the second derivative is negative, $y$ attains a maximum at $x = 0$.Maximum value: $y = \sqrt{9 - 4(0)^2} = \sqrt{9} = 3$.Answer:3