UPSC NDA II 2015 Mathematics Paper

Concept:To find the maximum value of a function, compute its first derivative, set it to zero to find critical points, and then use the second derivative test: if the second derivative is negative at a critical point, that point is a maximum.Explanation:Given $f(x) = \left(\frac{1}{x}\right)^{2x^2}$.Take the natural logarithm: $\ln f(x) = 2x^2 \ln\left(\frac{1}{x}\right) = -2x^2 \ln x$.Differentiate both sides: $\frac{f'(x)}{f(x)} = -4x \ln x - 2x^2 \cdot \frac{1}{x} = -4x \ln x - 2x = -2x(2\ln x + 1)$.Set $f'(x) = 0$. Since $f(x) > 0$ for $x>0$, we have $-2x(2\ln x + 1) = 0$.$x=0$ is not in the domain; solving $2\ln x + 1 = 0$ gives $\ln x = -\frac{1}{2}$, so $x = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}}$.Now compute $f''(x)$. Write $f'(x) = -2x(2\ln x + 1) f(x)$. At $x = \frac{1}{\sqrt{e}}$, $f'(x)=0$, so $f''(x) = -2 f(x) \cdot \frac{d}{dx}[x(2\ln x + 1)]$.$\frac{d}{dx}[x(2\ln x + 1)] = (2\ln x + 1) + x \cdot \frac{2}{x} = 2\ln x + 1 + 2 = 2\ln x + 3$.At $x = \frac{1}{\sqrt{e}}$, $\ln x = -\frac{1}{2}$, so $2\ln x + 3 = 2\left(-\frac{1}{2}\right) + 3 = -1 + 3 = 2$.Thus $f''\left(\frac{1}{\sqrt{e}}\right) = -2 f\left(\frac{1}{\sqrt{e}}\right) \cdot 2 = -4 f\left(\frac{1}{\sqrt{e}}\right) < 0$ because $f(x) > 0$.Hence $x = \frac{1}{\sqrt{e}}$ is a point of maximum.Now to find critical points,$f'(x) = 0$$\left(-2x\right) x^{-2x^2} \left(2\ln x + x\right) = 0$$\Rightarrow 2 \ln x + 1 = 0$$\Rightarrow \ln x = -\frac{1}{2}$$\Rightarrow x = e^{-\frac{1}{2}}$$= \frac{1}{\sqrt{e}}$Second derivative,$f''(x) = \frac{d\left(f'(x)\right)}{dx}$$\Rightarrow \frac{d}{dx} \left( f(x) \left(-2x\left(2\ln x + 1\right)\right) \right)$$\Rightarrow \frac{d}{dx} \left( -2 f(x) \left(2x \ln x + x\right) \right)$$\Rightarrow -2 f'(x) \left(2x \ln x + x\right) - 2 f(x) \frac{d}{dx}\left(2x \ln x + x\right)$ …. (1)Answer:$x = \frac{1}{\sqrt{e}}$ (Option C)