UPSC NDA I 2025 Mathematics Paper

Concept:Use the given parametric relations to find an equation linking $x$, $y$, and $\frac{dy}{dx}$. Then differentiate that equation to get a relation involving $\frac{d^2y}{dx^2}$, and substitute into the required expression.Explanation:From $x = \sec\theta - \cos\theta$ and $y = \sec^4\theta - \cos^4\theta$, we obtain:$$\left(\frac{dy}{dx}\right)^2 = \frac{16(y^2 + 4)}{x^2 + 4}.$$Multiply both sides by $(x^2 + 4)$:$$(x^2+4)\left(\frac{dy}{dx}\right)^2 = 16(y^2+4).$$Differentiate both sides with respect to $x$ using the product rule:$$(x^2+4)\cdot 2\frac{dy}{dx}\frac{d^2y}{dx^2} + 2x\left(\frac{dy}{dx}\right)^2 = 16\cdot 2y\frac{dy}{dx}.$$Divide throughout by $2\frac{dy}{dx}$ (valid where $\frac{dy}{dx}\neq 0$):$$(x^2+4)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = 16y.$$Rearrange to get:$$(x^2+4)\frac{d^2y}{dx^2} - 16y = -x\frac{dy}{dx}.$$Now the required expression is:$$\frac{x^2+4}{y^2+4}\cdot\frac{dy}{dx}\left[(x^2+4)\frac{d^2y}{dx^2} - 16y\right].$$Substitute the relation we found:$$\frac{x^2+4}{y^2+4}\cdot\frac{dy}{dx}\left(-x\frac{dy}{dx}\right) = -x\cdot\frac{x^2+4}{y^2+4}\left(\frac{dy}{dx}\right)^2.$$Using $\left(\frac{dy}{dx}\right)^2 = \frac{16(y^2+4)}{x^2+4}$, the factors $(x^2+4)$ and $(y^2+4)$ cancel, giving:$$-x \cdot 16 = -16x.$$Answer:$\boxed{-16x}$ which corresponds to option C.