UPSC NDA I 2023 Mathematics Solved Paper

Concept:Use properties of definite integrals to simplify $I_1$ and $I_2$, then compute the ratio.Explanation:Step 1: Evaluate $I_1 = \int_{0}^{\pi} \frac{x}{1+\cos^2 x} \, dx$. Using $\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a+b-x)\,dx$, replace $x$ by $\pi - x$: $I_1 = \int_{0}^{\pi} \frac{\pi - x}{1+\cos^2(\pi - x)}\,dx = \int_{0}^{\pi} \frac{\pi - x}{1+\cos^2 x}\,dx$. Add the two expressions for $I_1$: $2I_1 = \int_{0}^{\pi} \frac{\pi}{1+\cos^2 x}\,dx$. Since $\frac{\pi}{1+\cos^2 x}$ is symmetric about $\frac{\pi}{2}$, $2I_1 = 2\int_{0}^{\pi/2} \frac{\pi}{1+\cos^2 x}\,dx$. Thus $I_1 = \pi \int_{0}^{\pi/2} \frac{1}{1+\cos^2 x}\,dx = \pi \int_{0}^{\pi/2} \frac{\sec^2 x}{1+\sec^2 x}\,dx = \pi \int_{0}^{\pi/2} \frac{\sec^2 x}{2+\tan^2 x}\,dx$. Substitute $t = \tan x$, $dt = \sec^2 x\,dx$, limits: $x=0 \to t=0$, $x=\pi/2 \to t=\infty$. Then $I_1 = \pi \int_{0}^{\infty} \frac{dt}{2+t^2} = \frac{\pi}{\sqrt{2}} \left[ \tan^{-1}\frac{t}{\sqrt{2}} \right]_{0}^{\infty} = \frac{\pi}{\sqrt{2}} \cdot \frac{\pi}{2} = \frac{\pi^2}{2\sqrt{2}}$. Step 2: Evaluate $I_2 = \int_{0}^{\pi} \frac{1}{1+\sin^2 x}\,dx$. Rewrite: $I_2 = \int_{0}^{\pi} \frac{\sec^2 x}{\sec^2 x + \tan^2 x}\,dx = \int_{0}^{\pi} \frac{\sec^2 x}{1+2\tan^2 x}\,dx$. Using symmetry, $I_2 = 2\int_{0}^{\pi/2} \frac{\sec^2 x}{1+2\tan^2 x}\,dx$. Substitute $t = \tan x$, $dt = \sec^2 x\,dx$, same limits: $I_2 = 2\int_{0}^{\infty} \frac{dt}{1+2t^2}$. Compute: $I_2 = 2 \cdot \frac{1}{\sqrt{2}} \left[ \tan^{-1}(\sqrt{2}t) \right]_{0}^{\infty} = \frac{2}{\sqrt{2}} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$. Step 3: Compute $\frac{I_1 + I_2}{I_1 - I_2}$. $I_1 = \frac{\pi^2}{2\sqrt{2}}$, $I_2 = \frac{\pi}{\sqrt{2}}$. $\frac{I_1 + I_2}{I_1 - I_2} = \frac{\frac{\pi^2}{2\sqrt{2}} + \frac{\pi}{\sqrt{2}}}{\frac{\pi^2}{2\sqrt{2}} - \frac{\pi}{\sqrt{2}}} = \frac{\frac{\pi}{2} + 1}{\frac{\pi}{2} - 1} = \frac{\pi + 2}{\pi - 2}$.Answer:$\frac{I_{1}+I_{2}}{I_{1}-I_{2}} = \frac{\pi + 2}{\pi - 2}$