UPSC NDA I 2016 Mathematics Solved Paper

Concept:Use the property: $\displaystyle \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx$ to show that both sequences $\{a_{2n}\}$ and $\{a_{2n+1}\}$ are constant (equal to zero), hence each forms an arithmetic progression with common difference zero.Explanation:We are given $a_n = \int_{0}^{\pi} \frac{\sin^{2}((n+1)x)}{\sin 2x}\,dx$.First consider statement 1: put $n \to 2n$. Then $a_{2n} = \int_{0}^{\pi} \frac{\sin^{2}((2n+1)x)}{\sin 2x}\,dx$.Apply the property: $\int_{0}^{\pi} f(x)\,dx = \int_{0}^{\pi} f(\pi - x)\,dx$.Thus $a_{2n} = \int_{0}^{\pi} \frac{\sin^{2}[(2n+1)(\pi - x)]}{\sin 2(\pi - x)}\,dx = \int_{0}^{\pi} \frac{\sin^{2}[(2n+1)\pi - (2n+1)x]}{\sin(2\pi - 2x)}\,dx$.Since $\sin(2\pi - 2x) = -\sin 2x$ and $\sin[(2n+1)\pi - \theta] = \sin \theta$ (because $(2n+1)\pi$ is an odd multiple of $\pi$), we get $a_{2n} = \int_{0}^{\pi} \frac{\sin^{2}((2n+1)x)}{-\sin 2x}\,dx = -a_{2n}$.Hence $2a_{2n} = 0 \implies a_{2n} = 0$ for all $n$. So $\{a_{2n}\}$ is a constant sequence, which is an AP with common difference zero. Statement 1 is correct.Now statement 2: put $n \to 2n+1$. Then $a_{2n+1} = \int_{0}^{\pi} \frac{\sin^{2}((2n+2)x)}{\sin 2x}\,dx$.Apply the same property: $a_{2n+1} = \int_{0}^{\pi} \frac{\sin^{2}[(2n+2)(\pi - x)]}{\sin 2(\pi - x)}\,dx = \int_{0}^{\pi} \frac{\sin^{2}[(2n+2)\pi - (2n+2)x]}{-\sin 2x}\,dx$.Since $(2n+2)\pi$ is an even multiple of $\pi$, $\sin[(2n+2)\pi - \theta] = -\sin \theta$. Thus the numerator becomes $[-\sin((2n+2)x)]^{2} = \sin^{2}((2n+2)x)$. So $a_{2n+1} = \int_{0}^{\pi} \frac{\sin^{2}((2n+2)x)}{-\sin 2x}\,dx = -a_{2n+1}$.Hence $2a_{2n+1} = 0 \implies a_{2n+1} = 0$ for all $n$. Therefore $\{a_{2n+1}\}$ is also a constant sequence, an AP with common difference zero. Statement 2 is correct.Answer:Both statements 1 and 2 are correct.