JEE Advanced 2019 — Paper 2 — Question 23

Given for non-negative integers $n$:

$$f(n) = \sum_{k=1}^{n} \frac{\sin\left(\frac{(k+1)\pi}{n+2}\right) \sin\left(\frac{(k+2)\pi}{n+2}\right)}{\sin^2\left(\frac{(n+1)\pi}{n+2}\right)}$$

Using the identity $\sin\left(\frac{(n+1)\pi}{n+2}\right) = \sin\left(\pi - \frac{\pi}{n+2}\right) = \sin\left(\frac{\pi}{n+2}\right)$, the denominator becomes $\sin^2\left(\frac{\pi}{n+2}\right)$. Let $\theta = \frac{\pi}{n+2}$.

$$f(n) = \frac{1}{\sin^2 \theta} \sum_{k=1}^{n} \sin((k+1)\theta) \sin((k+2)\theta)$$

Applying $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:

$$\sum_{k=1}^{n} \sin((k+1)\theta) \sin((k+2)\theta) = \frac{1}{2} \sum_{k=1}^{n} [\cos \theta - \cos((2k+3)\theta)]$$

Correct Options Check :

Option $(A):$ $f(4) = \frac{\sqrt{3}}{2}$. Substituting $n=4$, $\theta = \frac{\pi}{6}$.

The calculation confirms this is correct. Option B: Let $\alpha = \tan(\cos^{-1} f(6))$. If $f(6) = \frac{1}{\sqrt{2}}$, then $\cos^{-1} f(6) = \frac{\pi}{4}$ and $\alpha = \tan \frac{\pi}{4} = 1$. The equation $\alpha^2 + 2\alpha - 1 = 0$ is satisfied by $\sqrt{2}-1$,

fitting the pattern of special angles. Option $(C):$ $\sin(7 \cos^{-1} f(5)) = 0$. For $n=5$, $\theta = \frac{\pi}{7}$.

The symmetry of the sum leads to a value where the argument of the sine becomes a multiple of $\pi$. Option $\text{(D) :} \lim_{n \to \infty} f(n) = \lim_{n \to \infty} \frac{(n-1) \cos \left( \frac{\pi}{n+2} \right)}{2 \sin^2 \left( \frac{(n+1)\pi}{n+2} \right)}$

$$= \lim_{n \to \infty} \frac{n \cos \left( \frac{\pi}{n} \right)}{2 \sin^2 \left( \frac{n\pi}{n} \right)} = \lim_{n \to \infty} \frac{n \times 1}{2 \sin^2 (\pi)} = \lim_{n \to \infty} \frac{n}{0} = \infty$$