jee-advanced 2019 Q3

jee-advanced · India · paper2 Reciprocal Trig & Identities
For non-negative integers $n$, let $$f(n) = \frac{\sum_{k=0}^{n} \sin\left(\frac{k+1}{n+2}\pi\right)\sin\left(\frac{k+2}{n+2}\pi\right)}{\sum_{k=0}^{n} \sin^2\left(\frac{k+1}{n+2}\pi\right)}$$
Assuming $\cos^{-1}x$ takes values in $[0, \pi]$, which of the following options is/are correct?
(A) $f(4) = \frac{\sqrt{3}}{2}$
(B) $\lim_{n\rightarrow\infty} f(n) = \frac{1}{2}$
(C) If $\alpha = \tan(\cos^{-1}f(6))$, then $\alpha^2 + 2\alpha - 1 = 0$
(D) $\sin(7\cos^{-1}f(5)) = 0$
For non-negative integers $n$, let
$$f(n) = \frac{\sum_{k=0}^{n} \sin\left(\frac{k+1}{n+2}\pi\right)\sin\left(\frac{k+2}{n+2}\pi\right)}{\sum_{k=0}^{n} \sin^2\left(\frac{k+1}{n+2}\pi\right)}$$

Assuming $\cos^{-1}x$ takes values in $[0, \pi]$, which of the following options is/are correct?

(A) $f(4) = \frac{\sqrt{3}}{2}$

(B) $\lim_{n\rightarrow\infty} f(n) = \frac{1}{2}$

(C) If $\alpha = \tan(\cos^{-1}f(6))$, then $\alpha^2 + 2\alpha - 1 = 0$

(D) $\sin(7\cos^{-1}f(5)) = 0$