QJEEAdv19Ma082c: JEE Advanced 2019 Mathematics - Class 11 JEEAdv19Ma082c Solution: For non-negative integers n, let Assuming cos x takes values in …

Question

Verified Solution • 2025–26 Session moderate

For non-negative integers n, let

f(n) = {{\sum\limits_{k = 0}^n {\sin \left( {{{k + 1} \over {n + 2}}\pi } \right)} \sin \left( {{{k + 2} \over {n + 2}}\pi } \right)} \over {\sum\limits_{k = 0}^n {{{\sin }^2}\left( {{{k + 1} \over {n + 2}}\pi } \right)} }}

Assuming cos^$-1$ x takes values in [0,

\pi

], which of the following options is/are correct?

Accepted Answer

It is given, that for non-negative integers 'n',

f(n) = {{\sum\limits_{k = 0}^n {\sin \left( {{{k + 1} \over {n + 2}}\pi } \right)} \sin \left( {{{k + 2} \over {n + 2}}\pi } \right)} \over {\sum\limits_{k = 0}^n {{{\sin }^2}\left( {{{k + 1} \over {n + 2}}\pi } \right)} }}

= {{\sum\limits_{k = 0}^n {\left( {\cos {\pi \over {n + 2}} - \cos \left( {{{2k + 3} \over {n + 2}}\pi } \right)} \right)} } \over {\sum\limits_{k = 0}^n {\left( {1 - \cos \left( {{{2k + 2} \over {n + 2}}\pi } \right)} \right)} }}

[

\because

2\sin A\sin B = \cos (A - B) - \cos (A + B)\,and\,2si{n^2}A = 1 - \cos 2A

]

={{\left( {\cos \left( {{\pi \over {n + 2}}} \right)} \right)\sum\limits_{k = 0}^n {1 - \left\{ {\cos {{3\pi } \over {n + 2}} + \cos {{5\pi } \over {n + 2}} + \cos {{7\pi } \over {n + 2}} + ... + \cos \left( {{{2n + 3} \over {n + 2}}\pi } \right)} \right\}} } \over {\sum\limits_{k = 0}^n {1 - \left\{ {\cos {{2\pi } \over {n + 2}} + \cos {{4\pi } \over {n + 2}} + \cos {{6\pi } \over {n + 2}} + ... + \cos \left( {{{2n + 2} \over {n + 2}}\pi } \right)} \right\}} }}

={{(n + 1)\cos \left( {{\pi \over {n + 2}}} \right) - {{\sin \left( {{{n\pi } \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {{{n + 3} \over {n + 2}}\pi } \right)} \over {(n + 1) - {{\sin \left( {{{n\pi } \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {{{n + 3} \over {n + 2}}\pi } \right)}}

[

\because

\cos (\alpha ) + \cos (\alpha + \beta ) + cos(\alpha + 2\beta ) + ... + cos(\alpha + (n - 1)\beta ) = {{\sin \left( {{{n\beta } \over 2}} \right)} \over {\sin \left( {{\beta \over 2}} \right)}}\cos \left. {\left( {{{2\alpha + (n - 1)\beta } \over 2}} \right)} \right]

= {{(n + 1)\cos \left( {{\pi \over {n + 2}}} \right) - {{\sin \left( {\pi - {\pi \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {\pi + {\pi \over {n + 2}}} \right)} \over {(n + 1) - {{\sin \left( {\pi - {\pi \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos (\pi )}}

= {{(n + 1)\cos \left( {{\pi \over {n + 2}}} \right) + {{\sin \left( {{\pi \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {{\pi \over {n + 2}}} \right)} \over {(n + 1) + {{\sin \left( {{\pi \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}}}

= {{(n + 2)\cos \left( {{\pi \over {n + 2}}} \right)} \over {(n + 2)}} = \cos \left( {{\pi \over {n + 2}}} \right)

\Rightarrow f(n) = \cos \left( {{\pi \over {n + 2}}} \right)

Now,

f(6) = \cos \left( {{\pi \over 8}} \right)

\because

\alpha = \tan ({\cos ^{ - 1}}f((6))) = \tan \left( {{\pi \over 8}} \right)

{\begin{matrix} \cos^{- 1} \cos x = x \\ i f x \in (0, \frac{π}{2}) \end{matrix}}

= \sqrt 2 - 1

\Rightarrow (\alpha + 1) = \sqrt 2 \Rightarrow {(\alpha + 1)^2} = 2 \Rightarrow {\alpha ^2} + 2\alpha + 1 = 2

\Rightarrow {\alpha ^2} + 2\alpha - 1 = 0

Now,

f(4) = \cos \left( {{\pi \over {4 + 2}}} \right) = \cos \left( {{\pi \over 6}} \right) = {{\sqrt 3 } \over 2}

,

Now,

\sin (7{\cos ^{ - 1}}f(5)) = \sin \left( {7{{\cos }^{ - 1}}\left( {\cos \left( {{\pi \over {5 + 2}}} \right)} \right)} \right) = \sin \left( {7\left( {{\pi \over 7}} \right)} \right) = \sin \pi = 0

and Now,

\mathop {\lim }\limits_{n \to \infty } f(x) = \mathop {\lim }\limits_{n \to \infty } \cos {\pi \over {n + 2}} = \cos 0 = 1

Hence, options (a), (b) and (c) are correct.

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