JEE Advance - Mathematics (2019 - Paper 2 Offline - No. 1)
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?
$$f(n) = {{\sum\limits_{k = 0}^n {\sin \left( {{{k + 1} \over {n + 2}}\pi } \right)} \sin \left( {{{k + 2} \over {n + 2}}\pi } \right)} \over {\sum\limits_{k = 0}^n {{{\sin }^2}\left( {{{k + 1} \over {n + 2}}\pi } \right)} }}$$
Assuming cos$$-1$$ x takes values in [0, $$\pi $$], which of the following options is/are correct?
If $$\alpha $$ = tan(cos$$-$$1 f(6)), then $$\alpha $$2 + 2$$\alpha $$ $$-$$1 = 0
$$f(4) = {{\sqrt 3 } \over 2}$$
sin(7 cos$$-$$1 f(5)) = 0
$$\mathop {\lim }\limits_{n \to \infty } \,f(n) = {1 \over 2}$$
Explanation
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, $$f(6) = \cos \left( {{\pi \over 8}} \right)$$
$$ \because $$ $$\alpha = \tan ({\cos ^{ - 1}}f((6))) = \tan \left( {{\pi \over 8}} \right)$$
$$\left\{ \matrix{ {\cos ^{ - 1}}\cos x = x \hfill \cr if\,x \in \left( {0,\,{\pi \over 2}} \right) \hfill \cr} \right\}$$
$$ = \sqrt 2 - 1$$
$$ \Rightarrow (\alpha + 1) = \sqrt 2 \Rightarrow {(\alpha + 1)^2} = 2 \Rightarrow {\alpha ^2} + 2\alpha + 1 = 2$$
$$ \Rightarrow {\alpha ^2} + 2\alpha - 1 = 0$$
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.
$$f(n) = {{\sum\limits_{k = 0}^n {\sin \left( {{{k + 1} \over {n + 2}}\pi } \right)} \sin \left( {{{k + 2} \over {n + 2}}\pi } \right)} \over {\sum\limits_{k = 0}^n {{{\sin }^2}\left( {{{k + 1} \over {n + 2}}\pi } \right)} }} $$
$$= {{\sum\limits_{k = 0}^n {\left( {\cos {\pi \over {n + 2}} - \cos \left( {{{2k + 3} \over {n + 2}}\pi } \right)} \right)} } \over {\sum\limits_{k = 0}^n {\left( {1 - \cos \left( {{{2k + 2} \over {n + 2}}\pi } \right)} \right)} }}$$
[$$ \because $$ $$2\sin A\sin B = \cos (A - B) - \cos (A + B)\,and\,2si{n^2}A = 1 - \cos 2A$$]
$$={{\left( {\cos \left( {{\pi \over {n + 2}}} \right)} \right)\sum\limits_{k = 0}^n {1 - \left\{ {\cos {{3\pi } \over {n + 2}} + \cos {{5\pi } \over {n + 2}} + \cos {{7\pi } \over {n + 2}} + ... + \cos \left( {{{2n + 3} \over {n + 2}}\pi } \right)} \right\}} } \over {\sum\limits_{k = 0}^n {1 - \left\{ {\cos {{2\pi } \over {n + 2}} + \cos {{4\pi } \over {n + 2}} + \cos {{6\pi } \over {n + 2}} + ... + \cos \left( {{{2n + 2} \over {n + 2}}\pi } \right)} \right\}} }}$$
$$={{(n + 1)\cos \left( {{\pi \over {n + 2}}} \right) - {{\sin \left( {{{n\pi } \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {{{n + 3} \over {n + 2}}\pi } \right)} \over {(n + 1) - {{\sin \left( {{{n\pi } \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {{{n + 3} \over {n + 2}}\pi } \right)}}$$
[$$ \because $$ $$\cos (\alpha ) + \cos (\alpha + \beta ) + cos(\alpha + 2\beta ) + ... + cos(\alpha + (n - 1)\beta ) = {{\sin \left( {{{n\beta } \over 2}} \right)} \over {\sin \left( {{\beta \over 2}} \right)}}\cos \left. {\left( {{{2\alpha + (n - 1)\beta } \over 2}} \right)} \right]$$
$$ = {{(n + 1)\cos \left( {{\pi \over {n + 2}}} \right) - {{\sin \left( {\pi - {\pi \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {\pi + {\pi \over {n + 2}}} \right)} \over {(n + 1) - {{\sin \left( {\pi - {\pi \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos (\pi )}}$$
$$ = {{(n + 1)\cos \left( {{\pi \over {n + 2}}} \right) + {{\sin \left( {{\pi \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {{\pi \over {n + 2}}} \right)} \over {(n + 1) + {{\sin \left( {{\pi \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}}}$$
$$ = {{(n + 2)\cos \left( {{\pi \over {n + 2}}} \right)} \over {(n + 2)}} = \cos \left( {{\pi \over {n + 2}}} \right)$$
$$ \Rightarrow f(n) = \cos \left( {{\pi \over {n + 2}}} \right)$$
Now, $$f(6) = \cos \left( {{\pi \over 8}} \right)$$
$$ \because $$ $$\alpha = \tan ({\cos ^{ - 1}}f((6))) = \tan \left( {{\pi \over 8}} \right)$$
$$\left\{ \matrix{ {\cos ^{ - 1}}\cos x = x \hfill \cr if\,x \in \left( {0,\,{\pi \over 2}} \right) \hfill \cr} \right\}$$
$$ = \sqrt 2 - 1$$
$$ \Rightarrow (\alpha + 1) = \sqrt 2 \Rightarrow {(\alpha + 1)^2} = 2 \Rightarrow {\alpha ^2} + 2\alpha + 1 = 2$$
$$ \Rightarrow {\alpha ^2} + 2\alpha - 1 = 0$$
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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