JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 11)
If y = $$\sum\limits_{k = 1}^6 {k{{\cos }^{ - 1}}\left\{ {{3 \over 5}\cos kx - {4 \over 5}\sin kx} \right\}} $$,
then $${{dy} \over {dx}}$$ at x = 0 is _______.
then $${{dy} \over {dx}}$$ at x = 0 is _______.
Answer
91
Explanation
Put, $$\cos \alpha = {3 \over 5},\sin \alpha = {4 \over 5}$$
$$ \therefore {3 \over 5}\cos kx - {4 \over 5}\sin \,kx$$
$$ = \cos \alpha .\cos kx - \sin \alpha .\sin kx$$
$$ = \cos \left( {\alpha + kx} \right)$$
So, $$y = \sum\limits_{k = 1}^6 {k{{\cos }^{ - 1}}\left( {\cos \left( {\alpha + kx} \right)} \right)} $$
$$ = \sum\limits_{k = 1}^6 {\left( {{k^2}x + kx} \right)} $$
$$ \Rightarrow {{dy} \over {dx}} = \sum\limits_{k = 1}^6 {{k^2}} $$
$$ = {{6 \times 7 \times 13} \over 6} = 91$$
$$ \therefore {3 \over 5}\cos kx - {4 \over 5}\sin \,kx$$
$$ = \cos \alpha .\cos kx - \sin \alpha .\sin kx$$
$$ = \cos \left( {\alpha + kx} \right)$$
So, $$y = \sum\limits_{k = 1}^6 {k{{\cos }^{ - 1}}\left( {\cos \left( {\alpha + kx} \right)} \right)} $$
$$ = \sum\limits_{k = 1}^6 {\left( {{k^2}x + kx} \right)} $$
$$ \Rightarrow {{dy} \over {dx}} = \sum\limits_{k = 1}^6 {{k^2}} $$
$$ = {{6 \times 7 \times 13} \over 6} = 91$$
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