Verified Solution • 2025–26 Session moderate

QJEEM20Ma2478: JEE Main 2020 Mathematics - Class 12 JEEM20Ma2478 Solution: If y = , then at x = 0 is _______.

If y = ∑k=16kcos⁡−1{35cos⁡kx−45sin⁡kx}\sum\limits_{k = 1}^6 {k{{\cos }^{ - 1}}\left\{ {{3 \over 5}\cos kx - {4 \over 5}\sin kx} \right\}} ,

then dydx{{dy} \over {dx}} at x = 0 is _______.

Step-by-Step Verified Solution

Put, cos⁡α=35,sin⁡α=45\cos \alpha = {3 \over 5},\sin \alpha = {4 \over 5}

∴35cos⁡kx−45sin⁡ kx\therefore {3 \over 5}\cos kx - {4 \over 5}\sin \,kx

=cos⁡α.cos⁡kx−sin⁡α.sin⁡kx= \cos \alpha .\cos kx - \sin \alpha .\sin kx

=cos⁡(α+kx)= \cos \left( {\alpha + kx} \right)

So, y=∑k=16kcos⁡−1(cos⁡(α+kx))y = \sum\limits_{k = 1}^6 {k{{\cos }^{ - 1}}\left( {\cos \left( {\alpha + kx} \right)} \right)}

=∑k=16(k2x+kx)= \sum\limits_{k = 1}^6 {\left( {{k^2}x + kx} \right)}

⇒dydx=∑k=16k2\Rightarrow {{dy} \over {dx}} = \sum\limits_{k = 1}^6 {{k^2}}

=6×7×136=91= {{6 \times 7 \times 13} \over 6} = 91

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