Question

JEE Mains Maths question is in the image

Answer 1

Given,

$\displaystyle\longrightarrow y=\sum_{k=1}^6k\cos^{-1}\left(\dfrac{3}{5}\cos(kx)-\dfrac{4}{5}\sin(kx)\right)$

Take $\theta=\cos^{-1}\left(\dfrac{3}{5}\right)=\sin^{-1}\left(\dfrac{4}{5}\right)$ which is constant.

Then,

$\displaystyle\longrightarrow y=\sum_{k=1}^6k\cos^{-1}\left(\cos\theta\cos(kx)-\sin\theta\sin(kx)\right)$

$\displaystyle\longrightarrow y=\sum_{k=1}^6k\cos^{-1}\left(\cos(\theta+kx)\right)$

$\displaystyle\longrightarrow y=\sum_{k=1}^6k(\theta+kx)$

$\displaystyle\longrightarrow y=\sum_{k=1}^6\left(k\theta+k^2x\right)$

$\displaystyle\longrightarrow y=\sum_{k=1}^6k\theta+\sum_{k=1}^6k^2x$

$\displaystyle\longrightarrow y=\sum_{k=1}^6k\theta+x\sum_{k=1}^6k^2$

Then its derivative wrt $x$ will be,

$\displaystyle\longrightarrow\dfrac{dy}{dx}=\dfrac{d}{dx}\left[\sum_{k=1}^6k\theta+x\sum_{k=1}^6k^2\right]$

Since $\displaystyle\sum_{k=1}^6k\theta$ and $\displaystyle\sum_{k=1}^6k^2$ are constants wrt $x,$

$\displaystyle\longrightarrow\dfrac{dy}{dx}=\sum_{k=1}^6k^2$

Since $\displaystyle\sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}{6},$

$\displaystyle\longrightarrow\dfrac{dy}{dx}=\dfrac{6\times7\times13}{6}$

$\displaystyle\longrightarrow\underline{\underline{\dfrac{dy}{dx}=91}}$

Hence (2) is the answer.