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If $f(x)=\sum_{p=1}^7 p^2 \sin ^{-1}\left(\frac{4}{5} \sin (p x)-\frac{3}{5} \cos (p x)\right)$ then the value of $\frac{d f}{d x}$ at $x=1$ is (Given that $\sin ^{-1}(\sin x)=x$ )
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Verified Answer
The correct answer is:
784
Given
$\begin{aligned}
& \mathrm{f}(\mathrm{x})=\sum_{\mathrm{p}=1}^7 \mathrm{p}^2 \sin ^{-1}\left(\frac{4}{5} \sin (\mathrm{px})-\frac{3}{5} \cos (\mathrm{px})\right) \\
& \Rightarrow \sin ^{-1}(\sin (\mathrm{px}) \cos \alpha-\cos (\mathrm{px}) \sin \alpha \\
& \left\{\begin{array}{l}
\text { Let } \sin (\alpha)=\frac{3}{5} \\
\cos (\alpha)=\frac{4}{5}
\end{array}\right\} \\
& \Rightarrow \sin ^{-1}(\sin (\mathrm{px}-\alpha)) \\
& =(\mathrm{px}-\alpha) \\
& \text { now } \mathrm{f}(\mathrm{x})=\sum_{\mathrm{p}=1}^7 \mathrm{p}^2(\mathrm{px}-\alpha)
\end{aligned}$
$\begin{aligned}
& =x \sum_{\mathrm{p}=1}^7 \mathrm{p}^3-\alpha \sum_{\mathrm{p}=1}^7 \mathrm{p}^2 \\
& \mathrm{f}(\mathrm{x})=\mathrm{x}\left[\frac{7(7+1)}{2}\right]^2-\alpha\left(\frac{7(7+1)(14+1)}{6}\right) \\
& \mathrm{f}^{\prime}(\mathrm{x})=(28)^2=784
\end{aligned}$
$\begin{aligned}
& \mathrm{f}(\mathrm{x})=\sum_{\mathrm{p}=1}^7 \mathrm{p}^2 \sin ^{-1}\left(\frac{4}{5} \sin (\mathrm{px})-\frac{3}{5} \cos (\mathrm{px})\right) \\
& \Rightarrow \sin ^{-1}(\sin (\mathrm{px}) \cos \alpha-\cos (\mathrm{px}) \sin \alpha \\
& \left\{\begin{array}{l}
\text { Let } \sin (\alpha)=\frac{3}{5} \\
\cos (\alpha)=\frac{4}{5}
\end{array}\right\} \\
& \Rightarrow \sin ^{-1}(\sin (\mathrm{px}-\alpha)) \\
& =(\mathrm{px}-\alpha) \\
& \text { now } \mathrm{f}(\mathrm{x})=\sum_{\mathrm{p}=1}^7 \mathrm{p}^2(\mathrm{px}-\alpha)
\end{aligned}$
$\begin{aligned}
& =x \sum_{\mathrm{p}=1}^7 \mathrm{p}^3-\alpha \sum_{\mathrm{p}=1}^7 \mathrm{p}^2 \\
& \mathrm{f}(\mathrm{x})=\mathrm{x}\left[\frac{7(7+1)}{2}\right]^2-\alpha\left(\frac{7(7+1)(14+1)}{6}\right) \\
& \mathrm{f}^{\prime}(\mathrm{x})=(28)^2=784
\end{aligned}$
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