Q107 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics

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 $\sin ^{-1}(\sin x)=x$ )

  • A
    0
  • B
    628
  • C
    1140
  • Correct answer
    D
    784

Solution

Given $\begin{aligned} & \mathrm{f}(\mathrm{x})=\sum_{\mathrm{p}=1}^7 \mathrm{p}^2 \sin ^{-1}\left(\frac{4}{5} \sin (\mathrm{px})-\frac{3}{5} \cos (\mathrm{px})\right) \\ & \Rightarrow \sin ^{-1}(\sin (\mathrm{px}) \cos \alpha-\cos (\mathrm{px}) \sin \alpha \\ & \left\{\begin{array}{l} \text { Let } \sin (\alpha)=\frac{3}{5} \\ \cos (\alpha)=\frac{4}{5} \end{array}\right\} \\ & \Rightarrow \sin ^{-1}(\sin (\mathrm{px}-\alpha)) \\ & =(\mathrm{px}-\alpha) \\ & \text { now } \mathrm{f}(\mathrm{x})=\sum_{\mathrm{p}=1}^7 \mathrm{p}^2(\mathrm{px}-\alpha) \end{aligned}$ $\begin{aligned} & =x \sum_{\mathrm{p}=1}^7 \mathrm{p}^3-\alpha \sum_{\mathrm{p}=1}^7 \mathrm{p}^2 \\ & \mathrm{f}(\mathrm{x})=\mathrm{x}\left[\frac{7(7+1)}{2}\right]^2-\alpha\left(\frac{7(7+1)(14+1)}{6}\right) \\ & \mathrm{f}^{\prime}(\mathrm{x})=(28)^2=784 \end{aligned}$

Hence, the correct answer is option: 'D'.

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