Q112 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
$\int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) \mathrm{dx}=f(x)+c$, where $c$ is the constant of integration. If $\frac{5 \pi}{2} \lt x \lt \frac{7 \pi}{2}$ and $f\left(\frac{8 \pi}{3}\right)=-2$, then $f^{\prime}\left(\frac{8 \pi}{3}\right)=$
- A1
- Correct answerB$\sqrt{3}$
- C0
- D-1
Solution
$\mathrm{I}=\int(\sqrt{1-\sin \mathrm{x}}+\sqrt{1+\sin \mathrm{x}}) \mathrm{dx}$
$\begin{aligned} & \Rightarrow I=\int\left(\sqrt{1-\cos \left(\frac{\pi}{2}-x\right)}+\sqrt{1+\cos \left(\frac{\pi}{2}-x\right)}\right) d x \\ & \left.\Rightarrow I=\sqrt{2} \int\left[\sin \left(\frac{\pi}{4}-\frac{x}{2}\right)\right] x+\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)\right] d x \\ & \Rightarrow I=2 \sqrt{2} \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)-2 \sqrt{2} \sin \left(\frac{\pi}{4}-\frac{x}{2}\right)+C\end{aligned}$
$\begin{aligned} & =4 \cos \left(\frac{x}{2}\right)+C \Rightarrow f(x)=4 \cos \left(\frac{x}{2}\right) \Rightarrow f\left(\frac{8 \pi}{3}\right)=-2 \\ & \Rightarrow f^{\prime}(x)=-2 \sin \frac{x}{2} \Rightarrow f^{\prime}\left(\frac{8 \pi}{3}\right)=\sqrt{3} .\end{aligned}$
Hence, the correct answer is option: 'B'.