Q105 / 160 · AP EAMCET 2023 (17 May Shift 1) · Mathematics
The ordinates of the points on the curve $y=\tan ^{-1}$ $(\sin \sqrt{x}), 0 \leq x \leq 8 \pi^2$, at which the tangent is parallel to $\mathrm{x}$-axis are
- A$\pm \frac{\pi}{3}$
- B$\pm \frac{\pi}{6}$
- Correct answerC$\pm \frac{\pi}{4}$
- D$\pm \frac{\pi}{2}$
Solution
$\begin{aligned}
& \text {} y=\tan ^{-1}(\sin \sqrt{x}) \\
& \Rightarrow \tan y=\sin \sqrt{x} \Rightarrow \sec ^2 y \cdot \frac{d y}{d x}=\frac{1}{\sqrt{x}} \cdot \frac{1}{2} \cos \sqrt{x} \\
& \Rightarrow \frac{d y}{d x}=\frac{\cos \sqrt{x}}{2 \sqrt{x} \sec ^2 y}
\end{aligned}$
Tangent is parallel to $x$-axis
$\begin{aligned}
& \therefore \frac{d y}{d x} \Rightarrow \frac{\cos \sqrt{x}}{2 \sqrt{x} \sec ^2 y}=0 \\
& \Rightarrow \cos \sqrt{x}=0 \Rightarrow \cos \sqrt{x}=\cos \frac{\pi}{2} \\
& \Rightarrow x=\frac{\pi^2}{4} \Rightarrow \sqrt{x}= \pm \frac{\pi}{2} \Rightarrow \sin \sqrt{x}= \pm 1 \\
& \therefore y=\tan ^{-1}(\sin \sqrt{x})= \pm \frac{\pi}{4}
\end{aligned}$
Hence, the correct answer is option: 'C'.