The ordinates of the points on the curve y = | vedclass

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(C) Given the curve $y = 	an^{-1}(\sin \sqrt{x})$.To find the points where the tangent is parallel to the $x$-axis,we set the derivative $\frac{dy}{d

Vedclass · Accepted Answer

$\pm \frac{\pi}{4}$

Vedclass · Answer

(C) Given the curve $y = 	an^{-1}(\sin \sqrt{x})$.To find the points where the tangent is parallel to the $x$-axis,we set the derivative $\frac{dy}{dx} = 0$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{1 + (\sin \sqrt{x})^2} \cdot \cos \sqrt{x} \cdot \frac{1}{2\sqrt{x}}$.Setting $\frac{dy}{dx} = 0$ implies $\cos \sqrt{x} = 0$ (since $x > 0$ for the derivative to be defined).Thus,$\sqrt{x} = (2n+1)\frac{\pi}{2}$ for $n = 0, 1, 2, \dots$.Given $0 \leq x \leq 8\pi^2$,we have $0 \leq \sqrt{x} \leq 2\sqrt{2}\pi \approx 8.88$.Possible values for $\sqrt{x}$ are $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}$.At these points,$\sin \sqrt{x} = \sin((2n+1)\frac{\pi}{2}) = \pm 1$.Therefore,$y = 	an^{-1}(\pm 1) = \pm \frac{\pi}{4}$.

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 the $x$-axis are

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