Question

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:

• A. \( \pm \dfrac{\pi}{3} \)
• B. \( \pm \dfrac{\pi}{6} \)
• C. \( \pm \dfrac{\pi}{4} \)
• D. \( \pm \dfrac{\pi}{2} \)

Hint:

For tangents parallel to the X-axis, find where the derivative of the function is zero.

Answer 1

The Correct Option is C

We are given \( y = \tan^{-1}(\sin(\sqrt{x})) \). To find the points where the tangent is parallel to the X-axis, we set \( \frac{dy}{dx} = 0 \). Using the chain rule: \[ \frac{dy}{dx} = \frac{1}{1 + (\sin(\sqrt{x}))^2} \cdot \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \] For \( \frac{dy}{dx} = 0 \), we must have \( \cos(\sqrt{x}) = 0 \), since the other terms are never zero in the domain. So, \( \sqrt{x} = \frac{\pi}{2}, \frac{3\pi}{2}, \dots, \frac{15\pi}{2} \Rightarrow x = \left( \frac{(2n + 1)\pi}{2} \right)^2 \) for \( 0 \le x \le 8\pi^2 \) At those points, \( \sin(\sqrt{x}) = \pm1 \Rightarrow y = \tan^{-1}(\pm1) = \pm \frac{\pi}{4} \) Thus, the ordinates are \( \pm \dfrac{\pi}{4} \)