Question

The slope of the tangent to curve $y={\int }_x^{x^2}{\cos }^{-1}{t}^{2}dt$ at $x=\frac{1}{\sqrt[4]{42}}$ is

• A. $\left(\frac{\sqrt[4]{48}}{2}-\frac{3}{4}\right)\pi$
• B. $\left(\frac{\sqrt[4]{48}}{3}-\frac{1}{4}\right)\pi$
• C. $\left(\frac{\sqrt[5]{58}}{4}-\frac{1}{3}\right)\pi$
• D. none of these

Answer 1

Answer: (B)

$\left(\frac{\sqrt[4]{48}}{3}-\frac{1}{4}\right)\pi$

Given, Curve equation as $y={\int }_x^{x^2}{\cos }^{-1}{t}^{2}dt$

Slope of the curve$=\frac{dy}{dx}$

On differentiating curve equation once gives,

$\Rightarrow \frac{dy}{dx}=({\cos }^{-1}{x}^4.\frac{d}{dx}(x^2)-{\cos }^{-1}{x}^2.\frac{d}{dx}(x\left)\right)$

$\Rightarrow \frac{dy}{dx}=(2x{\cos }^{-1}{x}^4-{\cos }^{-1}{x}^2)$

Given $x=\frac{1}{\sqrt[4]{42}}$

$\Rightarrow \frac{dy}{dx}=(2\left(\frac{1}{\sqrt[4]{42}}\right){\cos }^{-1}{\left(\frac{1}{\sqrt[4]{42}}\right)}^4-{\cos }^{-1}{\left(\frac{1}{\sqrt[4]{42}}\right)}^2)$

$\Rightarrow \frac{dy}{dx}=\left(2\sqrt[4]{42}\right){\cos }^{-1}\left(\frac{1}{2}\right)-{\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right))$

$\Rightarrow \frac{dy}{dx}=\left(\sqrt[4]{48}\right)\frac{\pi }{3}-\frac{\pi }{4}=(\frac{\sqrt[4]{48}}{3}-\frac{1}{4})\pi$