If y = 2sin^{-1} sqrt{1-x} + sin^{-1} (2sqrt{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $\sqrt{x} = \sin 	heta$,which implies $x = \sin^2 	heta$. Since $0 < x < \frac{1}{2}$,we have $0 < \sin^2 	heta < \frac{1}{2}$,so $0 < \sin

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $\sqrt{x} = \sin 	heta$,which implies $x = \sin^2 	heta$. Since $0 Then,$\sqrt{1-x} = \sqrt{1-\sin^2 	heta} = \cos 	heta$. Also,$2\sqrt{x(1-x)} = 2\sin 	heta \cos 	heta = \sin 2	heta$. Substituting these into the expression for $y$: $y = 2\sin^{-1}(\cos 	heta) + \sin^{-1}(\sin 2	heta)$. Since $0 Also,$\sin^{-1}(\cos 	heta) = \sin^{-1}(\sin(\frac{\pi}{2} - 	heta)) = \frac{\pi}{2} - 	heta$. Thus,$y = 2(\frac{\pi}{2} - 	heta) + 2	heta = \pi - 2	heta + 2	heta = \pi$. Since $y = \pi$ is a constant,$\frac{dy}{dx} = 0$.

If $y = 2\sin^{-1} \sqrt{1-x} + \sin^{-1} (2\sqrt{x(1-x)})$ for $0 < x < \frac{1}{2}$,then $\frac{dy}{dx}$ equals-

Similar Questions

$\tan^{-1} \left[ \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \right] = $

If $\sec^{-1} \left( \frac{x^2 + y^2}{x^2 - y^2} \right) = 2a$, such that $y \frac{dy}{dx} = x \cdot f(a)$, then the value of $f \left( \frac{2\pi}{3} \right)$ is

If $\tan ^{-1} x = \frac{\pi}{4} - \tan ^{-1} \left( \frac{1}{3} \right)$,then $x$ is

If $y = \tan^{-1} \frac{x}{1+2x^2} + \tan^{-1} \frac{x}{1+6x^2} + \tan^{-1} \frac{x}{1+12x^2}$,then $\left(\frac{dy}{dx}\right)_{x=\frac{1}{2}} = $

Suppose $\tan ^{-1} y = \tan ^{-1} x + \tan ^{-1} \left( \frac{2x}{1 - x^2} \right)$,where $|x| < \frac{1}{\sqrt{3}}$,then one of the values of $y$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module