If sqrt{x-xy} + sqrt{y-xy} = 1,then frac{dy}{ | vedclass

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

Question

(A) Given equation is $\sqrt{x(1-y)} + \sqrt{y(1-x)} = 1$. Let $x = \sin^2 	heta$ and $y = \sin^2 \phi$. Then $\sqrt{\sin^2 	heta (1-\sin^2 \phi)} +

Vedclass · Accepted Answer

$-\sqrt{\frac{y-y^2}{x-x^2}}$

Vedclass · Answer

(A) Given equation is $\sqrt{x(1-y)} + \sqrt{y(1-x)} = 1$. Let $x = \sin^2 	heta$ and $y = \sin^2 \phi$. Then $\sqrt{\sin^2 	heta (1-\sin^2 \phi)} + \sqrt{\sin^2 \phi (1-\sin^2 	heta)} = 1$. $\sin 	heta \cos \phi + \sin \phi \cos 	heta = 1$. $\sin(	heta + \phi) = 1$. $	heta + \phi = \frac{\pi}{2}$. $\phi = \frac{\pi}{2} - 	heta$. Since $y = \sin^2 \phi$,we have $y = \sin^2(\frac{\pi}{2} - 	heta) = \cos^2 	heta$. Thus $x = \sin^2 	heta$ and $y = \cos^2 	heta$. Adding these,$x+y = \sin^2 	heta + \cos^2 	heta = 1$,so $y = 1-x$. Differentiating with respect to $x$,$\frac{dy}{dx} = -1$. Alternatively,checking the options: $\frac{dy}{dx} = -\sqrt{\frac{y(1-y)}{x(1-x)}} = -\sqrt{\frac{(1-x)x}{x(1-x)}} = -\sqrt{1} = -1$. Thus,option $A$ is correct.

If $\sqrt{x-xy} + \sqrt{y-xy} = 1$,then $\frac{dy}{dx} = $

Similar Questions

If $y=1+xe^y$,then $\frac{dy}{dx}=$

If $y^{2}+\log _{e}\left(\cos ^{2} x\right)=y, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),$ then

If $\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0$ and $x \neq y$,then $\frac{dy}{dx}$ is equal to:

If $\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$,then $\frac{dx}{dy} = $

For $x \in R$,$f(x) = |\log 2 - \sin x|$ and $g(x) = f(f(x))$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module