Question

The area bounded by the curves $\sqrt{x}+\sqrt{y}=1$ and $x+y=1$ is ?

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{2}$
• D. $\frac{5}{6}$
• E. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{1}{3}$

REF.image

$\to$ As quantity inside square

root is positive $\Rightarrow x,y⩾0$

Now $\sqrt{x}+\sqrt{y}=1$

$\Rightarrow \sqrt{y}=1-\sqrt{x}$

with x + y = 1 , $\sqrt{1-x}=1-\sqrt{x}$

$1-x=1+x-2\sqrt{x}$

$\Rightarrow 2x-2\sqrt{x}=0$

$\Rightarrow 2\sqrt{x}\left(\sqrt{x-1}\right)=0$

either$\sqrt{x}=0or\sqrt{x}=1$

$\to x=0,x=1$

$\Rightarrow y=1,y=0$

$\Rightarrow$ (0,1),(1,0) are P.O.I

$\sqrt{x}+\sqrt{y}=1\Rightarrow \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\sqrt{\frac{y}{x}}$

$A={\int }_0^1(1-x)-(1-\sqrt{x}{)}^{2}dx$

$={\int }_0^{1}1-x-1-x+2\sqrt{x}dx$

$={\int }_0^1-2x+2\sqrt{x}dx$

$=\frac{-2x^2}{2}+\frac{2\times 2}{3}{x}^{3/2}{\int }_0^1$

$=-1+\frac{4}{3}=\frac{1}{3}\Rightarrow \left(A\right)$