Question

The graph of $\sqrt{x}+\sqrt{y}=\sqrt{2}$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Given, $\sqrt{x}+\sqrt{y}=\sqrt{2}\cdots \left(i\right)$
Clearly above curve is defined only when $x\ge 0,y\ge 0$ (in $1^{st}$ quadrant)
$x=0\Rightarrow y=2$ and
$y=0\Rightarrow x=2$
Differentiating $\left(i\right)$ w.r.t. $x,$ we get
$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}{y}^{\prime }=0$
$\Rightarrow y^{\prime }=-\frac{\sqrt{y}}{\sqrt{x}}<0$
$\therefore y$ decreases for all $x>0,y>0\left(\text{local maxima in}\right(0,2\left)\right)$
Now, $y^{\prime \prime }=\frac{\sqrt{x}+\sqrt{y}}{2x\sqrt{x}}>0$
$\Rightarrow$ curvature is concave up.
Hence option $\left(a\right)$ is correct.