Question

The function $y=\sqrt{2x-x^2}$

• A. increases in $(0,2)$
• B. increases in $(0,1)$ and decreases in $(1,2)$
• C. decreases in $(0,2)$
• D. increases in $(1,2)$ and decreases in $(0,1)$

Answer 1

The correct option is B increases in $(0,1)$ and decreases in $(1,2)$

$y=\sqrt{2x-x^2}$
$\frac{dy}{dx}=\frac{1-x}{2\sqrt{2x-x^2}}$

$y$ increases if $\frac{dy}{dx}>0$
and decreases if $\frac{dy}{dx}<0$
$\therefore y$ increases in $(0,1)$ and decreases in $(1,2)$