Question

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

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

Answer 1

The correct option is A increases in $(0,1)$ but decreases in $(1,2)$

Given
$y=\sqrt{2x-x^2}$
By differentiating both side w.r.t $x$ we get
$\frac{dy}{dx}=\frac{1}{2\sqrt{2x-x^2}}\times (2-2x)=\frac{1-x}{\sqrt{2x-x^2}}$
Here, $\frac{dy}{dx}$ is defined for
$2x-x^2>0$
i.e, $0<x<2$
Now, $\frac{dy}{dx}>0$
when $1-x>0$
($\because$ $\sqrt{2x-x^2}$ is always positive)
$\Rightarrow x<1$
$\therefore$ Given function increases in $0<x<1$
And $\frac{dy}{dx}<0$
When $1-x<0$

i.e. $x>1$

Thus, $y$ decreases in $1<x<2$ i.e. $(1,2)$
Hence, option A is correct.