Question

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

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

Answer 1

Answer: (B)

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,$\int -x<0$
$[\because \sqrt{2x-x^2}$ is always positive]
$x>1$
$\therefore$ Given function decreases in $1<x<2$.