Question

$y=\sqrt{x-x^2}$ increases in the interval

• A. $(0,\frac{1}{2})$
• B. $(-\frac{1}{2},0)$
• C. $[0,\frac{1}{2}]$
• D. $[0,\frac{1}{2})$

Answer 1

Answer: (A)

$(0,\frac{1}{2})$

$f\left(x\right)=\sqrt{x-x^2}$
$f^{\prime }\left(x\right)=\frac{1-2x}{\sqrt{x-x^2}}$
For f(x) to be increasing, f'(x)>0
$\Rightarrow \frac{1-2x}{\sqrt{x(1-x)}}>0$
$\Rightarrow 1-2x>0$ and $x>0,1-x>0$
$\Rightarrow x<\frac{1}{2}$ and $x>0,x<1$ (Since $\sqrt{(}x-x^2)$ must be non-negative)
$\Rightarrow x>0$ and $x<\frac{1}{2}$
$\Rightarrow x\in (0,\frac{1}{2})$