Question

Let $f\left(x\right)=x\sqrt{4ax-x^2},(a>0).$ Then $f\left(x\right)$ has range in-

• A. $(0,3a)$
• B. $(a,4a)$
• C. $(0,4a)$
• D. None of the above

Answer 1

The correct option is C $(0,4a)$

$f\left(x\right)=x\sqrt{4ax-x^2}$
Differentiate w.r.t ${}^{\prime }{x}^{\prime }$
$f^1\left(x\right)=\frac{x}{\alpha \sqrt{4ax-x^2}}\times (4a-2x)+\sqrt{4ax-x^2}$
$=\frac{x(4a-2x)}{\alpha \sqrt{4ax-x^2}}+\sqrt{4ax-x^2}$
$=\frac{x\alpha (2x-a)}{\alpha \sqrt{4ax-x^2}}+\sqrt{4ax-x^2}$
$=\frac{(2x-a)x}{\sqrt{4ax-x^2}}+\sqrt{4ax-x^2}$
$Equate{f}^1\left(x\right)=0$
$0=\frac{(2x-a)x}{\sqrt{4ax-x^2}}+\sqrt{4ax-x^2}$
$-\sqrt{4ax-x^2}=\frac{(2x-a)x}{\sqrt{4ax-x^2}}$
$-{\left(\sqrt{4ax-x^2}\right)}^2=(2x-a)x$
$4ax-x^2=2ax-x^2$
$2ax=0$
$x=0$
So factor is $x\sqrt{x(4a-x)}=0$
$x=0or\sqrt{x(4a-x)}=0$
$4ax-x^2=0$
$x^2=4ax$
$x=4a$
SInce the function has the range $(0,4a)$
$(0,4a)$ falls in this range