Mathematics - Application of Derivative Question with Solution | TestHub

$f\left(x\right)=x\sqrt{kx-x^2}$${\Rightarrow f}^{\text{'}}\left(x\right)=\frac{3kx-4x^2}{2\sqrt{kx-x^2}}$
As per the given condition$f^{\text{'}}\left(x\right)\ge 0$for$x\in \left[\mathrm{0,} 3\right]$
$\Rightarrow 3kx-4x^2\ge 0$for$x\in \left[\mathrm{0,} 3\right]$
$\Rightarrow 3k-4x \ge 0$for$x\in \left[\mathrm{0,} 3\right]$
$\Rightarrow k\ge \frac{4x}{3}$for$x\in \left[\mathrm{0,} 3\right]$
$\Rightarrow k\ge 4$. So minimum value of$k$is$m=4.$
Now$f\left(x\right)=x\sqrt{4x-x^2}$
Since given function in increasing hence maximum value will occur at$x=3$
$\Rightarrow f\left(3\right)=3\sqrt{4\times 3-3^2}=3\sqrt{3},M=3\sqrt{3}$
Hence$(m, M) = (4, 3\sqrt{3})$