Question

If $m$ is the minimum value of $k$ for which the function $f\left(x\right)=x\sqrt{kx-x^2}$ is increasing in the interval $[0,3]$ and $M$ is the maximum value of $f$ in $[0,3]$ when $k=m,$ then the ordered pair $(m,M)$ is equal to :

• A. $(4,3\sqrt{2})$
• B. $(5,3\sqrt{6})$
• C. $(4,3\sqrt{3})$
• D. $(3,3\sqrt{3})$

Answer 1

The correct option is C $(4,3\sqrt{3})$

$f\left(x\right)=x\sqrt{kx-x^2}$
$f^{\prime }\left(x\right)=\sqrt{kx-x^2}+\frac{x(k-2x)}{2\sqrt{kx-x^2}}=\frac{2kx-2x^2+kx-2x^2}{2\sqrt{kx-x^2}}=\frac{3kx-4x^2}{2\sqrt{kx-x^2}}$
$f\left(x\right)$ is increasing in the interval $[0,3]$
hence, $f^{\prime }\left(x\right)\ge 0\forall x\in [0,3]$
Therefore, $3kx-4x^2\ge 0\Rightarrow x(4x-3k)\le 0\forall x\in [0,3]$
$\Rightarrow 4x\le 3k$
$\Rightarrow x\le \frac{3k}{4}$
The maximum value of $x$ in which $f\left(x\right)$ should be increasing is $3.$
$\Rightarrow k\ge 4$
Also $kx-x^2\ge 0\Rightarrow x(x-k)\le 0\forall x\in [0,3]$
$\Rightarrow k\ge 3$
$\therefore k_{min}=m=4$
Maximum value of $f\left(x\right)$ when $k=4,M=f\left(3\right)=3\sqrt{4\cdot 3-3^2}=3\sqrt{3}$
$\therefore (m,M)=(4,3\sqrt{3})$