If m is the minimum value of k for which the function f(x)=x√kx-x^2 is increasing in the interval [0,3] and M is the maximum value of f in [0,3] when…

$f(x)=x\sqrt{kx-{x}^{2}}$f(x)=xkx−x2​ ${\Rightarrow f}^{'}(x)=\frac{3kx-4{x}^{2}}{2\sqrt{kx-{x}^{2}}}$⇒f′(x)=2kx−x2​3kx−4x2​

As per the given condition ${f}^{'}(x)\geq 0$f′(x)≥0 for $x\in [0,3]$x∈[0,3]

$\Rightarrow 3kx-4{x}^{2}\geq 0$⇒3kx−4x2≥0 for $x\in [0,3]$x∈[0,3]

$\Rightarrow 3k-4x \geq 0$⇒3k−4x≥0 for $x\in [0,3]$x∈[0,3]

$\Rightarrow k\geq \frac{4x}{3}$⇒k≥34x​ for $x\in [0,3]$x∈[0,3]

$\Rightarrow k\geq 4$⇒k≥4 . So minimum value of $k$k is $m=4.$m=4.

Now $f(x)=x\sqrt{4x-{x}^{2}}$f(x)=x4x−x2​

Since given function in increasing hence maximum value will occur at $x=3$x=3

$$\Rightarrow f(3)=3\sqrt{4\times 3-{3}^{2}}=3\sqrt{3}, M=3\sqrt{3}$$⇒f(3)=34×3−32​=33​,M=33​

Hence $(m, M) = (4, 3\sqrt{3})$(m,M)=(4,33​)