If m is the minimum value of k for which the | vedclass

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(D) Given $f(x) = x\sqrt{kx - x^2}$.For $f(x)$ to be defined,$kx - x^2 \geq 0$,which implies $x(k - x) \geq 0$. For $x \in [0, 3]$,this requires $k \g

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$(4, 3\sqrt{3})$

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(D) Given $f(x) = x\sqrt{kx - x^2}$.For $f(x)$ to be defined,$kx - x^2 \geq 0$,which implies $x(k - x) \geq 0$. For $x \in [0, 3]$,this requires $k \geq 3$.The derivative is $f'(x) = \sqrt{kx - x^2} + x \cdot \frac{k - 2x}{2\sqrt{kx - x^2}} = \frac{2(kx - x^2) + kx - 2x^2}{2\sqrt{kx - x^2}} = \frac{3kx - 4x^2}{2\sqrt{kx - x^2}}$.For $f(x)$ to be increasing on $[0, 3]$,we need $f'(x) \geq 0$ for all $x \in (0, 3)$.This implies $3kx - 4x^2 \geq 0$,or $x(3k - 4x) \geq 0$.Since $x > 0$,we require $3k - 4x \geq 0$,or $k \geq \frac{4x}{3}$.For this to hold for all $x \in [0, 3]$,$k$ must be at least the maximum value of $\frac{4x}{3}$ on $[0, 3]$,which is $\frac{4(3)}{3} = 4$.Thus,$m = 4$.Now,for $k = 4$,$f(x) = x\sqrt{4x - x^2}$.To find the maximum value $M$ on $[0, 3]$,we check the derivative $f'(x) = \frac{12x - 4x^2}{2\sqrt{4x - x^2}} = \frac{2x(3 - x)}{\sqrt{4x - x^2}}$.The critical point is $x = 3$. Since $f(0) = 0$ and $f(3) = 3\sqrt{4(3) - 3^2} = 3\sqrt{12 - 9} = 3\sqrt{3}$,the maximum value $M = 3\sqrt{3}$.Therefore,$(m, M) = (4, 3\sqrt{3})$.

If $m$ is the minimum value of $k$ for which the function $f(x) = x\sqrt{kx - x^2}$ 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

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