Key Concepts and Formulas

• Increasing Function: A function $f(x)$ is increasing in an interval $[a, b]$ if its derivative $f'(x) \ge 0$ for all $x$ in the interval $[a, b]$.
• Product Rule: The derivative of a product of two functions $u(x)$ and $v(x)$ is given by $(uv)' = u'v + uv'$.
• Finding Maximum Value: To find the maximum value of a function in a closed interval, we find the critical points (where the derivative is zero or undefined) and evaluate the function at these points and at the endpoints of the interval. The largest of these values is the maximum.

Step-by-Step Solution

Step 1: Find the derivative of f(x)

We are given $f(x) = x\sqrt{kx - x^2}$. We need to find $f'(x)$ using the product rule. Let $u(x) = x$ and $v(x) = \sqrt{kx - x^2} = (kx - x^2)^{1/2}$. Then $u'(x) = 1$ and $v'(x) = \frac{1}{2}(kx - x^2)^{-1/2}(k - 2x) = \frac{k - 2x}{2\sqrt{kx - x^2}}$ Therefore, $f'(x) = u'(x)v(x) + u(x)v'(x) = \sqrt{kx - x^2} + x\left(\frac{k - 2x}{2\sqrt{kx - x^2}}\right)$ $f'(x) = \frac{2(kx - x^2) + x(k - 2x)}{2\sqrt{kx - x^2}} = \frac{2kx - 2x^2 + kx - 2x^2}{2\sqrt{kx - x^2}} = \frac{3kx - 4x^2}{2\sqrt{kx - x^2}}$

Step 2: Determine the condition for f(x) to be increasing

For $f(x)$ to be increasing in the interval $[0, 3]$, we must have $f'(x) \ge 0$ for all $x \in [0, 3]$. Since $x \in [0, 3]$, the term $\sqrt{kx - x^2}$ must be real, which implies $kx - x^2 \ge 0$, or $x(k - x) \ge 0$. This means $k \ge x$. Since $x \in [0,3]$, we must have $k \ge 3$. Also, we need $f'(x) \ge 0$, which means $\frac{3kx - 4x^2}{2\sqrt{kx - x^2}} \ge 0$ Since the denominator is positive (as long as $kx - x^2 > 0$), we need the numerator to be non-negative: $3kx - 4x^2 \ge 0$ $x(3k - 4x) \ge 0$ Since $x \ge 0$, we need $3k - 4x \ge 0$, which means $3k \ge 4x$, or $k \ge \frac{4}{3}x$. Since we want $f(x)$ to be increasing in $[0, 3]$, we need $k \ge \frac{4}{3}x$ for all $x \in [0, 3]$. The maximum value of $\frac{4}{3}x$ in the interval $[0, 3]$ is $\frac{4}{3}(3) = 4$. Therefore, we must have $k \ge 4$.

Step 3: Find the minimum value of k

The minimum value of $k$ for which $f(x)$ is increasing in $[0, 3]$ is $m = 4$.

Step 4: Find the maximum value of f(x) when k = m = 4

Now we have $f(x) = x\sqrt{4x - x^2}$ and we want to find its maximum value $M$ in the interval $[0, 3]$. We already have $f'(x) = \frac{3(4)x - 4x^2}{2\sqrt{4x - x^2}} = \frac{12x - 4x^2}{2\sqrt{4x - x^2}} = \frac{4x(3 - x)}{2\sqrt{4x - x^2}} = \frac{2x(3 - x)}{\sqrt{4x - x^2}}$. Setting $f'(x) = 0$, we get $2x(3 - x) = 0$, which gives $x = 0$ or $x = 3$. Also, $f'(x)$ is undefined when $4x - x^2 = 0$, i.e., $x(4 - x) = 0$, which gives $x = 0$ or $x = 4$. Since $x=4$ is not in the interval $[0,3]$, we don't consider it.

Now we evaluate $f(x)$ at the critical points $x = 0$ and $x = 3$, and at the endpoint $x = 0$: $f(0) = 0\sqrt{4(0) - 0^2} = 0$ $f(3) = 3\sqrt{4(3) - 3^2} = 3\sqrt{12 - 9} = 3\sqrt{3}$

The maximum value of $f(x)$ in $[0, 3]$ is $M = 3\sqrt{3}$.

Step 5: Write the ordered pair (m, M)

The ordered pair is $(m, M) = (4, 3\sqrt{3})$.

Common Mistakes & Tips

• Remember to check the endpoints of the interval when finding the maximum value of a function.
• Be careful when simplifying expressions involving square roots.
• Make sure the derivative is greater than or equal to zero for the function to be increasing.

Summary

We found the derivative of the function $f(x)$ and determined the condition for it to be increasing in the interval $[0, 3]$. This allowed us to find the minimum value of $k$, which is $m = 4$. Then, we found the maximum value of $f(x)$ when $k = 4$ in the same interval, which is $M = 3\sqrt{3}$. Therefore, the ordered pair $(m, M)$ is $(4, 3\sqrt{3})$.

Final Answer

The final answer is $\boxed{(4, 3\sqrt{3})}$, which corresponds to option (B).