Consider `f(x)=xsqrt(kx-x^(2))` for `xepsilon [0, 3]`. Let `m` be the smallest value of `k` for which the function is increasing in the given interval and `M` be the largest value of `f(x) for k=m. Then `(m,M)` is

Options

A. `(4,3sqrt2)`
B. `(4,3sqrt3)`
C. `(3,3sqrt3)`
D. `(5,3sqrt6)`

Answer

To solve the problem, we need to analyze the function \( f(x) = x \sqrt{kx - x^2} \) for \( x \in [0, 3] \) and find the smallest value of \( k \) for which the function is increasing in this interval, as well as the largest value of \( f(x) \) when \( k \) is at its minimum.

### Step 1: Determine the domain of \( f(x) \)

The function \( f(x) \) is defined when the expression under the square root is non-negative:
\[ kx - x^2 \geq 0 \]
Factoring gives:
\[ x(k - x) \geq 0 \]
Since \( x \) is non-negative in the interval \([0, 3]\), we require:
\[ k - x \geq 0 \implies k \geq x \]
Thus, for \( x \in [0, 3] \), we need \( k \geq 3 \).

### Step 2: Differentiate \( f(x) \)

To find when \( f(x) \) is increasing, we compute the derivative \( f'(x) \):
Using the product and chain rule:
\[ f'(x) = \sqrt{kx - x^2} + x \cdot \frac{1}{2\sqrt{kx - x^2}}(k - 2x) \]
This simplifies to:
\[ f'(x) = \sqrt{kx - x^2} + \frac{x(k - 2x)}{2\sqrt{kx - x^2}} \]
Combining terms gives:
\[ f'(x) = \frac{2(kx - x^2) + x(k - 2x)}{2\sqrt{kx - x^2}} = \frac{(2k - 4)x^2 + 2kx}{2\sqrt{kx - x^2}} \]
Setting \( f'(x) \geq 0 \) for \( f(x) \) to be increasing:
\[ (2k - 4)x^2 + 2kx \geq 0 \]
Factoring out \( x \):
\[ x((2k - 4)x + 2k) \geq 0 \]

### Step 3: Analyze the inequality

The critical points occur when:
1. \( x = 0 \)
2. \( (2k - 4)x + 2k = 0 \) leads to \( x = \frac{2k}{4 - 2k} \)

For \( f(x) \) to be increasing for all \( x \in [0, 3] \), we need:
\[ 2k - 4 \leq 0 \implies k \leq 2 \]
However, since \( k \geq 3 \) from the domain condition, we find that \( k \) must be at least \( 4 \) for the function to be increasing.

### Step 4: Find the maximum value of \( f(x) \)

Now, we need to evaluate \( f(x) \) at \( k = 4 \):
\[ f(x) = x \sqrt{4x - x^2} \]
To find the maximum, we can evaluate \( f(x) \) at the endpoints and any critical points in the interval.

Calculating at \( x = 0 \):
\[ f(0) = 0 \]
Calculating at \( x = 3 \):
\[ f(3) = 3 \sqrt{4 \cdot 3 - 3^2} = 3 \sqrt{12 - 9} = 3 \sqrt{3} \]

### Conclusion

Thus, the smallest value of \( k \) for which \( f(x) \) is increasing is \( m = 4 \), and the largest value of \( f(x) \) when \( k = m \) is \( M = 3\sqrt{3} \).

The final answer is:
\[ (m, M) = (4, 3\sqrt{3}) \]