For any value of x the expression `2(k-x)(x+sqrt(x^2+k^2))` cannot exceed

Options

A. `k^(2)`
B. `2k^(2)`
C. `3k^(2)`
D. none of these

Explanation

To find the maximum value of the expression \(2(k-x)(x+\sqrt{x^2+k^2})\), we will follow these steps:

### Step 1: Define the function
Let \( f(x) = 2(k-x)(x+\sqrt{x^2+k^2}) \).

### Step 2: Differentiate the function
We need to find the derivative \( f'(x) \) to locate the critical points. Using the product rule, we differentiate:

\[ f'(x) = 2 \left[ (k-x) \frac{d}{dx}(x+\sqrt{x^2+k^2}) + (x+\sqrt{x^2+k^2}) \frac{d}{dx}(k-x) \right] \]

Calculating the derivatives:
- The derivative of \( k-x \) is \(-1\).
- The derivative of \( x+\sqrt{x^2+k^2} \) is \( 1 + \frac{x}{\sqrt{x^2+k^2}} \).

Thus, we have:

\[ f'(x) = 2 \left[ (k-x) \left( 1 + \frac{x}{\sqrt{x^2+k^2}} \right) - (x+\sqrt{x^2+k^2}) \right] \]

### Step 3: Set the derivative to zero
To find the critical points, we set \( f'(x) = 0 \):

\[ (k-x) \left( 1 + \frac{x}{\sqrt{x^2+k^2}} \right) - (x+\sqrt{x^2+k^2}) = 0 \]

### Step 4: Solve for \( x \)
Rearranging gives us:

\[ (k-x) + \frac{x(k-x)}{\sqrt{x^2+k^2}} - x - \sqrt{x^2+k^2} = 0 \]

This simplifies to:

\[ k - 2x - \sqrt{x^2+k^2} + \frac{x(k-x)}{\sqrt{x^2+k^2}} = 0 \]

### Step 5: Analyze the critical points
Now we need to analyze this equation to find the value of \( x \).

### Step 6: Substitute back into the original function
Once we find the critical points, we substitute back into \( f(x) \) to find the maximum value.

### Step 7: Evaluate at \( x = 0 \)
To check for maximum, we can evaluate \( f(0) \):

\[ f(0) = 2(k-0)(0+\sqrt{0^2+k^2}) = 2k \cdot k = 2k^2 \]

### Conclusion
Thus, the maximum value of the expression \( 2(k-x)(x+\sqrt{x^2+k^2}) \) cannot exceed \( 2k^2 \).

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