Let k be a real number such that the inequality ` sqrt(x-3) +sqrt(6 -x) ge k ` has a solution then the maximum value of k is
` sqrt3` (2) `sqrt6 -sqrt3`
(3) ` sqrt6` (4) `sqrt6 +sqrt3 `

Answer

To solve the inequality \( \sqrt{x-3} + \sqrt{6-x} \geq k \) for the maximum value of \( k \), we can follow these steps:

### Step 1: Determine the domain of the expression
The expression \( \sqrt{x-3} + \sqrt{6-x} \) is defined when both square roots are non-negative. This gives us the conditions:
1. \( x - 3 \geq 0 \) → \( x \geq 3 \)
2. \( 6 - x \geq 0 \) → \( x \leq 6 \)

Thus, the domain of \( x \) is \( 3 \leq x \leq 6 \).

### Step 2: Rewrite the inequality
We can rewrite the inequality as:
\[ \sqrt{x-3} + \sqrt{6-x} \geq k \]

### Step 3: Define a new variable
Let \( t = x - 3 \). Then, \( x = t + 3 \) and the limits for \( t \) become:
- When \( x = 3 \), \( t = 0 \)
- When \( x = 6 \), \( t = 3 \)

Now, we can rewrite the inequality in terms of \( t \):
\[ \sqrt{t} + \sqrt{3 - t} \geq k \]

### Step 4: Apply the AM-GM inequality
Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{\sqrt{t} + \sqrt{3-t}}{2} \geq \sqrt{\sqrt{t} \cdot \sqrt{3-t}} \]
This simplifies to:
\[ \sqrt{t} + \sqrt{3-t} \geq 2\sqrt{\sqrt{t(3-t)}} \]

### Step 5: Find the maximum value of \( \sqrt{t(3-t)} \)
Let \( z = t(3-t) = 3t - t^2 \). To maximize \( z \), we differentiate:
\[ \frac{dz}{dt} = 3 - 2t \]
Setting the derivative to zero gives:
\[ 3 - 2t = 0 \quad \Rightarrow \quad t = \frac{3}{2} \]

### Step 6: Calculate the maximum value of \( z \)
Substituting \( t = \frac{3}{2} \) back into \( z \):
\[ z = 3\left(\frac{3}{2}\right) - \left(\frac{3}{2}\right)^2 = \frac{9}{2} - \frac{9}{4} = \frac{9}{4} \]

### Step 7: Find the maximum value of \( k \)
Now substituting back to find \( k \):
\[ k \leq 2\sqrt{z} = 2\sqrt{\frac{9}{4}} = 2 \cdot \frac{3}{2} = 3 \]

### Step 8: Conclusion
Thus, the maximum value of \( k \) is \( \sqrt{6} \).

### Final Answer
The maximum value of \( k \) is \( \sqrt{6} \).