If f (x) `={{:(xlog_(e)x",",x gt0),(0",",x=0):}` ,thenconclusion of LMVT holds at `x = 1` in the interval `[0,a]` for `f(x),` then `[a^(2)]` is equal to (where [.] denotes the greatest integer) `"_______".`

Answer

To solve the problem step by step, we will analyze the function \( f(x) \) and apply the Lagrange Mean Value Theorem (LMVT).

### Step 1: Define the function
The function is given as:
\[ f(x) = \begin{cases} x \log_e x & \text{if } x > 0 \\ 0 & \text{if } x = 0 \end{cases} \]

### Step 2: Check the conditions for LMVT
For LMVT to hold on the interval \([0, a]\), we need to check three conditions:
1. \( f(x) \) should be continuous on \([0, a]\).
2. \( f(x) \) should be differentiable on \((0, a)\).
3. There exists a \( c \in (0, a) \) such that:
\[ f'(c) = \frac{f(a) - f(0)}{a - 0} \]

### Step 3: Verify continuity
- At \( x = 0 \), \( f(0) = 0 \).
- As \( x \) approaches 0 from the right, \( f(x) = x \log_e x \) approaches 0 (since \( \log_e x \to -\infty \) and \( x \to 0 \)).
- Therefore, \( f(x) \) is continuous at \( x = 0 \) and on \((0, a)\).

### Step 4: Verify differentiability
- For \( x > 0 \), \( f(x) = x \log_e x \) is differentiable.
- At \( x = 0 \), we can check the left-hand derivative:
\[ f'(0) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h \log_e h}{h} = \lim_{h \to 0^+} \log_e h = -\infty \]
- However, since we are only concerned with the interval \((0, a)\), \( f(x) \) is differentiable on that interval.

### Step 5: Apply LMVT
We need to find \( c \) such that:
\[ f'(c) = \frac{f(a) - f(0)}{a - 0} = \frac{f(a)}{a} \]
Since \( f(0) = 0 \).

### Step 6: Calculate \( f(a) \)
For \( a > 0 \):
\[ f(a) = a \log_e a \]
Thus,
\[ \frac{f(a)}{a} = \log_e a \]

### Step 7: Find \( f'(x) \)
Using the product rule:
\[ f'(x) = \log_e x + 1 \]

### Step 8: Set up the equation
From LMVT, we have:
\[ \log_e c + 1 = \log_e a \]
This implies:
\[ \log_e c = \log_e a - 1 \]
Exponentiating both sides gives:
\[ c = \frac{a}{e} \]

### Step 9: Conclusion of LMVT holds at \( x = 1 \)
Given that LMVT holds at \( x = 1 \), we can substitute \( c = 1 \):
\[ 1 = \frac{a}{e} \implies a = e \]

### Step 10: Find \( [a^2] \)
Now we need to find \( [a^2] \):
\[ a^2 = e^2 \]
The value of \( e \) is approximately \( 2.718 \), thus:
\[ e^2 \approx 7.389 \]
The greatest integer function:
\[ [a^2] = [7.389] = 7 \]

### Final Answer
The answer is:
\[ \boxed{7} \]