Suppose f, f' and f'' are continuous on [0, e] and that `f'(e )= f(e ) = f(1) = 1` and `int_(1)^(e )(f(x))/(x^(2))dx=(1)/(2)`, then the value of `int_(1)^(e ) f''(x)ln xdx` equals :

Options

A. 0
B. 1
C. 2
D. None of these

Answer

To solve the problem, we need to evaluate the integral \( \int_{1}^{e} f''(x) \ln x \, dx \) given the conditions on the function \( f \). We will use integration by parts to simplify the evaluation.

### Step-by-step Solution:

1. **Set Up Integration by Parts**:
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Here, let:
- \( u = \ln x \) (therefore, \( du = \frac{1}{x} \, dx \))
- \( dv = f''(x) \, dx \) (therefore, \( v = f'(x) \))

2. **Apply Integration by Parts**:
Applying the integration by parts, we get:
\[ \int_{1}^{e} f''(x) \ln x \, dx = \left[ \ln x \cdot f'(x) \right]_{1}^{e} - \int_{1}^{e} f'(x) \cdot \frac{1}{x} \, dx \]

3. **Evaluate the Boundary Terms**:
Now, we evaluate the boundary terms:
- At \( x = e \):
\[ \ln e \cdot f'(e) = 1 \cdot 1 = 1 \]
- At \( x = 1 \):
\[ \ln 1 \cdot f'(1) = 0 \cdot f'(1) = 0 \]
Thus, the boundary term evaluates to:
\[ \left[ \ln x \cdot f'(x) \right]_{1}^{e} = 1 - 0 = 1 \]

4. **Evaluate the Remaining Integral**:
Now we need to evaluate the integral:
\[ - \int_{1}^{e} f'(x) \cdot \frac{1}{x} \, dx \]
We can again use integration by parts on this integral. Let:
- \( u = f(x) \) (thus, \( du = f'(x) \, dx \))
- \( dv = \frac{1}{x} \, dx \) (thus, \( v = \ln x \))

Applying integration by parts again:
\[ \int f'(x) \cdot \frac{1}{x} \, dx = \left[ f(x) \ln x \right]_{1}^{e} - \int f(x) \cdot \frac{1}{x} \, dx \]

5. **Evaluate the Boundary Terms Again**:
- At \( x = e \):
\[ f(e) \ln e = 1 \cdot 1 = 1 \]
- At \( x = 1 \):
\[ f(1) \ln 1 = 1 \cdot 0 = 0 \]
Thus, the boundary term evaluates to:
\[ \left[ f(x) \ln x \right]_{1}^{e} = 1 - 0 = 1 \]

6. **Combine Results**:
Now substituting back, we have:
\[ -\left( 1 - \int_{1}^{e} f(x) \cdot \frac{1}{x} \, dx \right) \]
Therefore:
\[ \int_{1}^{e} f'(x) \cdot \frac{1}{x} \, dx = 1 - \int_{1}^{e} f(x) \cdot \frac{1}{x} \, dx \]

7. **Given Condition**:
We know from the problem statement:
\[ \int_{1}^{e} \frac{f(x)}{x^2} \, dx = \frac{1}{2} \]
This implies:
\[ \int_{1}^{e} f(x) \cdot \frac{1}{x} \, dx = 2 \cdot \frac{1}{2} = 1 \]

8. **Final Calculation**:
Putting everything together:
\[ \int_{1}^{e} f''(x) \ln x \, dx = 1 - 1 = 0 \]

### Conclusion:
Thus, the value of \( \int_{1}^{e} f''(x) \ln x \, dx \) is **0**.