A function `y=f(x)` satisfies A function `y=f(x)` satisfies` f" (x)= -1/x^2 - pi^2 sin(pix) ; f'(2) = pi+1/2` and` f(1)=0`. The value of `f(1/2)` then

Options

A. a.`f((1)/(2))=1-ln2`
B. b.`f((1)/(.2))=(pi)/(2)-ln2`
C. c.`f(x)=0`has exactly 3 roots
D. d.`f(x)=0` has exactly one root

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript.

### Step 1: Start with the second derivative
We are given that:
\[ f''(x) = -\frac{1}{x^2} - \pi^2 \sin(\pi x) \]

### Step 2: Integrate to find the first derivative
Integrate \( f''(x) \) with respect to \( x \):
\[ f'(x) = \int f''(x) \, dx = \int \left(-\frac{1}{x^2} - \pi^2 \sin(\pi x)\right) \, dx \]

The integral of \( -\frac{1}{x^2} \) is \( \frac{1}{x} \) and the integral of \( -\pi^2 \sin(\pi x) \) is \( \frac{\pi}{\pi^2} \cos(\pi x) = -\frac{1}{\pi} \cos(\pi x) \).

Thus,
\[ f'(x) = \frac{1}{x} + \frac{\pi}{\pi^2} \cos(\pi x) + C_1 \]
or simplifying,
\[ f'(x) = \frac{1}{x} + \frac{1}{\pi} \cos(\pi x) + C_1 \]

### Step 3: Use the condition \( f'(2) = \pi + \frac{1}{2} \)
Substituting \( x = 2 \) into the equation for \( f'(x) \):
\[ f'(2) = \frac{1}{2} + \frac{1}{\pi} \cos(2\pi) + C_1 \]
Since \( \cos(2\pi) = 1 \):
\[ f'(2) = \frac{1}{2} + \frac{1}{\pi} + C_1 = \pi + \frac{1}{2} \]

### Step 4: Solve for \( C_1 \)
Setting the equation:
\[ \frac{1}{2} + \frac{1}{\pi} + C_1 = \pi + \frac{1}{2} \]
Subtract \( \frac{1}{2} \) from both sides:
\[ \frac{1}{\pi} + C_1 = \pi \]
Thus,
\[ C_1 = \pi - \frac{1}{\pi} \]

### Step 5: Substitute \( C_1 \) back into \( f'(x) \)
Now we have:
\[ f'(x) = \frac{1}{x} + \frac{1}{\pi} \cos(\pi x) + \left(\pi - \frac{1}{\pi}\right) \]

### Step 6: Integrate to find \( f(x) \)
Integrate \( f'(x) \):
\[ f(x) = \int f'(x) \, dx = \int \left(\frac{1}{x} + \frac{1}{\pi} \cos(\pi x) + \pi - \frac{1}{\pi}\right) \, dx \]

This gives:
\[ f(x) = \ln |x| + \frac{1}{\pi^2} \sin(\pi x) + \pi x - \frac{1}{\pi} x + C_2 \]

### Step 7: Use the condition \( f(1) = 0 \)
Substituting \( x = 1 \):
\[ f(1) = \ln(1) + \frac{1}{\pi^2} \sin(\pi) + \pi(1) - \frac{1}{\pi}(1) + C_2 = 0 \]
Since \( \ln(1) = 0 \) and \( \sin(\pi) = 0 \):
\[ \pi - \frac{1}{\pi} + C_2 = 0 \]
Thus,
\[ C_2 = -\pi + \frac{1}{\pi} \]

### Step 8: Final expression for \( f(x) \)
Substituting \( C_2 \) back into \( f(x) \):
\[ f(x) = \ln |x| + \frac{1}{\pi^2} \sin(\pi x) + \pi x - \frac{1}{\pi} x - \pi + \frac{1}{\pi} \]

### Step 9: Calculate \( f(1/2) \)
Now we need to find \( f(1/2) \):
\[ f\left(\frac{1}{2}\right) = \ln\left(\frac{1}{2}\right) + \frac{1}{\pi^2} \sin\left(\frac{\pi}{2}\right) + \pi \left(\frac{1}{2}\right) - \frac{1}{\pi} \left(\frac{1}{2}\right) - \pi + \frac{1}{\pi} \]
\[ = -\ln(2) + \frac{1}{\pi^2} \cdot 1 + \frac{\pi}{2} - \frac{1}{2\pi} - \pi + \frac{1}{\pi} \]
\[ = -\ln(2) + \frac{1}{\pi^2} + \frac{\pi}{2} - \pi + \frac{1}{2\pi} \]

### Final Result
The value of \( f(1/2) \) is:
\[ f\left(\frac{1}{2}\right) = -\ln(2) + \frac{1}{\pi^2} + \frac{\pi}{2} - \pi + \frac{1}{2\pi} \]