The tangent to the graph of the function `y = f(x)` at the point with abscissa `x = a` makes with x-axis an angle of `pi//3` and at the abscissa `x = b` an angle of `pi/4`. The value of the integral `int_a^b f' (x) f'' (x) dx` is

Options

A. `1/2 (1 - sqrt(3))`
B. `1/2 (1 + sqrt(3))`
C. `-1`
D. none

Answer

To solve the problem, we need to find the value of the integral \( \int_a^b f'(x) f''(x) \, dx \). We are given that the tangent to the graph of the function \( y = f(x) \) at \( x = a \) makes an angle of \( \frac{\pi}{3} \) with the x-axis, and at \( x = b \) it makes an angle of \( \frac{\pi}{4} \).

### Step-by-Step Solution:

1. **Determine the slopes of the tangent lines:**
- The slope of the tangent line at \( x = a \) is given by:
\[ f'(a) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]
- The slope of the tangent line at \( x = b \) is given by:
\[ f'(b) = \tan\left(\frac{\pi}{4}\right) = 1 \]

2. **Substitute for the integral:**
- We need to evaluate the integral \( \int_a^b f'(x) f''(x) \, dx \).
- We can use the substitution \( t = f'(x) \). Then, \( dt = f''(x) \, dx \), which implies \( dx = \frac{dt}{f''(x)} \).

3. **Change the limits of integration:**
- When \( x = a \), \( t = f'(a) = \sqrt{3} \).
- When \( x = b \), \( t = f'(b) = 1 \).

4. **Rewrite the integral:**
- The integral becomes:
\[ \int_{\sqrt{3}}^1 t \, dt \]

5. **Evaluate the integral:**
- The integral \( \int t \, dt \) is:
\[ \frac{t^2}{2} \bigg|_{\sqrt{3}}^1 = \frac{1^2}{2} - \frac{(\sqrt{3})^2}{2} = \frac{1}{2} - \frac{3}{2} = \frac{1 - 3}{2} = -1 \]

6. **Final result:**
- Therefore, the value of the integral \( \int_a^b f'(x) f''(x) \, dx \) is:
\[ -1 \]

### Final Answer:
\[ \int_a^b f'(x) f''(x) \, dx = -1 \]