If the normal to curve `y = f (x)` at the point `(3,4)` makes an angle `(3pi)/(4)` with positive x-axis then `f'(3)` equals:

Options

A. `-1`
B. `-3//4`
C. `4//5`
D. 1

Explanation

To solve the problem, we need to determine the value of \( f'(3) \) given that the normal to the curve \( y = f(x) \) at the point \( (3, 4) \) makes an angle of \( \frac{3\pi}{4} \) with the positive x-axis.

### Step-by-Step Solution:

1. **Understanding the Slope of the Normal:**
The slope of the normal line to the curve at a point is the negative reciprocal of the slope of the tangent line at that point. If \( m_t \) is the slope of the tangent, then the slope of the normal \( m_n \) can be expressed as:
\[ m_n = -\frac{1}{m_t} \]

2. **Finding the Slope of the Normal from the Angle:**
The slope of a line that makes an angle \( \theta \) with the positive x-axis is given by \( \tan(\theta) \). Here, the angle is \( \frac{3\pi}{4} \):
\[ m_n = \tan\left(\frac{3\pi}{4}\right) \]
We know that:
\[ \tan\left(\frac{3\pi}{4}\right) = -1 \]
Therefore, the slope of the normal at the point \( (3, 4) \) is:
\[ m_n = -1 \]

3. **Relating the Slope of the Normal to the Slope of the Tangent:**
From step 1, we have:
\[ -\frac{1}{f'(3)} = -1 \]
This implies:
\[ \frac{1}{f'(3)} = 1 \]

4. **Solving for \( f'(3) \):**
To find \( f'(3) \), we take the reciprocal:
\[ f'(3) = 1 \]

### Final Answer:
Thus, the value of \( f'(3) \) is:
\[ \boxed{1} \]