If the normal to the curve `y =f(x)` at the point `(1, 2)` makes an angle `3pi//4` with the positive x-axis, then `f'(1)` is

Options

A. 1
B. `-1`
C. `0`
D. `2`

Answer

To solve the problem, we need to find the derivative \( f'(1) \) given that the normal to the curve \( y = f(x) \) at the point \( (1, 2) \) 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 can be determined from the angle it makes with the positive x-axis. The slope \( m' \) of the normal is given by:
\[ m' = \tan\left(\frac{3\pi}{4}\right) \]

2. **Calculating the Slope of the Normal:**
We know that:
\[ \tan\left(\frac{3\pi}{4}\right) = -1 \]
Therefore, the slope of the normal at the point \( (1, 2) \) is:
\[ m' = -1 \]

3. **Relating the Slope of the Normal to the Slope of the Tangent:**
The slope of the normal is related to the slope of the tangent \( m \) by the equation:
\[ m' = -\frac{1}{m} \]
Thus, we can write:
\[ -1 = -\frac{1}{f'(1)} \]

4. **Solving for \( f'(1) \):**
Rearranging the equation gives:
\[ 1 = \frac{1}{f'(1)} \]
Therefore, we can find \( f'(1) \):
\[ f'(1) = 1 \]

5. **Conclusion:**
The value of \( f'(1) \) is:
\[ \boxed{1} \]