Let `f` be a continuous, differentiable, and bijective function. If the tangent to `y=f(x)a tx=a` is also the normal to `y=f(x)a tx=b ,` then there exists at least one `c in (a , b)` such that `f^(prime)(c)=0` (b) `f^(prime)(c)>0` `f^(prime)(c)<0` (d) none of these

Options

A. `f'(c)=0`
B. `f'(c)gt0`
C. `f'(c)lt0`
D. none of these

Answer

To solve the given problem, we need to analyze the conditions provided regarding the function \( f \) and its derivatives.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a function \( f \) that is continuous, differentiable, and bijective. The tangent line at \( x = a \) is also the normal line at \( x = b \). We need to find the behavior of the derivative \( f'(c) \) for some \( c \) in the interval \( (a, b) \).

2. **Tangent and Normal Lines**:
- The slope of the tangent line at \( x = a \) is given by \( f'(a) \).
- The equation of the tangent line at \( x = a \) can be expressed as:
\[ y - f(a) = f'(a)(x - a) \]
- The slope of the normal line at \( x = b \) is the negative reciprocal of the slope of the tangent line at \( x = b \), which is \( -\frac{1}{f'(b)} \).

3. **Equating the Lines**:
Since the tangent at \( x = a \) is also the normal at \( x = b \), we can set the slopes equal:
\[ f'(a) = -\frac{1}{f'(b)} \]
This implies that the product of the slopes is:
\[ f'(a) f'(b) = -1 \]

4. **Analyzing the Sign of the Derivatives**:
- Since \( f \) is bijective, \( f'(x) \) must not change signs over the interval \( (a, b) \). Therefore, if \( f'(a) > 0 \), then \( f'(b) < 0 \), and vice versa.
- This indicates that the derivative \( f' \) must take a value of zero at least once in the interval \( (a, b) \) due to the Intermediate Value Theorem.

5. **Conclusion**:
By the Mean Value Theorem, since \( f' \) changes from positive to negative (or negative to positive), there must exist at least one point \( c \) in the interval \( (a, b) \) such that:
\[ f'(c) = 0 \]

### Final Answer:
Thus, the correct option is:
(a) \( f'(c) = 0 \)