Let `f(x)` be a function such that `f(x), f'(x) and f''(x)` are in G.P., then function f(x) is

Options

A. constant
B. logarithmic
C. exponential
D. linear

Answer

To solve the problem, we need to determine the function \( f(x) \) such that \( f(x) \), \( f'(x) \), and \( f''(x) \) are in geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding Geometric Progression (G.P.):**
If \( f(x) \), \( f'(x) \), and \( f''(x) \) are in G.P., then the relationship can be expressed as:
\[ (f'(x))^2 = f(x) \cdot f''(x) \]

2. **Assuming a General Form of \( f(x) \):**
Let's assume \( f(x) = e^{mx} \), where \( m \) is a constant. This is a common assumption because exponential functions often yield simple derivatives.

3. **Finding the Derivatives:**
- The first derivative \( f'(x) \) is:
\[ f'(x) = m e^{mx} \]
- The second derivative \( f''(x) \) is:
\[ f''(x) = m^2 e^{mx} \]

4. **Substituting into the G.P. Condition:**
Now we substitute \( f(x) \), \( f'(x) \), and \( f''(x) \) into the G.P. condition:
\[ (f'(x))^2 = f(x) \cdot f''(x) \]
This gives:
\[ (m e^{mx})^2 = e^{mx} \cdot (m^2 e^{mx}) \]
Simplifying both sides:
\[ m^2 e^{2mx} = m^2 e^{2mx} \]
This is true for all \( x \).

5. **Conclusion:**
Since our assumption \( f(x) = e^{mx} \) satisfies the condition that \( f(x) \), \( f'(x) \), and \( f''(x) \) are in G.P., we conclude that the function \( f(x) \) can be expressed as:
\[ f(x) = e^{mx} \quad \text{for some constant } m. \]

### Final Answer:
The function \( f(x) \) is of the form \( f(x) = e^{mx} \) where \( m \) is a constant.