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**:
For three terms \( a, b, c \) to be in G.P., the condition is:
\[ b^2 = ac \]
In our case, we have:
\[ (f'(x))^2 = f(x) \cdot f''(x) \]

2. **Assuming a Function**:
Let's assume \( f(x) \) is an exponential function of the form:
\[ f(x) = e^{kx} \]
where \( k \) is a constant.

3. **Finding Derivatives**:
Now, we calculate the first and second derivatives:
\[ f'(x) = k e^{kx} \]
\[ f''(x) = k^2 e^{kx} \]

4. **Substituting into the G.P. Condition**:
Now we substitute \( f(x) \), \( f'(x) \), and \( f''(x) \) into the G.P. condition:
\[ (f'(x))^2 = (k e^{kx})^2 = k^2 e^{2kx} \]
\[ f(x) \cdot f''(x) = e^{kx} \cdot (k^2 e^{kx}) = k^2 e^{2kx} \]

5. **Verifying the G.P. Condition**:
We see that:
\[ (f'(x))^2 = k^2 e^{2kx} \quad \text{and} \quad f(x) \cdot f''(x) = k^2 e^{2kx} \]
Since both sides are equal, the condition for G.P. is satisfied.

6. **Conclusion**:
Therefore, the function \( f(x) \) such that \( f(x) \), \( f'(x) \), and \( f''(x) \) are in G.P. is:
\[ f(x) = e^{kx} \]
for some constant \( k \).

### Final Answer:
The function \( f(x) \) is an exponential function.