If `f(x)= int_(0)^(x)(f(t))^(2) dt, f:R rarr R ` be differentiable function and `f(g(x))` is differentiable at `x=a`, then

Options

A. g (x) must be differentiable at `x=a`
B. g(x) may be non-differentiable at `x=a`
C. g(x) may be discontinous at `x=a`
D. None of the above

Answer

To solve the problem, we start with the given function and apply the necessary calculus concepts.

### Step-by-Step Solution:

1. **Define the Function**: We start with the function defined as:
\[ f(x) = \int_0^x (f(t))^2 \, dt \]

2. **Differentiate Using the Fundamental Theorem of Calculus**: To find \( f'(x) \), we differentiate both sides with respect to \( x \):
\[ f'(x) = \frac{d}{dx} \left( \int_0^x (f(t))^2 \, dt \right) = (f(x))^2 \]
Here, we used the Fundamental Theorem of Calculus which states that if \( F(x) = \int_a^x f(t) \, dt \), then \( F'(x) = f(x) \).

3. **Chain Rule for Composite Functions**: Now, we consider \( f(g(x)) \) and differentiate it using the chain rule:
\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \]

4. **Substituting \( f'(g(x)) \)**: From our earlier result, we know:
\[ f'(g(x)) = (f(g(x)))^2 \]
Therefore, we can substitute this into our derivative:
\[ \frac{d}{dx} f(g(x)) = (f(g(x)))^2 \cdot g'(x) \]

5. **Evaluating at \( x = a \)**: Since we are given that \( f(g(x)) \) is differentiable at \( x = a \), we evaluate:
\[ \frac{d}{dx} f(g(x)) \bigg|_{x=a} = (f(g(a)))^2 \cdot g'(a) \]

6. **Conclusion on Differentiability**: For \( f(g(a)) \) to be differentiable, \( g'(a) \) must exist. Thus, we conclude that:
\[ g(x) \text{ must be differentiable at } x = a \]

### Final Answer:
The correct conclusion is that \( g(x) \) must be differentiable at \( x = a \).