The following functions are differentiable on (-1,2)

Options

A. `underset(x)overset(2x)int (log t)^(2)dt`
B. `underset(x)overset(2x)int (sin t)/(t)dt`
C. `underset(x)overset(2x)int (1-t+t^(2))/(1+t+t^(2))dt`
D. None of these

Answer

To determine which of the given functions are differentiable on the interval (-1, 2), we will analyze each function step by step.

### Step 1: Analyze the first function
Let \( f(x) = \int_{0}^{x} (2x \log t)^2 dt \).

To find \( f'(x) \), we apply the Leibniz rule for differentiation under the integral sign:
\[ f'(x) = \frac{d}{dx} \int_{0}^{x} (2x \log t)^2 dt = (2x \log x)^2 + \int_{0}^{x} \frac{\partial}{\partial x} (2x \log t)^2 dt \]
We need to evaluate \( f'(x) \) at \( x = 0 \) to check differentiability.

However, \( \log(0) \) is undefined, which implies that \( f'(x) \) is not defined at \( x = 0 \). Therefore, \( f(x) \) is not differentiable at \( x = 0 \).

### Step 2: Analyze the second function
Let \( g(x) = \int_{0}^{x} \frac{\sin t}{t} dt \).

We find \( g'(x) \) using the Leibniz rule:
\[ g'(x) = \frac{\sin x}{x} \]
At \( x = 0 \), \( g'(0) \) is not defined since \( \frac{\sin 0}{0} \) is an indeterminate form. Therefore, \( g(x) \) is not differentiable at \( x = 0 \).

### Step 3: Analyze the third function
Let \( h(x) = \int_{0}^{x} \frac{1 - t + t^2}{1 + t + t^2} dt \).

We find \( h'(x) \) using the Leibniz rule:
\[ h'(x) = \frac{1 - x + x^2}{1 + x + x^2} \]
This function is defined for all \( x \) in the interval (-1, 2).

To check if \( h'(x) \) is continuous and differentiable, we observe that the denominator \( 1 + x + x^2 \) is never zero in this interval, ensuring that \( h'(x) \) is well-defined and continuous.

### Conclusion
From the analysis:
1. The first function \( f(x) \) is not differentiable at \( x = 0 \).
2. The second function \( g(x) \) is not differentiable at \( x = 0 \).
3. The third function \( h(x) \) is differentiable on the interval (-1, 2).

Thus, the correct answer is that only the third function is differentiable on (-1, 2).