Statement I The function `f(x) = int_(0)^(x) sqrt(1+t^(2) dt )` is an odd function and `g(x)=f'(x)` is an even function , then `f(x)` is an odd function.

Options

A. Statement I is true, Statement II is also true, Statement II is the correct explanation of Statement 1.
B. Statement I is true, Statement II is also true , Statement II is not the correct explanation of Statement II.
C. Statement I is true, Statement II is false
D. Statement I is false , Statement II is true

Answer

To solve the problem, we need to analyze the statements regarding the functions \( f(x) \) and \( g(x) \).

### Step 1: Define the function \( f(x) \)

Given:
\[ f(x) = \int_{0}^{x} \sqrt{1 + t^2} \, dt \]

### Step 2: Check if \( f(x) \) is an odd function

A function \( f(x) \) is odd if:
\[ f(-x) = -f(x) \]

Calculate \( f(-x) \):
\[ f(-x) = \int_{0}^{-x} \sqrt{1 + t^2} \, dt \]

Using the property of definite integrals, we can change the limits:
\[ f(-x) = -\int_{-x}^{0} \sqrt{1 + t^2} \, dt = -\int_{0}^{x} \sqrt{1 + (-u)^2} \, (-du) \quad \text{(where \( u = -t \))} \]
This simplifies to:
\[ f(-x) = -\int_{0}^{x} \sqrt{1 + u^2} \, du = -f(x) \]

Thus, \( f(-x) = -f(x) \), confirming that \( f(x) \) is indeed an odd function.

### Step 3: Find the derivative \( g(x) = f'(x) \)

Using the Fundamental Theorem of Calculus:
\[ g(x) = f'(x) = \sqrt{1 + x^2} \]

### Step 4: Check if \( g(x) \) is an even function

A function \( g(x) \) is even if:
\[ g(-x) = g(x) \]

Calculate \( g(-x) \):
\[ g(-x) = \sqrt{1 + (-x)^2} = \sqrt{1 + x^2} = g(x) \]

Thus, \( g(x) \) is an even function.

### Conclusion

From the analysis:
- \( f(x) \) is an odd function.
- \( g(x) = f'(x) \) is an even function.

Therefore, both statements are true:
1. \( f(x) \) is an odd function.
2. \( g(x) \) is an even function.

### Final Answer

Both statements are true.