Consider the function f : `(-oo , oo) to (-oo , oo)` defined by f(x) = `(x^(2) - ax + 1)/(x^(2) + ax + 1) , 0 lt a lt 2`
Let `g (x) = underset(0) overset(e^(x))(int) (f'(t))/(1 + t^(2)) `dt
Which of the following is true ?

Options

A. g'(x) is positive on `(-oo , 0)` and negative on `(0 , oo)`
B. g'(x) is negative on `(-oo, 0)` and positive on `(0,oo)`
C. g'(x) changes sign on both `(-oo , 0)` and `(0 , oo)`
D. g'(x) does not change sign on `(-oo , oo)`

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x^2 - ax + 1}{x^2 + ax + 1} \) and its derivative, and then evaluate the function \( g(x) \).

### Step 1: Find the derivative \( f'(x) \)

Using the quotient rule for differentiation, we have:

\[ f'(x) = \frac{(g(x)h'(x) - g'(x)h(x))}{(h(x))^2} \]

where \( g(x) = x^2 - ax + 1 \) and \( h(x) = x^2 + ax + 1 \).

Calculating \( g'(x) \) and \( h'(x) \):

- \( g'(x) = 2x - a \)
- \( h'(x) = 2x + a \)

Now applying the quotient rule:

\[ f'(x) = \frac{(2x - a)(x^2 + ax + 1) - (x^2 - ax + 1)(2x + a)}{(x^2 + ax + 1)^2} \]

### Step 2: Simplify the expression for \( f'(x) \)

Expanding the numerator:

1. \( (2x - a)(x^2 + ax + 1) = 2x^3 + 2ax^2 + 2x - ax^2 - a^2x - a \)
2. \( (x^2 - ax + 1)(2x + a) = 2x^3 + ax^2 - 2ax^2 - a^2x - 2x - a \)

Now combining these:

\[ f'(x) = \frac{(2x^3 + 2ax^2 + 2x - ax^2 - a^2x - a) - (2x^3 + ax^2 - 2ax^2 - a^2x - 2x - a)}{(x^2 + ax + 1)^2} \]

After simplifying, we get:

\[ f'(x) = \frac{(2x - a)(x^2 + ax + 1) - (x^2 - ax + 1)(2x + a)}{(x^2 + ax + 1)^2} \]

### Step 3: Evaluate \( g'(x) \)

Given \( g(x) = \int_0^{e^x} \frac{f'(t)}{1 + t^2} dt \), we can differentiate \( g(x) \) using the Fundamental Theorem of Calculus and the chain rule:

\[ g'(x) = \frac{d}{dx} \left( \int_0^{e^x} \frac{f'(t)}{1 + t^2} dt \right) = \frac{f'(e^x)}{1 + (e^x)^2} \cdot e^x \]

### Step 4: Analyze the sign of \( g'(x) \)

Since \( 0 < a < 2 \), we need to analyze \( f'(0) \):

\[ f'(0) = \frac{(2(0) - a)(0^2 + a(0) + 1) - (0^2 - a(0) + 1)(2(0) + a)}{(0^2 + a(0) + 1)^2} \]

Calculating this gives:

\[ f'(0) = \frac{-a \cdot 1 - 1 \cdot a}{1} = -2a \]

Since \( 0 < a < 2 \), \( f'(0) < 0 \). Thus, \( g'(x) \) is negative.

### Conclusion

Since \( g'(x) < 0 \) for all \( x \), \( g(x) \) is a decreasing function. Therefore, the correct option is that \( g'(x) \) does not change sign in \( x \in \mathbb{R} \).