Determine Monotonicity and Local Extrema of Rational Function

To determine the monotonicity and local extrema of the given function $f(x)$, we start by differentiating $f(x)$ with respect to $x$.

The function is given as:
$f(x) = \frac{x^2 - ax + 1}{x^2 + ax + 1}, \quad 0 < a < 2$

Using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$, where $u = x^2 - ax + 1$ and $v = x^2 + ax + 1$:
$u' = 2x - a$
$v' = 2x + a$

Now, calculate $f'(x)$:
$f'(x) = \frac{(2x - a)(x^2 + ax + 1) - (x^2 - ax + 1)(2x + a)}{(x^2 + ax + 1)^2}$

Expanding the numerator:
$(2x - a)(x^2 + ax + 1) = 2x^3 + 2ax^2 + 2x - ax^2 - a^2 x - a = 2x^3 + ax^2 + (2 - a^2)x - a$
$(x^2 - ax + 1)(2x + a) = 2x^3 + ax^2 - 2ax^2 - a^2 x + 2x + a = 2x^3 - ax^2 + (2 - a^2)x + a$

Subtracting the two expressions gives:
$\text{Numerator} = [2x^3 + ax^2 + (2 - a^2)x - a] - [2x^3 - ax^2 + (2 - a^2)x + a] = 2ax^2 - 2a = 2a(x^2 - 1)$

Thus, the first derivative is:
$f'(x) = \frac{2a(x^2 - 1)}{(x^2 + ax + 1)^2}$

Analysis of Monotonicity on $(-1, 1)$:

1. Since $0 < a < 2$, $2a > 0$.
2. The denominator $(x^2 + ax + 1)^2$ is strictly positive for all $x \in \mathbb{R}$ because the discriminant of the quadratic $x^2 + ax + 1$ is $\Delta = a^2 - 4 < 0$ (as $0 < a < 2$), meaning $x^2 + ax + 1 > 0$ for all real $x$.
3. For $x \in (-1, 1)$, we have $x^2 < 1 \implies x^2 - 1 < 0$.

Therefore, for all $x \in (-1, 1)$:
$f'(x) < 0$
This implies that $f(x)$ is strictly decreasing on the interval $(-1, 1)$.

Local Extrema at $x = 1$:

1. At $x = 1$, $f'(1) = \frac{2a(1^2 - 1)}{(1 + a + 1)^2} = 0$.
2. For $x$ slightly less than $1$ (in $(-1, 1)$), $f'(x) < 0$.
3. For $x$ slightly greater than $1$ (for $x > 1$), $x^2 - 1 > 0 \implies f'(x) > 0$.

Since $f'(x)$ changes sign from negative to positive as $x$ increases through $x = 1$, by the First Derivative Test, $f(x)$ has a local minimum at $x = 1$.

Hence, $f(x)$ is decreasing on $(-1, 1)$ and has a local minimum at $x = 1$.

Correct Option: A