Evaluate Relation Between Higher Derivatives of Rational Function

To determine which statement is true, we first calculate the first and second derivatives of the function $f(x)$.

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

Let $N(x) = x^2 - ax + 1$ and $D(x) = x^2 + ax + 1$.

Using the quotient rule, the first derivative $f'(x)$ is:
$f'(x) = \frac{N'(x)D(x) - N(x)D'(x)}{(D(x))^2}$

Substitute $N'(x) = 2x - a$ and $D'(x) = 2x + a$:
$N'(x)D(x) - N(x)D'(x) = (2x - a)(x^2 + ax + 1) - (x^2 - ax + 1)(2x + a)$
$= (2x^3 + ax^2 + (2-a^2)x - a) - (2x^3 - ax^2 + (2-a^2)x + a)$
$= 2ax^2 - 2a = 2a(x^2 - 1)$

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

Evaluating $f'(x)$ at $x = 1$ and $x = -1$:
$f'(1) = 0 \quad \text{and} \quad f'(-1) = 0$

Now, we differentiate $f'(x)$ to find the second derivative $f''(x)$:
$f''(x) = 2a \cdot \frac{2x(D(x))^2 - (x^2 - 1) \cdot 2D(x)D'(x)}{(D(x))^4} = 2a \cdot \frac{2xD(x) - 2(x^2 - 1)D'(x)}{(D(x))^3}$

Now, evaluate $f''(x)$ at $x = 1$:
$D(1) = 1^2 + a(1) + 1 = 2 + a$
Since $x^2 - 1 = 0$ at $x = 1$:
$f''(1) = 2a \cdot \frac{2(1)D(1) - 0}{(D(1))^3} = \frac{4a}{(D(1))^2} = \frac{4a}{(2+a)^2}$

Next, evaluate $f''(x)$ at $x = -1$:
$D(-1) = (-1)^2 + a(-1) + 1 = 2 - a$
Since $x^2 - 1 = 0$ at $x = -1$:
$f''(-1) = 2a \cdot \frac{2(-1)D(-1) - 0}{(D(-1))^3} = -\frac{4a}{(D(-1))^2} = -\frac{4a}{(2-a)^2}$

Now substitute $f''(1)$ and $f''(-1)$ into the expression in Option A:
$(2+a)^2 f''(1) + (2-a)^2 f''(-1) = (2+a)^2 \left( \frac{4a}{(2+a)^2} \right) + (2-a)^2 \left( -\frac{4a}{(2-a)^2} \right)$
$= 4a - 4a = 0$

Thus, statement (A) is correct.