The user wants me to analyze the function f(x) and determine its behavior regarding monotonicity and local extrema.

Step 1: Analyze the given function

The function is given by: $f(x) = \frac{x^2 - ax + 1}{x^2 + ax + 1}$ with the condition $0 < a < 2$. The domain is $(-\infty, \infty)$.

First, let's analyze the denominator, $D(x) = x^2 + ax + 1$. The discriminant of this quadratic is $\Delta = a^2 - 4(1)(1) = a^2 - 4$. Since $0 < a < 2$, we have $0 < a^2 < 4$. This implies that $\Delta = a^2 - 4 < 0$. Because the leading coefficient (1) is positive and the discriminant is negative, the denominator $x^2 + ax + 1$ is always positive for all real values of $x$. This means the function $f(x)$ is continuous and differentiable for all $x \in \mathbb{R}$.

Step 2: Find the derivative of the function

To find the intervals where the function is increasing or decreasing, we need to compute the first derivative, $f'(x)$, using the quotient rule: $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$.

Let $u(x) = x^2 - ax + 1$ and $v(x) = x^2 + ax + 1$. Then, $u'(x) = 2x - a$ and $v'(x) = 2x + a$.

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

Now, let's simplify the numerator:

Numerator = $( (2x)(x^2 + ax + 1) - a(x^2 + ax + 1) ) - ( (x^2 - ax + 1)(2x) + a(x^2 - ax + 1) )$ = $(2x^3 + 2ax^2 + 2x - ax^2 - a^2x - a) - (2x^3 - 2ax^2 + 2x + ax^2 - a^2x + a)$ = $(2x^3 + ax^2 + (2-a^2)x - a) - (2x^3 - ax^2 + (2-a^2)x + a)$ = $2x^3 + ax^2 + (2-a^2)x - a - 2x^3 + ax^2 - (2-a^2)x - a$ = $2ax^2 - 2a$ = $2a(x^2 - 1)$

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

Step 3: Analyze the sign of the derivative

The denominator, $(x^2 + ax + 1)^2$, is always positive. We are given that $0 < a < 2$, so the term $2a$ is also positive. Therefore, the sign of $f'(x)$ is determined solely by the sign of the term $(x^2 - 1)$.

• Case 1: $f'(x) < 0$ This occurs when $x^2 - 1 < 0$, which means $x^2 < 1$. This inequality holds for $-1 < x < 1$. So, $f(x)$ is decreasing on the interval $(-1, 1)$.

• Case 2: $f'(x) > 0$ This occurs when $x^2 - 1 > 0$, which means $x^2 > 1$. This inequality holds for $x < -1$ or $x > 1$. So, $f(x)$ is increasing on the intervals $(-\infty, -1)$ and $(1, \infty)$.

• Case 3: $f'(x) = 0$ This occurs when $x^2 - 1 = 0$, which means $x = -1$ or $x = 1$. These are the critical points of the function.

Step 4: Determine the nature of the critical points

We use the First Derivative Test to classify the critical points:

• At $x = 1$:

  • For $x$ slightly less than 1 (i.e., in $(-1, 1)$), $f'(x) < 0$ (function is decreasing).
  • For $x$ slightly greater than 1, $f'(x) > 0$ (function is increasing). Since the function changes from decreasing to increasing at $x=1$, there is a local minimum at $x=1$.

• At $x = -1$:

  • For $x$ slightly less than -1, $f'(x) > 0$ (function is increasing).
  • For $x$ slightly greater than -1 (i.e., in $(-1, 1)$), $f'(x) < 0$ (function is decreasing). Since the function changes from increasing to decreasing at $x=-1$, there is a local maximum at $x=-1$.

Step 5: Evaluate the given options

• A: $f(x)$ is decreasing on $(-1,1)$ and has a local minimum at $x=1$ This matches our findings from Step 3 and Step 4. This statement is true.

• B: $f(x)$ is increasing on $(-1,1)$ and has a local minimum at $x=1$ The first part is incorrect. The function is decreasing on $(-1,1)$.

• C: $f(x)$ is increasing on $(-1,1)$ but has neither a local maximum nor a local minimum at $x=1$ Both parts of this statement are incorrect.

• D: $f(x)$ is decreasing on $(-1,1)$ but has neither a local maximum nor a local minimum at $x=1$ The second part is incorrect. The function has a local minimum at $x=1$.

Therefore, the only true statement is option A.