Key Concepts and Formulas

• Limits: The limit $\mathop {\lim }\limits_{x \to a} \frac{f(x)}{g(x)} = L$ implies specific relationships between the behavior of $f(x)$ and $g(x)$ as $x$ approaches $a$. If $L$ is finite and non-zero, it suggests that the lowest degree term in $f(x)$ is related to the degree of $g(x)$.
• Extrema and Derivatives: At a local maximum or minimum (extreme value) of a differentiable function $f(x)$, the first derivative $f'(x)$ is equal to zero.
• Polynomial Representation: A polynomial of degree $n$ can be represented as $f(x) = a_n x^n + a_{n-1}x^{n-1} + ... + a_1 x + a_0$, where $a_n \neq 0$.

Step-by-Step Solution

Step 1: Analyze the Limit Condition

We are given that $\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$. We need to extract information about the polynomial $f(x)$ from this limit.

First, simplify the limit: $\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$ $1 + \mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 3$ $\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2$

Let $f(x) = Ax^4 + Bx^3 + Cx^2 + Dx + E$. Then $\mathop {\lim }\limits_{x \to 0} \frac{Ax^4 + Bx^3 + Cx^2 + Dx + E}{x^2} = 2$ For the limit to exist and equal 2, we must have $D = 0$ and $E = 0$. Therefore, $f(x)$ must be of the form $f(x) = Ax^4 + Bx^3 + Cx^2$. Then, $\mathop {\lim }\limits_{x \to 0} \frac{Ax^4 + Bx^3 + Cx^2}{x^2} = \mathop {\lim }\limits_{x \to 0} (Ax^2 + Bx + C) = C$ Since the limit is 2, we have $C = 2$. Thus, $f(x) = Ax^4 + Bx^3 + 2x^2$. Let $A = a_3$ and $B = a_2$. $f(x) = a_3 x^4 + a_2 x^3 + 2x^2$

Step 2: Utilize the Extrema Condition

We are given that $f(x)$ has extreme values at $x=1$ and $x=2$. This means $f'(1) = 0$ and $f'(2) = 0$. First, find $f'(x)$: $f'(x) = 4a_3 x^3 + 3a_2 x^2 + 4x$ Now, apply the conditions $f'(1) = 0$ and $f'(2) = 0$:

For $x = 1$: $f'(1) = 4a_3 (1)^3 + 3a_2 (1)^2 + 4(1) = 0$ $4a_3 + 3a_2 + 4 = 0 \quad \ldots(1)$

For $x = 2$: $f'(2) = 4a_3 (2)^3 + 3a_2 (2)^2 + 4(2) = 0$ $32a_3 + 12a_2 + 8 = 0$ Divide by 4: $8a_3 + 3a_2 + 2 = 0 \quad \ldots(2)$

Step 3: Solve the System of Linear Equations

We have the following system of equations:

1. $4a_3 + 3a_2 = -4$
2. $8a_3 + 3a_2 = -2$

Subtract Equation (1) from Equation (2): $(8a_3 + 3a_2) - (4a_3 + 3a_2) = -2 - (-4)$ $4a_3 = 2$ $a_3 = \frac{1}{2}$

Substitute $a_3 = \frac{1}{2}$ into Equation (1): $4\left(\frac{1}{2}\right) + 3a_2 = -4$ $2 + 3a_2 = -4$ $3a_2 = -6$ $a_2 = -2$

Step 4: Construct $f(x)$ and Calculate $f(2)$

Now we have $a_3 = \frac{1}{2}$ and $a_2 = -2$. So, $f(x) = \frac{1}{2}x^4 - 2x^3 + 2x^2$ Now calculate $f(2)$: $f(2) = \frac{1}{2}(2)^4 - 2(2)^3 + 2(2)^2$ $f(2) = \frac{1}{2}(16) - 2(8) + 2(4)$ $f(2) = 8 - 16 + 8$ $f(2) = 0$

Common Mistakes & Tips

• Remember that extrema imply $f'(x) = 0$, not $f(x) = 0$.
• When dealing with limits of the form $\lim_{x \to 0} \frac{f(x)}{x^n}$, the value of the limit provides direct information about the lowest degree terms of $f(x)$.
• Be careful with algebraic manipulations when solving systems of equations.

Summary

By analyzing the limit condition, we determined the form of the polynomial $f(x)$ to be $f(x) = a_3 x^4 + a_2 x^3 + 2x^2$. Using the extrema conditions, $f'(1) = 0$ and $f'(2) = 0$, we formed a system of two linear equations and solved for $a_2$ and $a_3$. Finally, we constructed the complete polynomial and evaluated $f(2)$, which resulted in 0.

Final Answer

The final answer is \boxed{0}, which corresponds to option (A).