Key Concepts and Formulas

• Polynomials: A polynomial of degree $n$ has the general form $f(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$, where $a_n \neq 0$.
• Extreme Values and Derivatives: If $f(x)$ has an extreme value at $x=c$, then $f'(c) = 0$.
• Limits: $\lim_{x \to 0} \frac{f(x)}{x^n} = L$ (where $L$ is a finite non-zero constant) implies that $f(x)$ has the form $f(x) = Lx^n +$ (terms of higher degree).
• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $r_1$ and $r_2$, $r_1 + r_2 = -\frac{b}{a}$ and $r_1 r_2 = \frac{c}{a}$.

Step-by-Step Solution

Step 1: Defining the general form of the polynomial

Since $f(x)$ is a polynomial of degree four, we can write it as $f(x) = ax^4 + bx^3 + cx^2 + dx + e$ where $a, b, c, d, e$ are constants and $a \neq 0$.

Step 2: Applying the limit condition

We are given that $\lim_{x \to 0} \frac{f(x)}{x^2} = 5$. Substituting the polynomial form, we have $\lim_{x \to 0} \frac{ax^4 + bx^3 + cx^2 + dx + e}{x^2} = 5$ For this limit to exist and be equal to 5, the terms with powers of $x$ less than 2 must be zero. Otherwise, the limit would be infinite. Therefore, $d = 0$ and $e = 0$. This simplifies the polynomial to $f(x) = ax^4 + bx^3 + cx^2$ Now, the limit becomes $\lim_{x \to 0} \frac{ax^4 + bx^3 + cx^2}{x^2} = \lim_{x \to 0} (ax^2 + bx + c) = 5$ As $x$ approaches 0, $ax^2 + bx$ approaches 0, so we must have $c = 5$. Thus, $f(x) = ax^4 + bx^3 + 5x^2$

Step 3: Finding the derivative and applying the extreme value condition

We are given that $f(x)$ has extreme values at $x = 4$ and $x = 5$. This means $f'(4) = 0$ and $f'(5) = 0$. First, we find the derivative: $f'(x) = 4ax^3 + 3bx^2 + 10x = x(4ax^2 + 3bx + 10)$ Since $f'(4) = 0$ and $f'(5) = 0$, and $4 \neq 0$ and $5 \neq 0$, $x=4$ and $x=5$ must be roots of the quadratic $4ax^2 + 3bx + 10 = 0$.

Step 4: Using Vieta's formulas to find a and b

Since $x=4$ and $x=5$ are roots of $4ax^2 + 3bx + 10 = 0$, we can use Vieta's formulas. The sum of the roots is $4 + 5 = 9 = -\frac{3b}{4a}$, which gives $36a = -3b$, or $b = -12a$. The product of the roots is $4 \times 5 = 20 = \frac{10}{4a}$, which gives $80a = 10$, or $a = \frac{1}{8}$. Then $b = -12a = -12 \left(\frac{1}{8}\right) = -\frac{3}{2}$.

Step 5: Constructing f(x) and calculating f(2)

Now we have $a = \frac{1}{8}$, $b = -\frac{3}{2}$, and $c = 5$. Therefore, $f(x) = \frac{1}{8}x^4 - \frac{3}{2}x^3 + 5x^2$ We want to find $f(2)$: $f(2) = \frac{1}{8}(2^4) - \frac{3}{2}(2^3) + 5(2^2) = \frac{1}{8}(16) - \frac{3}{2}(8) + 5(4) = 2 - 12 + 20 = 10$

Common Mistakes & Tips

• Remember to consider the implications of the limit as $x$ approaches 0. The powers of $x$ in the numerator must be at least as large as the power of $x$ in the denominator for the limit to exist and be finite.
• Vieta's formulas can be a quick way to find the coefficients of a quadratic if you know its roots.
• Be careful with arithmetic when substituting values to calculate $f(2)$.

Summary

We used the limit condition to determine the form of the polynomial $f(x)$, then used the extreme value condition to set up equations for the remaining coefficients. Solving these equations, we found the complete polynomial and calculated $f(2)$. The value of $f(2)$ is 10.

Final Answer

The final answer is \boxed{10}, which corresponds to option (B).