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Step-by-Step Solution
Step 1: Interpret the given limit condition
We are given
$ \lim_{x \to 0} \left[1 + \frac{f(x)}{x^2}\right] = 3.$
From this, we deduce
$ \lim_{x \to 0} \frac{f(x)}{x^2} = 2.$
This implies that near $x = 0,$ the polynomial $f(x)$ behaves like $2x^2$ (plus higher powers of $x$). Hence, the leading terms of $f(x)$ up to $x^4$ can be written as:
$ f(x) = a_1 x^2 + a_2 x^3 + a_3 x^4. $
From the limit, $a_1 = 2.$ So
$ f(x) = 2x^2 + a_2 x^3 + a_3 x^4. $
Step 2: Use the condition on extreme values (derivative = 0)
The problem states that $f(x)$ has extreme values at $x = 1$ and $x = 2$. For an extreme value, the derivative $f'(x)$ must be zero at those points. Let us compute $f'(x)$:
$ f'(x) = \frac{d}{dx}\bigl(2x^2 + a_2 x^3 + a_3 x^4\bigr)
= 4x + 3a_2 x^2 + 4a_3 x^3. $
The conditions for extreme values give us:
$ f'(1) = 4 + 3a_2 + 4a_3 = 0, $
$ f'(2) = 8 + 12a_2 + 32a_3 = 0. $
Step 3: Solve for the unknown coefficients
We have the system of linear equations:
(1) $4 + 3a_2 + 4a_3 = 0,$
(2) $8 + 12a_2 + 32a_3 = 0.$
To solve, subtract equation (1) multiplied by 2 from equation (2), or use any standard method. One convenient approach is:
Multiply equation (1) by 2 to get: $8 + 6a_2 + 8a_3 = 0.$
Subtract this from equation (2):
$ \bigl[8 + 12a_2 + 32a_3\bigr] - \bigl[8 + 6a_2 + 8a_3\bigr] = 0. $
This simplifies to:
$ (12a_2 - 6a_2) + (32a_3 - 8a_3) = 0 \quad \Longrightarrow \quad 6a_2 + 24a_3 = 0. $
But an even quicker route (as shown in the provided solution) can be used; either way, upon consistent manipulation, we find:
$ a_3 = \frac{1}{2}, \quad a_2 = -2. $
Step 4: Write down the polynomial $f(x)$
Having found $a_1, a_2,$ and $a_3,$ we can write:
$ f(x) = 2x^2 - 2x^3 + \tfrac{1}{2}x^4. $
Step 5: Evaluate $f(2)$
Substitute $x = 2$ into the polynomial:
$ f(2) = 2 \cdot (2)^2 \;-\; 2 \cdot (2)^3 \;+\; \tfrac{1}{2} \cdot (2)^4. $
Compute step-by-step:
$ (2)^2 = 4,$ so $2 \cdot 4 = 8.$
$ (2)^3 = 8,$ hence $-2 \cdot 8 = -16.$
$ (2)^4 = 16,$ so $\tfrac{1}{2} \cdot 16 = 8.$
Summing these terms: $ 8 - 16 + 8 = 0.$
Therefore, $ f(2) = 0. $
Answer: 0
