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(D) Given $\mathop {\lim }\limits_{x 	o 0} \left[ {1 + \frac{{f(x)}}{{{x^2}}}} ight] = 3$,this implies $\mathop {\lim }\limits_{x 	o 0} \frac{{f(x

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(D) Given $\mathop {\lim }\limits_{x 	o 0} \left[ {1 + \frac{{f(x)}}{{{x^2}}}} ight] = 3$,this implies $\mathop {\lim }\limits_{x 	o 0} \frac{{f(x)}}{{{x^2}}} = 2$.Since $f(x)$ is a polynomial of degree $4$,let $f(x) = ax^4 + bx^3 + cx^2 + dx + e$.For the limit to exist and equal $2$,we must have $e=0$ and $d=0$,and $c=2$.Thus,$f(x) = ax^4 + bx^3 + 2x^2$.Then $f'(x) = 4ax^3 + 3bx^2 + 4x$.Since $f(x)$ has extreme values at $x=1$ and $x=2$,$f'(1) = 0$ and $f'(2) = 0$.$f'(1) = 4a + 3b + 4 = 0 \Rightarrow 4a + 3b = -4$.$f'(2) = 4a(8) + 3b(4) + 4(2) = 32a + 12b + 8 = 0 \Rightarrow 8a + 3b = -2$.Subtracting the first equation from the second: $(8a + 3b) - (4a + 3b) = -2 - (-4) \Rightarrow 4a = 2 \Rightarrow a = \frac{1}{2}$.Substituting $a = \frac{1}{2}$ into $4a + 3b = -4$: $4(\frac{1}{2}) + 3b = -4 \Rightarrow 2 + 3b = -4 \Rightarrow 3b = -6 \Rightarrow b = -2$.So,$f(x) = \frac{1}{2}x^4 - 2x^3 + 2x^2$.Calculating $f(2)$: $f(2) = \frac{1}{2}(16) - 2(8) + 2(4) = 8 - 16 + 8 = 0$.

Let $f(x)$ be a polynomial of degree four having extreme values at $x=1$ and $x=2$. If $\mathop {\lim }\limits_{x \to 0} \left[ {1 + \frac{{f(x)}}{{{x^2}}}} \right] = 3$,then $f(2)$ is equal to:

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