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(B) Given $\lim _{x \rightarrow 0} \frac{f(x)}{x^2}=5$. Since $f(x)$ is a polynomial of degree four,let $f(x) = ax^4 + bx^3 + cx^2 + dx + e$. For the

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$10$

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(B) Given $\lim _{x ightarrow 0} \frac{f(x)}{x^2}=5$. Since $f(x)$ is a polynomial of degree four,let $f(x) = ax^4 + bx^3 + cx^2 + dx + e$. For the limit to exist and equal $5$,we must have $e=0$,$d=0$,and $c=5$. Thus,$f(x) = ax^4 + bx^3 + 5x^2$. The derivative is $f'(x) = 4ax^3 + 3bx^2 + 10x = x(4ax^2 + 3bx + 10)$. Since $f(x)$ has extreme values at $x=4$ and $x=5$,$f'(4)=0$ and $f'(5)=0$. $f'(4) = 4(16a) + 3(4b) + 10 = 64a + 12b + 10 = 0 \implies 32a + 6b = -5$. $f'(5) = 4(25a) + 3(5b) + 10 = 100a + 15b + 10 = 0 \implies 20a + 3b = -2$. Solving these equations: $b = \frac{-2 - 20a}{3}$. Substituting into the first: $32a + 2(-2 - 20a) = -5 \implies 32a - 4 - 40a = -5 \implies -8a = -1 \implies a = \frac{1}{8}$. Then $3b = -2 - 20(\frac{1}{8}) = -2 - 2.5 = -4.5 \implies b = -1.5 = -\frac{3}{2}$. Now,$f(2) = \frac{1}{8}(2^4) - \frac{3}{2}(2^3) + 5(2^2) = \frac{16}{8} - \frac{3 	imes 8}{2} + 5(4) = 2 - 12 + 20 = 10$.

Let $f: R \rightarrow R$ be a polynomial function of degree four having extreme values at $x=4$ and $x=5$. If $\lim _{x \rightarrow 0} \frac{f(x)}{x^2}=5$,then $f(2)$ is equal to:

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