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

The limit $$\lim_{x \to 0}\frac{f(x)}{x^{2}} = 5$$ means that near $$x = 0$$ the polynomial behaves like $$5x^{2}$$.
Therefore $$x^{2}$$ must be a factor of $$f(x)$$ and the remaining factor must equal $$5$$ at $$x = 0$$.

Write $$f(x) = x^{2}\,g(x)$$ where $$g(x)$$ is a quadratic: $$g(x)=ax^{2}+bx+5$$.
Hence

$$f(x)=a x^{4}+b x^{3}+5x^{2}\qquad -(1)$$

Differentiate to locate stationary points:

$$f'(x)=4ax^{3}+3bx^{2}+10x\qquad -(2)$$

The extreme values occur at $$x=4$$ and $$x=5$$, so

$$f'(4)=0,\; f'(5)=0 \qquad -(3)$$

Substitute $$x=4$$ into $$(2)$$:

$$4a(4)^{3}+3b(4)^{2}+10(4)=0$$
$$256a+48b+40=0\qquad -(4)$$

Substitute $$x=5$$ into $$(2)$$:

$$4a(5)^{3}+3b(5)^{2}+10(5)=0$$
$$500a+75b+50=0\qquad -(5)$$

Solve the linear system $$(4),(5)$$.
Multiply $$(4)$$ by $$75$$ and $$(5)$$ by $$48$$ to eliminate $$b$$:

$$19200a+3600b=-3000$$
$$24000a+3600b=-2400$$

Subtract the first from the second:

$$(24000-19200)a=600 \;\Longrightarrow\; 4800a=600\;\Longrightarrow\; a=\frac{1}{8}$$

Insert $$a=\tfrac{1}{8}$$ into $$(4)$$:

$$256\left(\frac{1}{8}\right)+48b+40=0$$
$$32+48b+40=0$$
$$48b=-72 \;\Longrightarrow\; b=-\frac{3}{2}$$

Now evaluate $$f(2)$$ using $$(1)$$:

$$f(2)=2^{2}\Bigl(a\cdot2^{2}+b\cdot2+5\Bigr)$$
$$\;=4\Bigl(a\cdot4+2b+5\Bigr)$$

Insert $$a=\tfrac{1}{8},\; b=-\tfrac{3}{2}$$:

$$a\cdot4=\frac{1}{2},\quad 2b=-3$$
$$a\cdot4+2b+5=\frac{1}{2}-3+5=\frac{5}{2}$$

Therefore $$f(2)=4\left(\frac{5}{2}\right)=10$$.

Hence $$f(2)=10$$, which matches Option B.