$$\begin{aligned} & \lim _{x \rightarrow 0} \frac{f(x)}{x^2}=5 \\ & \lim _{x \rightarrow 0} \frac{\left.a x^4+b x^3+c x^2+d x+e\right)}{x^2}=5 \\ & c=5 \text { and } d=e=0 \\ & f(x)=a x^4+b x^3+5 x^2 \\ & f^{\prime}(x)=4 a x^3+3 b x^2+10 x \\ & =x\left(4 a x^2+3 b x+10\right) \end{aligned}$$

has extremes at 4 and so $f^{\prime}(4)=0 \& f^{\prime}(5)=0$

so $\mathrm{a}=\frac{1}{8} \& \mathrm{~b}=\frac{-3}{2}$

so $f(2)=\frac{1}{8} \times 2^4-\frac{3}{2} \times 2^3+5 \times 2^2$

$$=2-12+20=10$$