Consider $\mathrm{p}\left(\mathrm{x}\right)$ to be a polynomial of degree 5 having extremum at x=–1,1and $\underset{\mathrm{x}\to 0}{\lim }\left(\frac{\mathrm{p}\left(\mathrm{x}\right)}{{\mathrm{x}}^3}-2\right)=4$. Then the value of [P(1)] is (where [.] represents greatest integer function)

$$\begin{gathered} \mathrm{p}\left(\mathrm{x}\right)={\mathrm{ax}}^5+{\mathrm{bx}}^4+{\mathrm{cx}}^3+{\mathrm{dx}}^2+\mathrm{ex}+\mathrm{f} \\ \underset{\mathrm{x}\to 0}{\mathrm{lt}} \frac{\mathrm{p}\left(\mathrm{x}\right)}{{\mathrm{x}}^3}-2=4 \\ \underset{\mathrm{x}\to 0}{\mathrm{lt}} {\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}+\frac{\mathrm{d}}{\mathrm{x}}+\frac{\mathrm{e}}{{\mathrm{x}}^2}+\frac{\mathrm{f}}{{\mathrm{x}}^3}-2=4 \end{gathered}$$

if exist 
$$\begin{gathered} \Rightarrow \mathrm{d}=\mathrm{e}=\mathrm{f}=0 \\ \mathrm{c}-2=4 \\ \mathrm{c}=6 \end{gathered}$$

$$\begin{gathered} \mathrm{p}\left(\mathrm{x}\right)={\mathrm{ax}}^5+{\mathrm{bx}}^4+6{\mathrm{x}}^3 \\ {\mathrm{p}}^1\left(\mathrm{x}\right)=5{\mathrm{ax}}^4+4{\mathrm{bx}}^3+18{\mathrm{x}}^2 \\ \mathrm{p}\left(\mathrm{x}\right) \mathrm{is} \min \mathrm{or} \max \\ \Rightarrow {\mathrm{p}}^1\left(\mathrm{x}\right)=0 \\ 5{\mathrm{ax}}^2+4\mathrm{bx}+18=0 \\ -1, 1 \mathrm{are} \mathrm{root} \mathrm{no} \mathrm{area} \end{gathered}$$

$$\begin{gathered} -1+1=\frac{-4\mathrm{b}}{5\mathrm{a}}\Rightarrow \mathrm{b}=0 \\ -1\times 1=\frac{18}{5\mathrm{a}}\Rightarrow \mathrm{a}=\frac{-18}{5} \\ \mathrm{p}\left(\mathrm{x}\right)=\frac{-18}{5}{\mathrm{x}}^5+6{\mathrm{x}}^3 \\ \mathrm{p}\left(1\right)=\frac{-18}{5}+6=\frac{12}{5} \\ =204 \\ \left[\mathrm{p}\left(1\right)\right]=2 \end{gathered}$$