Let \(f: R \rightarrow R\) be a function such that \(\mathrm{f}(x)=x^3+x^2 \mathrm{f}^{\prime}(1)+x \mathrm{f}^{\prime \prime}(2)+6, x \in \mathrm{R}\), then \(\mathrm{f}(2)\) equals

\(\mathrm{f}(x)=x^3+x^2 \mathrm{f}^{\prime}(1)+x \mathrm{f}^{\prime \prime}(2)+6 \)
\( \therefore \mathrm{f}^{\prime}(x)=3 x^2+2 x \mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime \prime}(2) \)
\( \therefore \mathrm{f}^{\prime \prime}(x)=6 x+2 \mathrm{f}^{\prime}(1)\)
Substituting \(x=1\) in (i), we get
\(
\begin{aligned}
& \mathrm{f}^{\prime}(1)=3(1)^2+2(1) \mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime \prime}(2) \\
& \Rightarrow \mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime \prime}(2)=-3
\end{aligned}
\)
Substituting \(x=2\) in (ii), we get
\(
\begin{aligned}
& \mathrm{f}^{\prime \prime}(2)=6(2)+2 \mathrm{f}^{\prime}(1) \\
& \Rightarrow \mathrm{f}^{\prime \prime}(2)=12+2 \mathrm{f}^{\prime}(1)
\end{aligned}
\)
From (iii) and (iv), we get
\(
\begin{aligned}
& \mathrm{f}^{\prime}(1)+12+2 \mathrm{f}^{\prime}(1)=-3 \\
& \Rightarrow 3 \mathrm{f}^{\prime}(1)=-15 \\
& \Rightarrow \mathrm{f}^{\prime}(1)=-5 \\
& \text { From (iii) },-5+\mathrm{f}^{\prime \prime}(2)=-3 \\
& \Rightarrow \mathrm{f}^{\prime \prime}(2)=2
\end{aligned}
\)
\(\therefore \mathrm{f}(2)=2^3+2^2(-5)+2(2)^{\prime}+6\)
\(=8-20+4+6\)
\(=-2\)