Let $f: R \rightarrow R$ be a function such that $\mathrm{f}(x)=x^3+x^2 \mathrm{f}^{\prime}(1)+x…
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
$30$
$-4$
$-2$
$8$
Solution
$\begin{array}{ll} & \mathrm{f}(x)=x^3+x^2 \mathrm{f}^{\prime}(1)+x \mathrm{f}^{\prime \prime}(2)+6 \\ \therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2+2 x \mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime \prime}(2) \\ \therefore \quad & \mathrm{f}^{\prime \prime}(x)=6 x+2 \mathrm{f}^{\prime}(1)\end{array}$
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
$\begin{aligned}
& \mathrm{f}^{\prime}(1)+12+2 \mathrm{f}^{\prime}(1)=-3 \\
& \Rightarrow 3 \mathrm{f}^{\prime}(1)=-15 \\
& \Rightarrow \mathrm{f}^{\prime}(1)=-5 \\
& \text { From (iii) },-5+\mathrm{f}^{\prime \prime}(2)=-3 \\
& \Rightarrow \mathrm{f}^{\prime \prime}(2)=2
\end{aligned}$
$\begin{aligned} \therefore \quad \mathrm{f}(2) & =2^3+2^2(-5)+2(2)^{\prime}+6 \\ & =8-20+4+6 \\ & =-2\end{aligned}$