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(C) Given $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6$. $(i)$ Differentiating with respect to $x$,we get $f^{\prime}(x)=3 x^2+2 x f^{\prime}(

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-$2$

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(C) Given $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6$. $(i)$ Differentiating with respect to $x$,we get $f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2)$. (ii) Differentiating again,we get $f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$. (iii) Substituting $x=1$ in (ii),we get $f^{\prime}(1)=3(1)^2+2(1) f^{\prime}(1)+f^{\prime \prime}(2) \Rightarrow f^{\prime}(1)+f^{\prime \prime}(2)=-3$. (iv) Substituting $x=2$ in (iii),we get $f^{\prime \prime}(2)=6(2)+2 f^{\prime}(1) \Rightarrow f^{\prime \prime}(2)=12+2 f^{\prime}(1)$. $(v)$ Substituting $(v)$ into (iv),we get $f^{\prime}(1)+12+2 f^{\prime}(1)=-3 \Rightarrow 3 f^{\prime}(1)=-15 \Rightarrow f^{\prime}(1)=-5$. From $(v)$,$f^{\prime \prime}(2)=12+2(-5)=2$. Now,$f(2)=2^3+2^2(-5)+2(2)+6 = 8-20+4+6 = -2$.

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

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