# Solve this following

Question:

Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a function such that

$f(x)=x^{3}+x^{2} f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3), x \in R$

Then $f(2)$ equal :

1. 8

2. $-2$

3. $-4$

4. 30

Correct Option: , 2

Solution:

$f(x)=x^{3}+x^{2} f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3)$

$\Rightarrow f^{\prime}(x)=3 x^{2}+2 x f^{\prime}(1)+f^{\prime \prime}(x)$    $\ldots \ldots(1)$

$\Rightarrow f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$  $\ldots \ldots(2)$

$\Rightarrow f^{\prime \prime \prime}(x)=6$   $\ldots . .(3)$

put $x=1$ in equation (1):

$f^{\prime}(1)=3+2 f^{\prime}(1)+f^{\prime \prime}(2)$   $\ldots . .(4)$

put $x=2$ in equation (2):

$f^{\prime \prime}(2)=12+2 f^{\prime}(1)$   $\ldots . .(5)$

from equation (4) \& (5) :

$-3-f^{\prime}(1)=12+2 f^{\prime}(1)$

$\Rightarrow 3 f^{\prime}(1)=-15$

$\Rightarrow f^{\prime}(1)=-5 \Rightarrow f^{\prime \prime}(2)=2 \quad \ldots .(2)$

put $x=3$ in equation (3):

$f^{\prime \prime \prime}(3)=6$

$\therefore f(x)=x^{3}-5 x^{2}+2 x+6$

$f(2)=8-20+4+6=-2$