Question

Let f $:$ R $\to$ R be a function such that $f\left(x\right)=x^3+x^2{f}^{\prime }\left(1\right)+x{f}^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right),x\in R$. Then $f\left(2\right)$ equal?

• A. $8$
• B. $-2$
• C. $-4$
• D. $30$

Answer 1

The correct option is B $-2$

$f\left(x\right)=x^3+x^2{f}^{\prime }\left(1\right)+x{f}^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)$
$\Rightarrow f^{\prime }\left(x\right)=3x^2+2x{f}^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)$ .$\left(1\right)$
$\Rightarrow f^{\prime \prime }\left(x\right)=6x+2f^{\prime }\left(1\right)$ ..$\left(2\right)$
$\Rightarrow f^{\prime \prime \prime }\left(x\right)=6$ .$\left(3\right)$
Put $x=1$ in equation $\left(1\right)$:
$f^{\prime }\left(1\right)=3+2f^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)$ ..$\left(4\right)$
Put $x=2$ in equation $\left(2\right)$:
$f^{\prime \prime }\left(2\right)=12+2f^{\prime }\left(1\right)$ .$\left(5\right)$
from equation $\left(4\right)$ & $\left(5\right)$:
$-3-f^{\prime }\left(1\right)=12+2f^{\prime }\left(1\right)$
$\Rightarrow 3f^{\prime }\left(1\right)=-15$
$\Rightarrow f^{\prime }\left(1\right)=-5$
$\Rightarrow f^{\prime \prime }\left(2\right)=2$ ..$\left(2\right)$
put $x=3$ in equation $\left(3\right)$:
$f^{\prime \prime \prime }\left(3\right)=6$
$\therefore f\left(x\right)=x^3-5x^2+2x+6$
$f\left(2\right)=8-20+4+6=-2$.