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 $f\left(2\right)$ equals :

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

Answer 1

The correct option is D $-2$

$f\left(x\right)=x^3+x^2{f}^{\prime }\left(1\right)+x{f}^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)$
$\Rightarrow f^{\prime }\left(x\right)=3x^2+2x{f}^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)\cdots \left(i\right)$
$\Rightarrow f^{\prime \prime }\left(x\right)=6x+2f^{\prime }\left(1\right)\cdots \left(ii\right)$
$\Rightarrow f^{\prime \prime \prime }\left(x\right)=6\cdots \left(iii\right)$
​​​​​​$\Rightarrow f^{\prime \prime \prime }\left(3\right)=6$
Now from eqn $\left(i\right)$
$f^{\prime }\left(1\right)=3+2f^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)$
$\Rightarrow f^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)+3=0\cdots \left(iv\right)$
from eqn $\left(ii\right)$
$f^{\prime \prime }\left(2\right)=12+2f^{\prime }\left(1\right)$
$\Rightarrow 2f^{\prime }\left(1\right)-f^{\prime \prime }\left(2\right)+12=0\cdots \left(v\right)$
from eqn. $\left(iv\right)$ and $\left(v\right)$
$3f^{\prime }\left(1\right)+15=0$
$\Rightarrow f^{\prime }\left(1\right)=-5$ and $f^{\prime \prime }\left(2\right)=2$
So,
$f\left(x\right)=x^3-5x^2+2x+6$
$\Rightarrow f\left(2\right)=8-20+4+6=-2$