Question

If $f:R\to R$ is a function such that $f\left(x\right)=x^3+x^2{f}^{\prime }\left(1\right)+x{f}^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)$ $\forall x\in R,$ then $f\left(2\right)-f\left(1\right)=$

• A. $f\left(0\right)$
• B. $-f\left(0\right)$
• C. $f^{\prime }\left(0\right)$
• D. $-f^{\prime }\left(0\right)$

Answer 1

The correct option is B $-f\left(0\right)$

$f\left(x\right)=x^3+x^2{f}^{\prime }\left(1\right)+x{f}^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)$
$f^{\prime }\left(x\right)=3x^2+2x{f}^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)$
Put $x=1$
$f^{\prime }\left(1\right)=3+2f^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)$
$\Rightarrow f^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)=-3\cdots \left(i\right)$

$f^{\prime \prime }\left(x\right)=6x+2f^{\prime }\left(1\right)$
Put $x=2$
$\Rightarrow f^{\prime \prime }\left(2\right)=12+2f^{\prime }\left(1\right)\cdots \left(ii\right)$

$f^{\prime \prime \prime }\left(x\right)=6$
put $x=3$
$f^{\prime \prime \prime }\left(3\right)=6\cdots \left(iii\right)$

from $\left(i\right)$ and $\left(ii\right)$,
$f^{\prime }\left(1\right)=-5,$ and $f^{\prime \prime }\left(2\right)=2$
$\therefore f\left(x\right)=x^3-5x^2+2x+6$

$\Rightarrow f\left(2\right)=2^3-5\cdot 2^2+2\cdot 2+6=-2$
$\Rightarrow f\left(1\right)=1^3-5\cdot 1^2+2\cdot 1+6=4$
$\Rightarrow f\left(0\right)=6$

$\therefore f\left(2\right)-f\left(1\right)=-f\left(0\right)$