Question

If $f\left(x\right)=x^3+x^2{f}^{{}^{\prime }}\left(1\right)+x{f}^{\prime \prime }\left(2\right)+f^{{}^{\prime \prime \prime }}\left(3\right)$ for all $xϵR$ then

• A. $f\left(0\right)+f\left(2\right)=f\left(1\right)$
• B. $f\left(0\right)+f\left(3\right)=0$
• C. $f\left(1\right)+f\left(3\right)=f\left(2\right)$
• D. none of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f\left(0\right)+f\left(2\right)=f\left(1\right)$
B $f\left(0\right)+f\left(3\right)=0$
C $f\left(1\right)+f\left(3\right)=f\left(2\right)$

$f\left(x\right)=x^3+x^2{f}^{{}^{\prime }}\left(1\right)+x{f}^{\prime \prime }\left(2\right)+f^{{}^{\prime \prime \prime }}\left(3\right)$ .....(1)
$f^{\prime }\left(x\right)=3x^2+2x{f}^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)$ ....(2)
$f^{\prime \prime }\left(x\right)=6x+2f^{\prime }\left(1\right)$ .....(3)
$f^{\prime \prime \prime }\left(x\right)=6$
$\Rightarrow f^{\prime \prime \prime }\left(3\right)=6$ .....(4)
Substitute $x=1$ in (2)
$f^{\prime }\left(1\right)=3+2f^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)$
$\Rightarrow f^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)+3=0$ .....(5)
Substitute $x=2$ in (3)
$f^{\prime \prime }\left(2\right)=12+2f^{\prime }\left(1\right)$ .....(6)
Solving (5) and (6), we get
$f^{\prime \prime }\left(2\right)=2,f^{\prime }\left(1\right)=-5$
Substitute these values and value from (4) in (1)
$f\left(x\right)=x^3-5x^2+2x+6$
Now, $f\left(0\right)=6$
$f\left(1\right)=4$
$f\left(2\right)=-2$
$f\left(3\right)=-6$
Hence, $f\left(0\right)+f\left(2\right)=f\left(1\right)$
$f\left(0\right)+f\left(3\right)=0$
$f\left(1\right)+f\left(3\right)=f\left(2\right)$