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 which of the following is false?

• A. f(0) + f(2) = f(1)
• B. f(0) + f(3) = f(0)
• C. f(1) + f(3) = f(2)
• D. f(1) + f(3) = f(0)

Answer 1

The correct option is D f(1) + f(3) = f(0)

$f\left(x\right)=x^3+x^2{f}^{\prime }\left(1\right)+x{f}^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)$
$f^{\prime }\left(x\right)=3x^2+2x{f}^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)\Rightarrow f^{\prime }\left(1\right)=3+2f^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)\Rightarrow f^{\prime }\left(1\right)=-3-f^{\prime \prime }\left(2\right)$ ...(1)
$f^{\prime \prime }\left(x\right)=6x+2f^{\prime }\left(1\right)\Rightarrow f^{\prime \prime }\left(2\right)=6\left(2\right)+2f^{\prime }\left(1\right)=12+2f^{\prime }\left(1\right)$ ...(2)
$f^{\prime \prime \prime }\left(x\right)=6$ Substituting f''(2) from (2) in (1)
$\Rightarrow f^{\prime \prime \prime }\left(x\right)=6$
$f^{\prime }\left(1\right)=-3-f^{\prime \prime }\left(2\right)=-3-12-2f^{\prime }\left(1\right)3f^{\prime }\left(1\right)=-15f^{\prime }\left(1\right)=-5f^{\prime \prime }\left(2\right)=12+2f^{\prime }\left(1\right)=12+2(-5)=12-10=2$
$f\left(0\right)=0^3+0^2{f}^{\prime }\left(1\right)+0f^{\prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)=f^{\prime \prime \prime }\left(3\right)=6$
$f\left(1\right)=1^3+1^2{f}^{\prime }\left(1\right)+1f^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)=1+1(-5)+1\left(2\right)+6=1-5+2+6=4$
$f\left(2\right)=2^3+2^2{f}^{\prime }\left(1\right)+2f^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)=8+4(-5)+2\left(2\right)+6=8-20+4+6=-2$
$f\left(3\right)=3^3+3^2{f}^{\prime }\left(1\right)+3f^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)=27+9(-5)+3\left(2\right)+6=27-45+6+6=-6f\left(1\right)+f\left(3\right)=-2\ne f\left(0\right)\Rightarrow D$