Question

If $f:R\to R$ is a function such that $f\left(x\right)=x^3+x^2{f}^{\prime }\left(1\right)+x{f}^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)$ for $xϵR$ then the value of f(5) is equal to

• A. $33$
• B. $32$
• C. $16$
• D. $12$

Answer 1

Answer: (B), (D)

$32$
$12$

Substitute $x=0$ in the given equation,
We have $f\left(0\right)=f^{\prime \prime \prime }\left(3\right)$ and putting $x=1$, we get $f\left(1\right)=1+f^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)$.
Thus $f\left(1\right)-f\left(0\right)=1+f^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)$. Also differentiating the given equation, we have
$f^{\prime }\left(x\right)=3x^2+2x{f}^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)$ ... (i)
$f^{\prime \prime }\left(x\right)=6x+2f^{\prime }\left(1\right)$, $f^{\prime \prime \prime }\left(x\right)=6$
Thus $f^{\prime \prime \prime }\left(3\right)=6$ and $f^{\prime }\left(2\right)=12+2f^{\prime }\left(1\right)$.

Subsitute $x=1$ in (i), we have
$f^{\prime }\left(1\right)=3+2f^{\prime }\left(1\right)+f^{\prime \prime }\left(2\right)=3+2f^{\prime }\left(1\right)+12+2f^{\prime }\left(1\right)=15+4f^{\prime }\left(1\right)$
$\Rightarrow$ $f^{\prime }\left(1\right)=-5$ and so $f^{\prime \prime }\left(2\right)=12-10=2$
$\therefore f\left(2\right)=2^3+2^2{f}^{\prime }\left(1\right)+2f^{\prime \prime }\left(2\right)+f^{\prime \prime \prime }\left(3\right)$
$=8+4(-5)+2\left(2\right)+6=18-20=-2$