Question

Let $f\left(x\right)$ be a polynomial of degree three with $f\left(2\right)=1$,$f^{\prime }\left(2\right)=1$ ,$f^{\prime \prime }\left(2\right)=2$ and $f^{\prime \prime \prime }\left(2\right)=6$ then

• A. $f\left(0\right)=-5$
• B. $f^{\prime }\left(0\right)=9$
• C. $f^{\prime \prime }\left(0\right)=-10$
• D. None of these

Answer 1

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

$f\left(0\right)=-5$
$f^{\prime }\left(0\right)=9$
$f^{\prime \prime }\left(0\right)=-10$

It is a polynomial of degree 3, so it will be $a{x}^3+b{x}^2+cx+d=f\left(x\right)$
$f\left(2\right)=1=>8a+4b+2c+d=1$
$f^{\prime }\left(x\right)=3a{x}^2+2bx+c=>f^{\prime }\left(2\right)=12a+4b+c=1$
$f^{\prime \prime }\left(x\right)=6ax+2b=>f^{\prime \prime }\left(2\right)=12a+2b=2$
$f^{\prime \prime \prime }\left(x\right)=6a=>f^{\prime \prime \prime }\left(2\right)=6a=6$
$a=1$ $b=-5,c=9$ and $d=-5$
$Hence,f\left(0\right)=-5$
$f^{\prime }\left(0\right)=9$ $f^{\prime \prime }\left(0\right)=-10$
So, options A, B, C are all correct.