Question

Let $P\left(x\right)$ be a polynomial of degree 4, with $P\left(2\right)=1,$ $P^{\prime }\left(2\right)=P^{\prime \prime }\left(2\right)=0$, $P^{\prime \prime \prime }\left(2\right)=12$ and $P^{\prime \prime \prime \prime }\left(2\right)=24$. Then the value of $P^{\prime \prime }\left(1\right)$ is

• A. $0$
• B. $26$
• C. $2$
• D. $12$

Answer 1

The correct option is A $0$

Let $P\left(x\right)=a(x-2{)}^4+b(x-2{)}^3+c(x-2{)}^2+d(x-2)+e$

Now using given conditions,
$1=P\left(2\right)=e$

$0=P^{\prime }\left(2\right)=d$

$0=P^{\prime \prime }\left(2\right)=2c\Rightarrow c=0$

$12=P{}^{\prime \prime \prime }\left(2\right)=6b\Rightarrow b=2$

$24=P^{iv}\left(2\right)=24a\Rightarrow a=1$

Thus $P\left(x\right)=(x-2{)}^4+2(x-2{)}^3+1$

$\Rightarrow P^{\prime \prime }\left(x\right)=12(x-2{)}^2+12(x-2)$

$\Rightarrow P^{\prime \prime }\left(1\right)=12+12(-1)=0$

Hence, option A.