If $p(x)$ is a polynomial of degree 3 which satisfies $p^{\prime \prime}(1)=0$ and $p^{\prime \prime…
If $p(x)$ is a polynomial of degree 3 which satisfies $p^{\prime \prime}(1)=0$ and $p^{\prime \prime \prime}(1)=6$, then $p^{\prime \prime}(0)$ is equal to
6
-6
0
3
Solution
p(x) is a polynomial of degree 3.
$\begin{aligned} & \text { Let } p(x)=a x^3+b x^2+c x+d \\ & p^{\prime}(x)=3 a x^2+2 b x+c \\ & p^{\prime \prime}(x)=6 a x+2 b \\ & p^{\prime \prime \prime}(x)=6 a \\ & \because p^{\prime \prime}(\mathrm{l})=0 \text { and } p^{\prime \prime \prime}(\mathrm{l})=6 \\ & \end{aligned}$
$\Rightarrow \quad 6 a+2 b=0$...(i)
$6 a=6 \Rightarrow a=1$
From Eq. (i), we get
$\begin{aligned} 6 \times 1+2 b & =0 \\ 2 b & =-6 \Rightarrow b=-3 \\ \therefore \quad p^{\prime \prime}(0) & =6 a(0)+2 b \\ & =0+2(-3)=-6\end{aligned}$