A) \[p(-2)=19\]
B) \[p(2)=19\]
C) \[p(-2)=11\]
D) \[p(2)=11\]
Correct Answer: A
Solution :
\[p(x)=a{{x}^{2}}+bx+c\] \[p(0)=1=c=1\] \[\left. \begin{align} & p(1)=4 \\ & p(-1)=6 \\ \end{align} \right]\] \[\left. \begin{align} & a+b+c=4 \\ & a-b+c=6 \\ \end{align} \right]=\left. \begin{align} & a=4 \\ & b=-1 \\ \end{align} \right]\] \[p(x)=4{{x}^{2}}-x+1\] \[p(-2)=16+2+1=19\]
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