Let \(\mathrm{P}(x)\) be a polynomial of degree 2 , with \(\mathrm{P}(2)=-1, \mathrm{P}^{\prime}(2)=0, \mathrm{P}^{\prime \prime}(2)=2\), then \(\mathrm{P}(1.001)\) is

\(\begin{aligned}
& \text { Let } \mathrm{P}(x)=\mathrm{a} x^2+\mathrm{b} x+\mathrm{c} \\
& \Rightarrow \mathrm{P}^{\prime}(x)=2 \mathrm{a} x+\mathrm{b} \\
& \Rightarrow \mathrm{P}^{\prime \prime}(x)=2 \mathrm{a} \\
& \mathrm{P}^{\prime \prime}(2)=2 \mathrm{a} \\
& \Rightarrow 2=2 \mathrm{a} \\
& \Rightarrow \mathrm{a}=1 \\
& \mathrm{P}^{\prime}(2)=2 \mathrm{a}(2)+\mathrm{b} \\
& \Rightarrow 0=4 \mathrm{a}+\mathrm{b} \\
& \Rightarrow 0=4(1)+\mathrm{b} \\
& \Rightarrow \mathrm{b}=-4 \\
& \mathrm{P}(2)=\mathrm{a}(2)^2+\mathrm{b}(2)+\mathrm{c} \\
& \Rightarrow-1=4 \mathrm{a}+2 \mathrm{~b}+\mathrm{c} \\
& \Rightarrow-1=4(1)+2(-4)+\mathrm{c} \\
& \Rightarrow \mathrm{c}=3 \\
& \mathrm{P}(x)=x^2-4 x+3 \\
& \Rightarrow \mathrm{P}^{\prime}(x)=2 x-4 \\
& x=1.001=1+0.001=\mathrm{a}+\mathrm{h}
\end{aligned}\)
Here, \(\mathrm{a}=1, \mathrm{~h}=0.001\)
\(\begin{gathered}
\mathrm{P}(\mathrm{a})=\mathrm{P}(1)=1-4+3=0 \\
\mathrm{P}^{\prime}(\mathrm{a})=\mathrm{P}^{\prime}(1)=2-4=-2
\end{gathered}\)
\(\therefore \quad \mathrm{P}(1.001)=0+(0.001)(-2)=-0.002\)