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If $P (x)$ is a polynomial of 2nd degree $P (2) = - 1, P' (2) = 0, P'' (2) = 2$ and then $P' (1) =$

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Detailed Solution

$P (x) = a x^{2} + b x + c \Rightarrow P' (x) = 2 a x + b; P " (x) = 2 a \Rightarrow p " (2) = 2 \Rightarrow a = 1$

$\begin{array}{l} P^{'} (2) = 0 \Rightarrow 4 + b = 0 \Rightarrow b = - 4 \\ P^{'} (x) = 2 x - 4 \Rightarrow P^{'} (1) = 2 - 4 = - 2 \end{array}$

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