If the minimum value of $f (x) = a x^{2} + 2 x + 5, a > 0$ is equal to the maximum value of $g (x) = 3 + 2 x - x^{2}$ , then the value of $^{'} a^{'}$ is

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Solution

The correct option is B $1$
$f (x)_{min} = g (x)_{max}$
$\Rightarrow \frac{20 a - 4}{4 a} = \frac{- 12 - 4}{- 4} \Rightarrow 20 a - 4 = 16 a \Rightarrow a = 1$

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