If $f : R \to (- \infty, a]$ defined by $f (x) = - x^{2} + 6 x + 15$ is surjective, then the value of $a$ is

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Solution

$y = f (x) = - x^{2} + 6 x + 15$
$\Rightarrow x^{2} - 6 x - (15 - y) = 0$
Since $x$ is real, $D \geq 0$
$\Rightarrow 36 + 4 (15 - y) \geq 0$
$\Rightarrow 4 y \leq 96$
$\Rightarrow y \leq 24$
$∴$ Range of $f$ is $(- \infty, 24]$
Given, $f$ is surjective.
$\Rightarrow$ Range $=$ Codomain
$∴ a = 24$

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