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${⨍ (x)=ae}^{x} {+be}^{- x} (ab>0)$ has maximum value when

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$a = 0$

$a < 0$

$b > 0$

$a > 0$

answer is A.

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Detailed Solution

$f (x) = {ae}^{x} + {be}^{- x} (ab > 0) f^{I} (x) = {ae}^{x} - {be}^{- x}$

$f (x)$ is min or max

$\Rightarrow f^{I} (x) = 0$

${ae}^{x} = {be}^{- x} = b / e^{x} {(e^{x})}^{2} = b / a e^{x} = \sqrt{b / a}$

given $ab > 0 a > 0, b > 0$

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