If AM of the numbers $5^{1 + x}$ and $5^{1 - x}$ is 13 then the set of possible real values of x is

{5, \frac{1}{5}}

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{1, - 1}

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{x | x^{2} - 1 = 0, x ϵ R}

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none of these

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Solution

The correct options are
A ${x | x^{2} - 1 = 0, x ϵ R}$
D ${1, - 1}$
Let $5^{x} = t$
we get, $a = 5 t a n d b = \frac{5}{t}$

$A M = \frac{5 t + \frac{5}{t}}{2} = 13$
$\Rightarrow 5 t^{2} - 26 t + 5 = 0$
$\Rightarrow (5 t - 1) (t - 5) = 0$
$\Rightarrow 5^{x} = \frac{1}{5}, 5$
$∴ x = - 1, 1$
$∴$ Options B and C are correct.

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