If $y = 3^{x - 1} + 3^{- x - 1}$ (x real), then the least value of y is

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\frac{2}{3}

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None of these

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Solution

The correct option is C $\frac{2}{3}$
We have
$3^{x - 1} + 3^{- x - 1} = \frac{1}{3} (3^{x} + 3^{- x}) \geq \frac{1}{3} . 2 \sqrt{3^{x} . 3^{- x}}$ ,
$[∵ A . M . \geq G . M .]$
or $3^{x - 1} + 3^{- x - 1} \geq \frac{2}{3}$
Hence the least value of $3^{x - 1} + 3^{- x - 1}$ is $\frac{2}{3}$

Suggest Corrections

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