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For 0<a<x , the minimum value of the fun...

For `0

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The correct Answer is:
False
Since, `(log_(a) x + (1)/(log_(a) x))/(2) gt 1, " using " AM ge GM`
Here, equality holds only when `x =a` which is not possible. So, `log_(a) x + log_(x) a` is greater than 2
Hence, it is a false statement
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