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| Forum Index -> Differential Calculus -> View Full Question |
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f(x) = ln (xx + 1)
f '(x) = xx (1+lnx) / (xx + 1)
Clearly f '(x) = at x = 1/e. f '(x) < 0 when x < 1/e and f '(x) > 0 when x > 1/e
so minima occurs at x = 1/e
now check limiting values of f(x) when x tends to 0 and 1 and select the one with higher value for maxima. Both x -> 0 and x -> 1 gives f(x) = ln2
so the range is [ ln ((1/e)1/e+1) , ln2 )
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Bipin Kumar Dubey Chemical Dept. IIT Kharagpur |
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