f(x)=xloge⁡xmonotonically decreases in

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(0,1/e)

(1/e,1)

(1,∞)

(1/e,∞)

answer is A.

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Detailed Solution

f(x)=xloge⁡x For g(x)=x logex,g′(x)=x1x+loge⁡x=1+loge⁡xThus, g (x) increases for 1e,∞and decreases for 0,1e. From the graph,f(x)=xloge⁡x decreases in 0, 1e.

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