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f(x)=|x loge x| monotonically decreases ...

`f(x)=|x log_e x|` monotonically decreases in `(0,1/e)` (b) `(1/e ,1)` `(1,oo)` (d) `(1/e ,oo)`

A

`(0,1//e)`

B

`(1//e,1)`

C

`(1.oo)`

D

`(1//e,oo)`

Text Solution

Verified by Experts

The correct Answer is:
2
`f(x)=|x log_(e)x|`
For `g(x) =x log_(e)x`
`g(X)=x(1)/(x)+log_(e)x=1 + log_(e)x`
Thus g(X) increases for `((1)/(e ),oo)` and decrease for `(0,(1)/(e ))`

From the graph `f(x) =|x log+_(e)x| decreases in ((1)/(e),1)`
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