The set of all x for which `1+logx lt x`, is

Options

A. `(1,oo)`
B. (0,1)
C. `(0,oo)`
D. none of these

Answer

C

Explanation

Let f)x)=1+log x-x.
Clearly f(x) is difined for `x gt 0`
Now
`f'(x)=1/x-1=(1-x)/x`
`rArr f'(x)gt0" for "0ltxlt1 " and " f'(x) gt0 " for " xgt1`
`rArr` f(x) is increasing on (0,1) and decreasing as `(1/oo)`
`rArr f(x) ltf(1) " for all " x in (0,1) and f(x) ltf(1) " for all " x in (1,oo)`
`rArr f(x) ltf(1) " for all " x in (0,oo)`
`rArr 1+logxltx" for all " x in (0,oo)`
`Hence , 1+logxltx " for all " xgt0`