Question

Question #6ed7e

Answer 1

Given `f(x) = 1/logx and g(x) = 1/8 +x`

We have

`(fxxg)(x) = (1/logx)xx(1/8+x)`

`=>(fxxg)(x) = (1/8+x)/logx`

`(fxxg)(x)` will be defined when

`logx!=0=>x!=1`
and `x>0 and x in R`

So option (c) is valid here.