Let `f(x)=sinx` and `g(x)=(log)_e|x|dot` If the ranges of the composition functions `fog` an ` dgof` are `R_1` and` R_2,` respectively, then `R_1-{u :-1lt=u<1},R_2={v :-oo `R_1-{u :-oo `R_1-{u :-1 `R_1-{u :-1lt=ult=1},R_2={v :-oo

Options

A. `R_(1)={u:-1 le u lt 1}.R_(2)={v:-oo lt v lt 0}`
B. `R_(1)={u: -oo lt u lt 0}.R_(2)={v:-oo lt v lt 0}`
C. `R_(1)={u: -1 lt u lt 1}.R_(2)={v:-oo lt v lt 0}`
D. `R_(1)={u: -1 le u le 1}.R_(2)={v:-oo lt v le 0}`

Answer

D

Explanation

We have `fog(x)=f(g(x))=sin(log_(e)|x|)`.
`log_(e)|x| ` has range R, for which `sin(log_(e)|x|) in [-1,1].`
Therefore, `R_(1)={u:-1 le u le 1}.`
Also, ` gof(x)=g(f(x))=log_(e)|sinx|.`
` :' 0 le |sinx| le 1`
`-oo lt log_(e)|sinx| le 0`
`or R_(2)={v: -oo lt v le 0 }`