Let `E_(1)={x in R :x ne 1 and (x)/(x-1) gt 0} and E_(2)={x in E_(1):sin^(-1)(log_(e)((x)/(x-1))) " is real number"}`
(Here, the inverse trigonometric function `sin^(-1)x` assumes values in `[-(pi)/(2),(pi)/(2)]`.)
Let `f:E_(1) to R` be the function defined by `f(x)=log_(e)((x)/(x-1)) and g:E_(2) to R` be the function defined by `g(x)=sin^(-1)(log_(e)((x)/(x-1))) .`

The correct option is

Options

A. `a to s, b to q, c to p, d to p`
B. `a to r, b to r, c to u, d to t`
C. `a to s, b to q, c to p, d to u`
D. `a to s, b to r, c to u, d to t`

Answer

A

Explanation

`E_(1):(x)/(x-1) gt 0`
`implies x in (-oo ,0) cup (1,oo)`
Also `E_(2) : -1 lt log_(e)((x)/(x-1)) le 1`
`implies (1)/(e) le (x)/(x-1) le e`
`implies (1)/(e) le 1 +(1)/(x-1) le e`
`implies (1)/(e)-1 le (1)/(x-1) le e-1`
`implies (x-1) in (-oo, (e)/(1-e)] cup [(1)/(e-1),oo)`
`implies x in (-oo, (e)/(1-e)] cup [(1)/(e-1),oo)`
Now, `(x)/(x-1) in (0, oo)-{1} AA x in E_(1)`
`implies log_(e) ((x)/(x-1)) in (-oo,oo)-{0}`
`implies sin^(-1)(log_(e)((x)/(x-1))) in [-(pi)/(2),(pi)/(2)]-{0}`
`("considering "log_(e) ((x)/(x-1)) in [-1,1]-{0})`