Range of the function : `f(x)=log_2((pi+2sin^(- 1)((3-x)/7))/pi)`

Explanation

`f(x)=log_(2)((pi+2"sin"^(-1)((3-x)/(7)))/(pi))`
Least value of ` 2"sin"^(-1)((3-x)/(7)) ` is '-2' for which
`(pi+2"sin"^(-1)((3-x)/(7)))/(pi)` is positive.
So, for domain we must have ` -1 le (3-x)/(7) le 1`
` :. -7le 3-x le7`
`implies -10 le -x le4`
`implies -4 le x le10`
Thus, domain of f(x) is `[-4,10]`.
Also, `-(pi)/(2) le "sin"^(-1)((3-x)/(7)) le (pi)/(2)`
`implies -pi le 2"sin"^(-1)((3-x)/(7)) le pi`
`implies 0 le pi +2"sin"^(-1)((3-x)/(7)) le2pi`
`implies 0 lt (pi+2"sin"^(-1)((3-x)/(7)))/(pi) le 2` (for logarithm to get defined)
`implies - oo lt log_(2)((pi+2"sin"^(-1)((3-x)/(7)))/(pi)) le 1`
Hence range is `(- oo,1]`