Solve `sin^(-1)[((2x^2+4)/(1+x^2))]

Options

A. `x in [-1,0]`
B. `x in [0,1]`
C. `x in (-1,1)`
D. `x in (1,oo)`

Explanation

We have
`(2x^(2)+4)/(x^(2)+1)=(2(x^(2)+1)+2)/(x^(2)+1)=2+(2)/(x^(2)+1)`
`therefore 2lt(2x^(2)+4)/(x^(2)+1)le 4 forall x in R`
`rarr pi-4ltpi-(2x^(2)+45)/(x^(2)+1)ltpi-2 forall x in R`
`rarr -(pi)/(2)ltpi-(2x^(2)+4)/(x^(2)+1 lt(pi)/(2) forall x in R`
`=pi-(2x^(2)+4)/(x^(2)+1)`
Hence `sin^(-1){sin(2x^(2)+4)/(x^(2)+1)}ltpi-3`
`rarr pi-(2x^(2)+4)/(x^(2)+1)ltpi-3`
`rarr (2x^(2)+4)/(x^(2)+1)gt3`
`rarr 2+(2)/(x^(2)+1)rarr2gt3gtx^(2)+1rarrx^(2)1lt0rarrx in (-1,1)`