The domain of the function `f(x)=(log)_(3+x)(x^2-1)` is `(-3,-1)uu(1,oo)` `(-3,-1)uu(1,oo)` `(-3,-2)uu(-2,-1)uu(1,oo)` `(-3,-2)uu(-2,-1)uu(1,oo)`

Options

A. `(-3,-1) cup (1, oo)`
B. `[-3,-1) cup [1,oo)`
C. `(-3,-2) cup (-2,-1) cup (1,oo)`
D. `(-3,-2) cup (-2,-1) cup (1,oo)`

Answer

C

Explanation

`f(x)` is to be defined when `x^(2) -1 gt 0 " and " 3+x gt 0`
and `3+x ne 1,` i.e.,
`x^(2) gt 1" and " x gt -3 " and " x ne -2,`
i.e., `x lt -1 " or " x gt 1 " and " x gt -3 " and " x ne -2`
` :. D_(f)=(-3,-2) cup (-2, -1) cup (1,oo)`