If `f(x)=(1)/(2)-tan((pix)/(2)),-1 lt xlt1` and `g(x)=sqrt(3+4x-4x^(2))`, then domain `(f+g)` is given by

Options

A. `[(1)/(2),1]`
B. `[(1)/(2),-1]`
C. `[-(1)/(2),1]`
D. `[-(1)/(2),-1]`

Answer

C

Explanation

Given, `f(x)=(1)/(2)-tan((pix)/(2))," where "-1ltxlt1`
Given, domain of f(x) is `d_(1)=(-1,1)`
For domain of `g(x),3+4x-4x^(2)ge0`
`implies" "(2x-3)(2x+1)le0`
`:." Domain of g(x) is "d_(2)=[-(1)/(2),(3)/(2)]`
Hence, domain of `(f+g)=d_(1) nn d_(2)=[-(1)/(2),1]`