Let `f(x)=f_(1)(x)-2f_(2)(x),` where
where `f(x)={(min{x^(2)","|x|}",",|x| le 1),(max{x^(2)","|x|}",",|x| gt 1):}`
and `f_(2)(x)={(min{x^(2)","|x|}",",|x| gt 1),(max{x^(2)","|x|}",",|x| le 1):}`
and let `g(x)={(min{f(t):-3 le t le x", "-3 le x lt 0}),(max{f(t): 0le t le x", " 0 le x le 3}):}`
For `-3 le x le -1,` the range of g(x) is

Options

A. `x^(2)-2x+1`
B. `x^(2)+2x-1`
C. `x^(2)+2x+1`
D. `x^(2)-2x-1`

Answer

B

Explanation

For `x in (-1,0),f(x)+g(x)=x^(2)+2x-1.`