Let `f(x)=f_1(x)-2f_2 (x)`, where ,where `f_1(x)={((min{x^2,|x|},|x|le 1),(max{x^2,|x|},|x| le 1))` and `f_2(x)={((min{x^2,|x|},|x| lt 1),({x^2,|x|},|x| le 1))` and let `g(x)={ ((min{f(t):-3letlex,-3 le x le 0}),(max{f(t):0 le t le x,0 le x le 3}))` for `-3 le x le -1` the range of `g(x)` is

Options

A. `[-1, 3]`
B. `[-1,-15]`
C. `[-1, 9]`
D. None of these

Answer

A

Explanation

`f_(1)(x)=x^(2) and f_(2)(x)=|x|`
`or f(x)=f_(1)(x)-2f_(2)(x)=x^(2)-2|x|`
Graph of `f(x)`

`g(x)={(f(x)",",-3le x lt -1),(-1",",-1 le x lt 0),(0",",0le x le 2),(f(x)",",2 lt x le 3):}={(x^(2)+2x",",-3le x lt -1),(-1",",-1 le x lt 0),(0",",0le x le 2),(x^(2)-2x",",2 lt x le 3):}`
The range of `g(x)` for [-3, -1] is [-1, 3].