Home
Class 12
MATHS
Let f(x)=f1(x) where where f1(x) ={min...

Let `f(x)=f_1(x)` where where `f_1(x) ={min{x^2,|x|}, |x| leq 1 max {x^2,|x|}, |x|> 1 and f_2(x)= min {x^2,|x|}, |x|>1 max {x^2,|x|}, |x|leq1 and let g(x)= {min {f(t):-3 leqt leq x, -3 leq x <0} max {f(t):o leq t leq x, o leq x leq 3}` For `-3 leq x leq -1` ,the range of `g(x)` is

Promotional Banner

Similar Questions

Explore conceptually related problems

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

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

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

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

Let f(x)=f_(1)(x)-2f_(2)(x), where f_(1)(x)={{:(min{x^(2),|x|}",",|x|le1),(max{x^(2),|x|}",",|x|gt1):} "and "f_(2)(x)={{:(min {x^(2),|x|}",",|x|gt1),(max{x^(2),|x|}",",|x|le1):} "and let "g(x)={{:(min{f(t),-3letlex,-3lexlt0}),(max{f(t),0letltx,0lexle3}):} For x in(-1,00),f(x)+g(x) is

Let f(x)=f_(1)(x)-2f_(2)(x), where f_(1)(x)={{:(min{x^(2),|x|}",",|x|le1),(max{x^(2),|x|}",",|x|gt1):} "and "f_(2)(x)={{:(min {x^(2),|x|}",",|x|gt1),(max{x^(2),|x|}",",|x|le1):} "and let "g(x)={{:(min{f(t),-3letlex,-3lexlt0}),(max{f(t),0letltx,0lexle3}):} For x in(-1,00),f(x)+g(x) is

Let f(x)=f_(1)(x)-2f_(2)(x), where f_(1)(x)={{:(min{x^(2),|x|}",",|x|le1),(max{x^(2),|x|}",",|x|gt1):} "and "f_(2)(x)={{:(min {x^(2),|x|}",",|x|gt1),(max{x^(2),|x|}",",|x|le1):} "and let "g(x)={{:(min{f(t),-3letlex,-3lexlt0}),(max{f(t),0letltx,0lexle3}):} The graph of y=g(x) in its domain is broken at

Let f(x)=f_(1)(x)-2f_(2)(x), where f_(1)(x)={{:(min{x^(2),|x|}",",|x|le1),(max{x^(2),|x|}",",|x|gt1):} "and "f_(2)(x)={{:(min {x^(2),|x|}",",|x|gt1),(max{x^(2),|x|}",",|x|le1):} "and let "g(x)={{:(min{f(t),-3letlex,-3lexlt0}),(max{f(t),0letltx,0lexle3}):} The graph of y=g(x) in its domain is broken at