Let` f(x)=min{e^(x), (3)/(2), 1+e^(-x)}` if f(x)=k has at least 2 solutions then k can not be

Let f(x)=2+x, x>=0 and f(x)=4-x , x < 0 if f(f(x)) =k has at least one solution, then smallest value of k is

Let f(x)=(e^(x))/(x^(2))x in R-{0}. If f(x)=k has two distinct real roots,then the range of k, is

Let f(x)=[2+x,,xgt=0 , 4-x,,xlt0, If f(f(x))=k has at least one solution,then smallest value of k is

Let f(x)=sin^(-1)x+|sin^(-1)x|+sin^(-1)|x| If the equation f(x) = k has two solutions, then true set of values of k is

Let f(x)=x+2|x+1|+2|x-1|* If f(x)=k has exactly one real solution,then the value of k is (a) 3 (b) 0(c)1(d)2

Let f(x)=(x^(2)-1)(pi-e^(x)) then

Let f(x)=(x^(2)-1)(pi-e^(x)) then

f(x)=(sin^(2)x)e^(-2sin^(2)x)*max f(x)-min f(x) =

Let f(x)=x^(3)-3x^(2)+2x. If the equation f(x)=k has exactly one positive and one negative solution then the value of k equals.-(2sqrt(3))/(9)(b)-(2)/(9)(2)/(3sqrt(3)) (d) (1)/(3sqrt(3))