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

Let f:R rarr R be defined by, f(x)={[k-x;x le-1],[x^(2)+3,,xgt-1] .If f(x) has least value at x=-1 , then the possible value of k is

Let f:""R vec R be defined by f(x)={k-2x , if""xlt=-1 (-2x+3),x >-1} . If f has a local minimum at x=-1 , then a possible value of k is (1) 0 (2) -1/2 (3) -1 (4) 1