Let `f(x)=||x-1|+a|-4, if f(x)=0` has three real solution, then the values of lies in

Let f(x)=|x-1|+a|-4,quad if f(x)=0 has three real solution,then the values of lies in

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

Let g(x)=x^(2)-2x-2 and f(x)=x^(2)+x+alpha . If f(g(x))=0 has no real solution then the set of values of alpha 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