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|* 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)=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 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| . Find the values of k if f(x)=k (i) has exactly one real solution, (ii) has two negative solutions, (iii) has two solutions of opposite sign.

Let f(x)=x + 2|x+1|+2| x-1| . Find the values of k if f(x)=k (i) has exactly one real solution, (ii) has two negative solutions, (iii) has two solutions of opposite sign.

Let f(x)=x + 2|x+1|+2| x-1| . Find the values of k if f(x)=k (i) has exactly one real solution, (ii) has two negative solutions, (iii) has two solutions of opposite sign.

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)=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))