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

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)=(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)=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))

If the equation x^(log_(a)x^(2))=(x^(k-2))/a^(k),a ne 0 has exactly one solution for x, then the value of k is/are

The equation |x+2|-|x+1|+|x1|=K,x in R has one solution,then find value of K .

If x^2 + 5x - 2 k , is exactly divisible by (x-1), then the value of K is (A) 1 (B) 2 (C ) 3 (D) 4