" Let "f(x)=[|x|-|x-1|]^(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| .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)=[e^(x) (x-1)(x-2)] . Then f decrease in the interval :

Let f(x)=|x-1|+|x-2|+|x-3|+|x-4|AA x in R Then

Let f(x)=|x-1|+|2x-1|+|3x-1|+......+|119x-1| If 'm' denotes the minimum value of f(x) the sqrt(m)sqrt(m) equals is

Let f(x)=|x-1|+|2x-1|+|3x-1|+......+|119x-1| If 'm' denotes the minimum value of f(x) the sqrt(m)sqrt(m) equals is

Let f(x)=x(x-1)(x-2) , x in [0,2] Find f(0) and f(2)