InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The equation `(x)^(2)=[x]^(2)+2x` where `[x]` and `(x)` are the integers just less than or equal to `x` and just greater than or equal to `x` respectively, then number of values of `x` satisfying the given equation |
|
Answer» Case I If `x epsilonI` then `x=[x]=(x)` The given equation reduces to `x^(2)=x^(2)+2x` `implies2x=0` or `x=0`….i Case II If `x!inI`, then `(x)=[x]+1` The given equation reduces to `([x]+1)^(2)=[x]^(2)+2x` `implies1=2(x-[x])` or `{x}=1/2` `:.x=[x]+1/2=n+1/2,n epsilonI`............ii ,brgt Hence the solution of the original equation is `x=0, n+1/2, n epsilon I`. |
|