1.

The solution set of the equation `(x)^(2)+[x]^(2)=(x-1)^(2)+[x+1]^(2)`, where (x) denotes the least integer greater than or equal to x and [x] denotes the greatest integer less than or equal to x, isA. RB. R-ZC. R-ND. None of these

Answer» Correct Answer - B
We have,
`(x-1)=(x)-1" and "[x+1]=[x]+1`
`:.(x)^(2)+[x]^(2)=(x-1)^(2)+[x+1]^(2)`
`implies(x)^(2)+[x]^(2)={(x)-1}^(2)+{(x]+1)^(2)`
`implies(x)^(2)+[x]^(2)=(x)^(2)-2(x)+1+[x]^(2)+2[x]+1`
`implies[x]-(x)+1=0`
Now, two cases arise.
CASE I When `c in Z`
In this case,we have
[x]=x and `(x)=x implies[x]-(x)+ne0`
So, the equation [x]-(x)+=0 has no solution.
CASE II When `x in Z`
In this case, we have
x=n+k, where `n in Z` and `0 lt k lt 1`
`:. [x]=n` and (x)=n+1
`:. [x]-x+1=0` is true for all x
Hence, the solution set of the given equation is R-Z.


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