Find the solution set of `(x)^(2)+(x+1)^(2)=25` where `(x)` is the least integer greater than or equal to `x`.

Options

A. (2,4)
B. `(-5,-4] cup (2,3]`
C. `[-4,-3) cup [3,4)`
D. None of these

Answer

B

Explanation

We have the following cases:
CASE I When `x in Z`
In this case, we have
`(x)=x and (x+1)=x+1`
`:. (x^(2))+(x+1)^(2)=25`
`implies 2x^(2)+2x+1=25`
`implies 2x^(2)+2x-24=0`
`implies x^(2)+x-12=0 implies (x+4)(x-3)=0 implies x=3,-4`
CASE II When this case, we have
x=n+k, where ` in Z` and `0 lt k lt 1`
`:. (x)=n+1` and (x+1)=n+2
Now,
`(x)^(2)+(x+1)^(2)=25`
`implies n^(2)+2n+1+n^(2)+4n+4=25`
`implies 2n^(2)+6n-20=0`
`implies n^(2)+3n-10=0 implies(n+5)(n-2)=0 implies n=2,-5`
`:. x=2+k,-5+k`, where `0 lt k lt 1`.
Hence, `x in (2,3] cup (-5,-4]`.