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Solve the equation (x)^(2)=[x]^(2)+2x ...

Solve the equation
`(x)^(2)=[x]^(2)+2x`
where `[x]` and `(x)` are integers just less than or equal to `x` and just greater than or equal to `x` respectively.

A

`0`

B

`1`

C

`2`

D

infinite

Text Solution

Verified by Experts

The correct Answer is:
D
Case -I : If `xepsilonI` then `(x)=[x]`
`:.` equationn becomes `x=0`
Case II: If `x!inI`
`:.` then `(x)=[x]+1` then equation becomes `x=[x]+1/2`
`=n+1/2,"n" epsilonI`
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