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Let f(x) = [x]^(2) + [x+1] - 3, where [....

Let f(x) `= [x]^(2) + [x+1] - 3`, where [.] denotes the greatest integer function. Then

A

`f(x) ne 0` for all real values of x

B

f (x) = 0 for only two real value of x

C

f (x) = 0 for infinite values of x

D

f (x) = 0 for no real value of x

Text Solution

Verified by Experts

The correct Answer is:
C
`f(x)=[x]^(2) +[x+1]-3=0`
`[x]^(2) +[x]+1-3=0`
`[x]^(2) +[x]-2=0`
`{[x] +2} {[x] -1}=0`
`rArr [x] = -2, [x]=1`
`rArr x in [-2, -1) cup [1,2)` Hence f (x) = 0 for infinite values of x.
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