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Question: The point ([P + 1], [P]) (where [x] is the greatest integer less than or equal to x), lying inside t...

The point ([P + 1], [P]) (where [x] is the greatest integer less than or equal to x), lying inside the region bounded by the circle x2 + y2 –2x –15 = 0 and x2 + y2 –2x – 7 = 0. Then

A

PĪ [1, 0) Č (0, 1) Č (1, 2)

B

P Ī (0, 1)

C

P Ī (1, 2)

D

None of these

Answer

None of these

Explanation

Solution

Since the ([P +1], [P]) lies inside the circle

x2 + y2 –2x –15 = 0

\ ([P + 1]2 + [P]2 –2 ([P + 1]) –15 < 0

([P] + 1)2 + [P]2 –2 ([P] + 1) – 15 < 0
[Q [x + n] = [x] + n, n Ī I]

[P]2 + 1 + 2[P] + [P]2 –2[P] –2 –15 < 0,

Ž 2[P]2 –16 < 0, [P]2 < 8 ... (i)

From the second circle ([P] + 1)2 + [P]2 – 2

([P[ + 1) – 7 > 0

Ž 2[P]2 – 8 > 0, [P]2 > 4 ... (ii)

From (i) and (ii), 4 < [P]2 < 8, which is not possible.

\ for no values of 'P' the point will be within the region.