Question

The point ([p + 1], [p]) lying inside the circle
x² + y² – 2x – 15 = 0, then set of all values of p is (where [.] represent greatest integer function)

• A. [–2, 3)
• B. (–2, 3)
• C. [–2, 0) И (0, 3)
• D. [0, 3)

Answer 1

Answer: (A)

[–2, 3)

Answer 2

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

x² + y² – 2x – 15 = 0

But [x + n] = [x] + n, n О N

Hence [p + 1]² + [p]² – 2[p + 1] – 15 < 0

([p] + 1)² + [p]² – 2 ([p] + 1) – 15 < 0

[p]² + 1 + 2[p] + [p]² – 2[p] – 2 – 15 < 0

2[p]² – 16 < 0 Ю [p]² < 8

Ю – 2$\sqrt{2}$ < [p] < 2$\sqrt{2}$

Ю – 2 Ј p < 3