Conic Sections Question 210
Question: 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 $ x^{2}+y^{2}-2x-15=0 $ and $ x^{2}+y^{2}-2x-7=0, $ then
Options:
A) $ P\in [-1,0)\cup [0,1)\cup [1,2) $
B) $ P\in [-1,2)-{0,1} $
C) $ P\in (-1,2) $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Since the $ ([P+1],[P]) $ lies inside the circle $ x^{2}+y^{2}-2x-15=0 $
$ [But[x+n]=[x]+n,n\in N] $
$ \therefore {{[P+1]}^{2}}+{{[P]}^{2}}-2[P+1]-15<0 $
$ {{([P]+1)}^{2}}+{{[P]}^{2}}-2([P]+1)-15<0 $
$ 2{{[P]}^{2}}-16<0,{{[P]}^{2}}<8 $
From the second circle $ {{([P]+1)}^{2}}+{{[P]}^{2}}=-2([P]+1)-7<0 $ $ \Rightarrow 2{{[P]}^{2}}-8>0,{{[P]}^{2}}>4 $
…… (2) From (1) & (2), $ 4<{{[P]}^{2}}<8, $ which is not possible
$ \therefore $ For no values of -P- the point will be within the region.
BETA
