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The point ([P+1],[P]), (where [.] denotes the greatest integer function) lying inside the region bounded by the circle x^2+y^2-2 x-15=0 \: and\: x^2+y^2-2 x-7=0then

Option: 1

P \in[-1,0) \cup[0,1) \cup[1,2)


Option: 2

P \in[-1,2)-\{0,1\}

 


Option: 3

P \in(-1,2)


Option: 4

 none of these


Answers (1)

best_answer

\because The\: point\: ([P+1],[P])\: lies\: inside \: the\: circle\: x^2+y^2-2 x-15=0, then

[P+1]^2+[P]^2-2[P+1]-15<0

\Rightarrow \quad([P]+1)^2+[P]^2-2([P]+1)-15<0

\Rightarrow \quad 2[P]^2-16<0 \Rightarrow[P]^2<8           \cdots \text { (I) }                 

\\\because \: Circles\: are \: Concentric\\ \therefore Point ([P+1],[P]) out\: side\: the \: circle

\begin{array}{rr} x^2+y^2-2 x-7=0 \\ \therefore\left([P+1]^2+[P]^2-2([P+1])-7>0\right. \\ \Rightarrow([P]+1]^2+[P]^2-2([P]+1)-7>0 \\ \Rightarrow 2[P]^2-8>0 \end{array}

\therefore                                    \quad[P]^2>4 \quad                        \cdots \text { (II) }

From Eqs. (i) and (ii), we get
4<[P]^2<8 which is impossible

\therefore For no value of ' P ' the point will be within the region.

 

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Gunjita

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