Question

The point $\left(\right[P+1],[P\left]\right)$ (where $\left[x\right]$ is the greatest integer less then or equal to x) lying inside the region bounded by the circle $x^2+y^2+2x-15=0and{x}^2+y^2-2x-7=0$, then

• A. $Pϵ[-1,0)\cup (0,1)\cup [1,2)$
• B. $Pϵ[-1,2)-\{0,1\}$
• C. $Pϵ(-1,2)$
• D. $Pϵ(-2,2)$

Answer 1

The correct option is D $Pϵ(-2,2)$

Given, circles $x^2+y^2+2x-15=0,x^2+y^2-2x-7=0$

$(x+1{)}^2+y^2=4^2$ and $(x-1{)}^2+y^2=(2\sqrt{2}{)}^2$

Centre $(-1,0)$ Centre $(1,0)$

Let point $Q\left(\right[P+1],[P\left]\right)$ lies inside the region bounded by the two circle

Point$\left(\right[P+1],[P\left]\right)$ must lies inside both circles

${(x+1)}^2+y^2-16<0{\left(\right[P+1]+1)}^2+{\left(\right[P\left]\right)}^2-16<0{[P+1]}^2+{\left[P\right]}^2+2[P+1]-15<0\left(1\right)\text{Also},{(x-1)}^2+y^2-8<0{\left(\right[P+1]-1)}^2+{\left(\right[P\left]\right)}^2-8<0{[P+1]}^2+{\left[P\right]}^2-2[P+1]-7<0\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right)$

$4[P+1]-8<0[P+1]<2\Rightarrow P<1$

Intersection of two circle $\left(2x\right)\left(2\right)=8x=2,y=\pm \sqrt{7}$

From curve,

$-2\sqrt{2}+1<[P+1]<3P\in (-2,2)$