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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Mathematics
Chapter
Area Under Curves

BITSAT 2019 · Mathematics · Area Under Curves

The point ( [ P + 1 ] , [ P ] ) (where, [ x ] is the greatest...

The point ([P+1],[P])( [ P + 1 ] , [ P ] ) (where, [x][ x ] is the greatest integer function) lying inside the region bounded by the circle x2+y2−2x−15=0x^{2} + y^{2} - 2x - 15 = 0 and x2+y2−2x−7=0x^{2} + y^{2} - 2x - 7 = 0, then

Options

  1. AP∈[−1,2)−0,1P \in [ - 1, 2 ) - { 0, 1 }
  2. BP∈[−1,0)∪(0,1)∪(1,2]P \in [ - 1, 0 ) \cup (0,1) \cup (1,2]
  3. CP∈(−1,2)P \in ( - 1, 2 )
  4. DNone of these

Correct answer

D. None of these

Step-by-step solution

[P+1]=[P]+1[ P + 1 ] = [ P ] + 1, Let [P]=n[ P ] = n, then nn is integer

∴([P+1],[P])=(n+1,n)∴ ( [ P + 1 ] , [ P ] ) = ( n + 1, n ) lie inside the region of circles

S1=x2+y2−2x−15=0,C1=(1,0),r1=4{S}_{1} = x^{2} + y^{2} - 2x - 15 = 0, {C}_{1} = (1,0) , {r}_{1} = 4

and S2=x2+y2−2x−7=0,C2=(1,0),r2=22{S}_{2} = x^{2} + y^{2} - 2x - 7 = 0, {C}_{2} = (1,0) , {r}_{2} = 2\sqrt{2}

Both circles are concentric.

∴(n+1)2+n2−2(n+1)−7>0∴ ( n + 1{)}^{2} + n^{2} - 2 ( n + 1 ) - 7 > 0

and (n+1)2+n2−2(n+1)−15<0⇒4<n2<8( n + 1{)}^{2} + n^{2} - 2 ( n + 1 ) - 15 < 0 ⇒ 4 < n^{2} < 8

Which is not possible for any integer.

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