Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
BITSAT2019MathematicsArea Under CurvesActual

The point ( [ P + 1 ] , [ P ] ) (where, [ x ] is 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 = 0 , then

Options

  1. AP ∈ [ - 1 , 2 ) - 0 , 1
  2. BP ∈ [ - 1 , 0 ) ∪ ( 0 , 1 ) ∪ ( 1 , 2 ]
  3. CP ∈ ( - 1 , 2 )
  4. DNone of these

Correct answer

D. None of these

Step-by-step solution

[ P + 1 ] = [ P ] + 1 , Let [ P ] = n , then n is integer ∴   ( [ P + 1 ] , [ P ] ) = ( n + 1 , n ) lie inside the region of circles S 1 = x 2 + y 2 - 2 x - 15 = 0 , C 1 = ( 1 , 0 ) , r 1 = 4 and S 2 = x 2 + y 2 - 2 x - 7 = 0 , C 2 = ( 1 , 0 ) , r 2 = 2 2 Both circles are concentric. ∴   ( n + 1 ) 2 + n 2 - 2 ( n + 1 ) - 7 > 0 and ( n + 1 ) 2 + n 2 - 2 ( n + 1 ) - 15 < 0 ⇒ 4 < n 2 < 8 Which is not possible for any integer.

Practice Area Under Curves on Quantrex Academy →

More from Area Under Curves

The line y=m x bisects the area unclosed by lines x=0, y=0 and x= 3 2 and the curve y=1+4 x-x^2 . Then, the value of m is 2024If a, c, b are in GP, then the area of the triangle formed by the lines a x+b y+c=0 with the coordinates axes is equal to 2024The area enclosed by the curves y=x^3 and y= x is 2024The area of the region bounded by the curves x=y^2-2 and x=y is 2024If the area bounded by the curves y=a x^2 and x=a y^2,(a 0) is 3 sq units, then the value of a is 2024The area of the region R = (x, y): 5 x^2 y 2 x^2+9 is : 2023The area enclosed between the curve y= _e(x+e) and the coordinate axes is 2023The area of the region bounded by the parabola (y-2)^2=(x-1) , the tangent to the parabola at the point (2,3) and the X -axis is 2022 Full Area Under Curves list All BITSAT PYQs