Q 51 :

Let A={(x,y)2:y0, 2xy4-(x-1)2}

and B={(x,y)×:0ymin{2x,4-(x-1)2}}.

Then the ratio of the area of A to the area of B is                             [2023]

  • ππ+1    

     

  • π-1π+1    

     

  • π+1π-1    

     

  • ππ-1

     

(2)

Let A={(x,y)2:y0, 2xy4-(x-1)2}

and B={(x,y)×:0ymin{2x,4-(x-1)2}}

We have the following diagram y2+(x-1)2=4

Shaded portion A

Area A=area of quadrant of circle-area of (OAB)

              =π(4)4-12(2)(1)=(π-1)                      ...(i)

Now, for finding the area of portion B, we have

Area B=area of (AOB)+area of quadrant of circle

              =12·(1)(2)+π(2)24=(π+1)                     ...(ii)

Thus, according to the question, ratio of area A to area

B=π-1π+1



Q 52 :

Let Δ be the area of the region {(x,y)2:x2+y221, y24x, x1}. Then 12(Δ-21sin-127) is equal to          [2023]

  • 3-23

     

  • 23-23

     

  • 23-13

     

  • 3-43

     

(4)

If Δ be the area of region {(x,y)2:x2+y221, y24x, x1}. Then we have,

Area=2132xdx+2321(21-x2)dx

Δ=83·(33-1)+21sin-1(27)-63

[Δ-21sin-1(27)]=23-832

12[Δ-21sin-127]=3-43



Q 53 :

The area of the region A={(x,y):|cosx-sinx|ysinx, 0xπ2} is               [2023]

  • 5-22+1

     

  • 5+22-4.5

     

  • 1-32+45

     

  • 35-32+1

     

(1)

The area of the region, A={(x,y):|cosx-sinx|ysinx, 0xπ2}

|cosx-sinx|ysinx

For finding the intersecting point we must have  

cosx-sinx=sinx

tanx=12

Let ϕ=tan-112

So, tanϕ=12, sinϕ=15, cosϕ=25

Area=ϕπ/2(sinx-|cosx-sinx|)dx

=ϕπ/4(sinx-(cosx-sinx))dx+π/4π/2(sinx-(sinx-cosx))dx

=ϕπ/4(2sinx-cosx)dx+π/4π/2cosxdx

=[-2cosx-sinx]ϕπ/4+[sinx]π/4π/2

=5-22+1



Q 54 :

Let q be the maximum integral value of p in [0,10] for which the roots of the equation x2-px+54p=0 are rational. Then the area of the region 
{(x,y):0y(x-q)2, 0xq} is            [2023]

  • 164

     

  • 243

     

  • 1253

     

  • 25

     

(2)

Given equation is x2-px+54p=0

D=p2-5p, which is a perfect square when p=9.

  q=9

Required area=09(x-9)2dx

=09(x2-18x+81)dx=[x33-18×x22+81x]09

=243 sq. units.



Q 55 :

If the area of region S={(x,y): 2y-y2x22y, xy} is equal to n+2n+1-πn-1, then the natural number n is equal to _______ .    [2023]



(5)

Given, x2+y2-2y0; x2-2y0; xy

Area of segment OAB=90°360°×π(1)2-12×1×1

=(π4-12)

Hence, required area =02(x-x22)dx-(π4-12)

=(x22-x36)02 -(π-24)=2-43-π4+12=76-π4

Now, n+2n+1-πn-1=76-π4

For n=5n+2n+1-πn-1=76-π4         n=5



Q 56 :

Let the area enclosed by the lines x+y=2, y=0, x=0 and the curve f(x)=min{x2+34,1+[x]}, where [x] denotes the greatest integer x, be A. Then the value of 12A is____________ .             [2023]



(17)

 



Q 57 :

Let y=p(x) be the parabola passing through the points (-1,0),(0,1) and (1,0). If the area of the region

 {(x,y):(x+1)2+(y-1)21, yp(x)} is A, then 12(π-4A) is equal to ______ .         [2023]



(16)

There can be infinite parabola through given points.

In question, it must be given that axis of parabola is parallel to y-axis.

Equation of parabola passing through (-1, 0), (0, 1) and (1, 0) is x2=-(y-1)                     ...(i)

  Required area, A=-10[(1-x2)-(1-1-(x+1)2)]dx

=-10(-x2+1-(x+1)2)dx

=[-x33+x+121-(x+1)2+12sin-1(x+11)]-10

=12(π2)-13=π4-13

 12(π-4A)=12(π-4(π4-13))=12(43)=16



Q 58 :

If the area of the region {(x,y):|x2-2|yx} is A, then 6A+162 is equal to _______ .           [2023]


 



(27)

We have,   

Region={(x,y):|x2-2|yx}

The graph of the region is as shown in figure.

Required area, A=12[x-{-(x2-2)}]dx+22{x-(x2-2)}dx

=12(x2+x-2)dx+22(-x2+x+2)dx

=(x33+x22-2x)12+(-x33+x22+2x)22

=(223+1-22)-(13+12-2)+(-83+2+4)-(-223+1+22)

=27-1626       6A+162=27-162+162=27



Q 59 :

If A is the area in the first quadrant enclosed by the curve C: 2x2-y+1=0, the tangent to C at the point (1, 3) and the line x+y=1, then the value of 60A is ______ .    [2023]


 



(16)

Given, C:2x2-y+1=0

4x-dydx=0 dydx=4x [dydx](1,3)=4

Equation of tangent to the curve C is

     y-3=4(x-1)

y-4x+1=0    ...(i)

Also, given line is x+y=1    ...(ii)

Solving (i) and (ii), we get x=0.4, y=0.6

Given curve and equation (ii) intersect each other at (0,1) and (-0.5,1.5).

Required area,  

A=01(2x2+1)dx-area of EOA-area of BED+area of EDC

=[2x33+x]01-12×1×1-12(1-14)×3+12×(1-14)×0.6

=(23+1)-12-98+940=1660 sq. units

   60A=1660×60=16



Q 60 :

If the area bounded by the curve 2y2=3x, lines x+y=3, y=0 and outside the circle (x-3)2+y2=2 is A, then 4(π+4A) is equal to _____ .       [2023]


 



(42)

Given, y2=3x2, x+y=3, y=0, and (x-3)2+y2=2

2y2=3(3-y)

2y2+3y-9=0 2y2-3y+6y-9=0

(2y-3)(y+3)=0y=32,-2

Required area, A=032((3-y)-2y23)dy-π8(2)

=(3y-y22-2y39)03/2-π4=3×32-98-29(278)-π4

=92-98-34-π4

 4A+π=4[92-98-34]=212=10.50

  4(4A+π)=42