The area (in square units) of the region enclosed by the two circ

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 Multiple Choice QuestionsMultiple Choice Questions

241.

The area (in square units) of the region bounded by x2 = 8y, x = 4 and X-axis, is

  • 23

  • 43

  • 83

  • 103


242.

The area bouned by y = x2 + 2, x-axis, x = 1 and x = 2 is

  • 163

  • 173

  • 133

  • 203


243.

The area (in square units) bounded by the curves y2 = 4x and x2 = 4y in the plane is

  • 83

  • 163

  • 323

  • 643


244.

The area (in square unit) of the region enclosed by the curves y = x and y = x3 is

  • 112

  • 16

  • 13

  • 1


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245.

The area (in sq unit) of the region bounded by the curves 2x = y2 - 1 and x = 0 is

  • 13

  • 23

  • 1

  • 2


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246.

The area (in square units) of the region enclosed by the two circles x2 + y2 = 1 and (x - 1)2 + y2 = 1 is

  • 2π3 + 32

  • π3 + 32

  • π3 - 32

  • 2π3 - 32


D.

2π3 - 32

Intersection point of two circles

 

x2 + y2 = 1              . . . ix - 12 + y2 = 1   . . . iiis given by   x - 12 + 1 - x2 = 1 x2 +1 - 2x - x2 = 0 x = 12From eq. i, 14 + y2 = 1y2 = 1 - 14  y = ± 32Point A12, 32 and C12, - 32So, Area of region OABDO = 2 × Area of region OABDO  . . . iiiArea of OABDO = Area of OADO + Area of ABDA

= 0121 - x - 12dx + 1211 - x2dx= 12 . x - 11 - x - 12 + 12sin-1x - 11120 + 12x1 - x2 + 12sin-1x1112=- 12 12 32 + 12sin-1- 12 - 12 0 -  12sin-1- 1 + 12 . 0 +  12sin-11 - 1432 -  12sin-112

= - 38 - 12sin-112 + 12sin-11 + 12sin-11 - 38 - 12sin-112

= sin-11 - sin-112 - 34= π2 - π6 - 34= π3 - 34From eq. iiiArea of region OABCO = 2 × π3 - 34                                           = 2π3 - 32

 

 


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247.

The values of a function f(x) at different values of x are as follows
x 0 1 2 3 4 5
f(x) 2 3 6 11 18 27

Then, the approximate area (in square units) bounded by the curve y = f(x) and x-axis between x = 0 and 5, using the Trapezoidal rule, is

  • 50

  • 75

  • 52.5

  • 62.5


248.

The area (in square units) of the region bounded by the curves x = y2 and x = 3 - 2y2 is

  • 32

  • 2

  • 3

  • 4


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249.

The area (in sq units) bounded by the curves y2 = 4x and x2 = 4y is

  • 643

  • 163

  • 83

  • 23


250.

The area of the region described by(x, y)/x2 + y  1 and y  1 - x is

  • π2 - 23

  • π2 + 23

  • π2 + 43

  • π2 - 43


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