Let the population of rabbits surviving at a time t be governed

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The area (in sq. units) enclosed between the curves y = x2 and y = x is

  • 23

  • 16

  • 13

  • 1


2.

If one of the diameters of the circle, given by the equation, x2+y2−4x+6y−12=0, is a chord of a circle S, whose centre is at (−3, 2), then the radius of S is:

  • 5√2

  • 5√3

  • 5

  • 10

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3.

Let P be the point on the parabola, y2=8x which is at a minimum distance from the centre C of the circle, x2+(y+6)2=1. Then the equation of the circle, passing through C and having its centre at P is:

  • x2+y2−4x+8y+12=0

  • x2+y2−x+4y−12=0

  • x2+y2− 4 x +2y−24=0

  • x2+y2− 4 x +2y−24=0

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4.

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is:

  • 4/3

  • 4/√3

  • 2/√3

  • √3

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5.

The area (in sq units) of the region described by {x,y): y2 ≤ 2x and y ≥ 4x-1} is

  • 7/32

  • 5/64

  • 15/64

  • 15/64

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6.

The area (in sq units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipsestraight x squared over 9 plus straight y squared over 5 space equals space 1 space is space

  • 27/4

  • 18

  • 27/2

  • 27/2

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7.

Let O be the vertex and Q be nay point on the parabola x2 = 8y. If the point P divides the line segment OQ internally in the ratio 1:3 then the locus of P is 

  • x2= y

  • y2 =x

  • y2 =2x

  • y2 =2x

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8.

If =-1 and x =2 are extreme points of f(x) =α log|x| + βx2 +x, then

  • α = -6, β = 1/2

  • α = -6, β = -1/2

  • α = 2, β = -1/2

  • α = 2, β = -1/2

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9.

The area of the region described by A = {(x,y): x2 +y2 ≤ 1 and y2 ≤1-x} is

  • straight pi over 2 plus 4 over 3
  • straight pi over 2 minus 2 over 3
  • straight pi over 2 minus 2 over 3
  • straight pi over 2 minus 2 over 3
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10.

Let the population of rabbits surviving at a time t be governed by the differential equation.fraction numerator dp left parenthesis straight t right parenthesis over denominator dt end fraction space equals space 1 half straight p left parenthesis straight t right parenthesis space minus 200 If p(0) = 100 then p(t) is equal to 

  • 400 minus 300 straight e to the power of straight t over 2 end exponent
  • 300 minus 200 straight e to the power of negative straight t over 2 end exponent
  • 600 minus 500 straight e to the power of straight t over 2 end exponent
  • 600 minus 500 straight e to the power of straight t over 2 end exponent


A.

400 minus 300 straight e to the power of straight t over 2 end exponent

Given differential equation dp over dt minus 1 half straight p left parenthesis straight t right parenthesis space equals space minus 200 is a linear differential equation
Here, p(t) = fraction numerator negative 1 over denominator 2 end fraction comma space straight Q space left parenthesis straight t right parenthesis space equals space minus space 200
 If space equals space straight e to the power of integral negative open parentheses 1 half close parentheses dt end exponent space equals space straight e to the power of negative 1 half end exponent
Hence, solution is 
p(t), IF = ∫Q(t)IF dt
straight p left parenthesis straight t right parenthesis. straight e to the power of negative straight t over 2 end exponent space equals space integral negative 200. straight e to the power of negative straight t over 2 end exponent dt
straight p left parenthesis straight t right parenthesis. straight e to the power of negative straight t over 2 end exponent space equals integral negative 200 space straight e to the power of negative straight t over 2 end exponent. dt
straight p left parenthesis straight t right parenthesis. straight e to the power of negative straight t over 2 end exponent space equals space 400 space straight e to the power of negative straight t over 2 end exponent space plus space straight K
rightwards double arrow space straight p left parenthesis straight t right parenthesis space equals space 400 space plus ke to the power of negative 1 half end exponent
If space straight p left parenthesis 0 right parenthesis space equals space 100 space comma then space straight k space equals negative 300
rightwards double arrow space straight p left parenthesis straight t 0 space equals space 400 space minus 300 to the power of straight t over 2 end exponent

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