The solution set of 5 + 4cosθ

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

71.

1 + x + x + x2x + 1 + xdx is equal to

  • 121 + x + C

  •  231 + x32 + C

  • 1 + x + C

  • 21+ x32 +C


72.

 1 + x - x- 1ex + x- 1dx is equal to˸

  • 1 + xex + x- 1 + C

  • x - 1ex + x- 1 + C

  • - xex + x- 1 + C

  • xex + x- 1 + C


73.

- 22xdx is equal to

  • 1

  • 2

  • 3

  • 4


74.

01sin2tan-11 + x1 - xdx is equal to

  • π6

  • π4

  • π2

  • π


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75.

033x + 1x2 + 9dx is equal to :

  • log22 + π12

  • log22 + π2

  • log22 + π6

  • log22 + π3


76.

If [2, 6] is divided into four intervals of equal length, then the approximate value of 261x2 - xdx using Simpson's rule, is

  • 0.3222

  • 0.2333

  • 0.5222

  • 0.2555


77.

The differential equation of the family of parabola with focus as the origin and the axis as X-axis, is

  • ydydx2 + 4xdydx = 4y

  • - ydydx2 = 2xdydx - y

  • ydydx2 + y = 2xydydx

  • ydydx2 + 2xydydx + y = 0


78.

Soution of dydx = xlogx2 + xsiny + ycosy is

  • ysiny = x2logx + C

  • ysiny = x2 + C

  • ysiny = x2 + logx

  • ysiny = xlogx + C


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79.

The general solution of y2dx + x2 - xy + y2dy = 0 is :

  • tan-1yx = logy + C

  • 2tan-1xy + logx + C = 0

  • logy + x2 + y2 + logy + C = 0

  • sinh-1xy + logy + C = 0


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80.

The solution set of 5 + 4cosθ2cosθ + 1 = 0 in the interval 0, 2π, is :

  • π3, 2π3

  • π3, π

  • 2π3, 4π3

  • 2π3, 5π3


C.

2π3, 4π3

We have, 5 + 4cosθ2cosθ + 1 = 0     ...i                                cosθ  = 1 - tan2θ21 +  tan2θ2,                              cosθ = 1 - t21 + t2      put  tanθ2 = tThen Eq.i beecomes             5 + 41 - t21 + t221 - t21 + t2 +1 = 0  5 + 5t2 + 4 - 4t22 - 2t2 + 1 + t2 = 0                                       t2 + 93 - t2 = 0                                                               t = ± 3 tanθ2 = 3 or tanθ2 = - 3 

 θ2 = π3 or θ2 = 2π3        θ = 2π3 or  4π3


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