Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

71.

If the direction cosines of two lines are such that l + m + n = 0, l2 + m2 - n2 = 0, then the angle between them is :

  • π

  • π3

  • π4

  • π6


72.

If xa3 - x3dx = g(x) +c, then gx is equal to :

  • 23cos-1x

  • 23sin-1x3a3

  • 23sin-1x3a3

  • 23cos-1xa


73.

If dxx2 + 2x + 2 = fx + c, then f(x) is equal to

  • tan-1x + 1

  • 2tan-1x + 1

  • - tan-1x + 1

  • 3tan-1x + 1


74.

Observe the following statement:

A : x2 - 1x2ex2 + 1x2dx = ex2 + 1x2 + cR : f'xefxdx = fx + c

Then which of the followmg is true ?

  • Both A and R are true and R is not thecorrect reason for A

  • Both A and R are true and R is the correct reason for A

  • A is true, R is false

  • A is false, R is true


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75.

Dividing the interval [0, 6] into 6 equal parts and by using trapezoidal rule the value of 06x3dx is approximately

  • 330

  • 331

  • 332

  • 333


76.

0π2 dx1 + tan3x = ?

  • π

  • π2

  • π4

  • 3π2


77.

- 11 coshx1 +e2xdx = ?

  • 0

  • 1

  • e2 - 12e

  • e2 + 22e


78.

The solution of x2 + y2dx = 2xydy is :

  • cx2 - y2 = x

  • cx2 + y2 = x

  • cx2 - y2 = y

  • cx2 + y2 = y


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79.

The solution of 1 + x2dydx + 2xy - 4x2 = 0 is :

  • 3x1 + y2 = 4y3 + c

  • 3y(1 + x2) = 4x3 + c

  • 3x(1 + y2) = 4y3 + c

  • 3y(1 + y2) = 4x3 + c


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80.

The solution of dxdy + xy = x2 is :

  • 1y = cx - xlogx

  • 1x = cy - ylogy

  • 1x = cx + xlogy

  • 1y = cx - ylogx


B.

1x = cy - ylogy

dxdy + xy = x2 1x2dxdy + 1xy = 1Put 1x = t  - 1x2dxdy = dtdy - dtdy + ty = 1  dtdy - ty = - 1On compairing with dydx + Py = Q, we getP = - 1y, Q = - 1I.F. = epdx = e- 1ydy = 1yThe solution ist(I.F.) = QI.F.dy + c  t . 1y = - 1 .  1ydy +c  1x1y = - logy + c       1x = cy - y logy


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