Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

21.

The general solution of the equation dydx = y2 - x2yx + 1 is

  • y2 = (1 + x)log(1 + x) - c

  • y2 = 1 + xlogc1 - x - 1

  • y2 = 1 - xlogc1 - x - 1

  • y2 = 1 + xlogc1 + x - 1


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22.

The general solution of the differential equation dydx + sinx + y2 = sinx - y2 is

  • logetany2 = - 2sinx2 +C

  • logetany4 = 2sinx2 +C

  • logetany2 = - 2sinx2 +C

  • logetany4 = - 2sinx2 +C


D.

logetany4 = - 2sinx2 +C

We have,dydx + sinx + y2 = sinx - y2 dydx = sinx - y2 - sinx + y2 dydx = 2cosx - y2 + x + y22sinx - y2 - x + y22          sinC - sinD = 2cosC + D2 . sinC - D2 dydx = 2cosx2sin- y2 dydx = - cosx2siny2 dysiny2 = - 2cosx2dx        Variables are separated cscy2dy = - 2cosx2dxOn integrating both sides, we get

        cscy2dy = - 2cosx2dx 2logetany4 = - 4sinx2 +C   logetany4 = - 2sinx2 +C


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23.

If a, b, c are three vectors such that [a b c] = 5, then the value of [a x b, b x c, c x a] is

  • 15

  • 25

  • 20

  • 10


24.

The point of inflection of the function y = 0xt2 - 3t + 2dt is

  • 32, 34

  • - 32, - 34

  • - 12, - 32

  • 12, 32


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25.

The value of integral dxxx2 - a2

  • c - 1asin-1ax

  • c - 1acos-1ax

  • sin-1ax + c

  • c + 1asin-1ax


26.

The function y specified implicitly by the relation 0yetdt + 0xcostdt = 0 satisfies the differential equation

  • e2yd2ydx2 + dydx2 = sinx

  • eyd2ydx2 + dydx2 = sin2x

  • ey2d2ydx2 + dydx2 = sinx

  • eyd2ydx2 + dydx2 = sinx


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