Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

31.

The value of tan2tan-115 - π4 is

  • 177

  • - 177

  • 717

  • - 717


32.

The solution of the equation dydx = 1x + y + 1 is

  • x + y = Cey - 2

  • x + y = Clog(y) - 4

  • log(x+ y + 2) = Cy

  • log(x + y + 2) = C + y


33.

Image point of (1, 3, 4) in the plane 2x - y + z + 3 = 0 will be

  • (3, 5, 2)

  • (3, 5, - 2)

  • (- 3, 5, 2)

  • None of these


34.

The number of vectors of unit length perpendicular to vectors a = i^ + j^ and b = j^ + k^

  • infinite

  • one

  • two

  • three


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35.

By trapezoidal rule, the approximate value of the integral 06dx1 + x2 is

  • 1.3128

  • 1.4108

  • 1.4218

  • None of these


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36.

The solution of differential equation (ylog(x) - 1)ydx = xdy is

  • ylogex + Cx = 1

  • logxe + Cxx = y

  • logCx2 + ex2y = x

  • None of these


D.

None of these

Given differential equation is   ylogx - 1ydx = xdy                    dydx = ylogx - 1yx                              = y2logxx - yx           dydx + yx = y2logxx 1y2dydx + y-1x = logxx       ...iPut y-1 = v - y-2dydx = dvdx     y-2dydx = - dvdxFrom Eq. (ii), we have - dvdx + vx = logxx dvdx - vx = logxx             ...iiTlus is linear differential equationHere, IF = e- 1xdx = e- logx = 1xSo, soluton is v . IF = IF . Q dx +C

v . 1x = 1x- logxxdx + C v . 1x = - logxx2dx +C               = - logx- 1x + 1x . 1xdx + C 1x . y = logxx + 1x +C     v = 1y       1 = ylogx +1 +Cx       1 = ylogx + loge + Cx                        1 = loge      1 = yloge . x + Cx


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37.

If m things are distributed among a men and b women. Then, the chance that the number of things received by men is odd, is

  • b - am - b + am2b + am

  • b + am - b - am2b + am

  • b + am - b - amb + am

  • None of these


38.

The value of the integral I = tanx + cotxdx, where x  0, π2, is

  • 2sin-1cosx - sinx + C

  • 2sin-1sinx - cosx + C

  • 2sin-1cosx + sinx + C

  • - 2sin-1sinx + cosx + C


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39.

The solution of the differential equation a + xdydx + xy = 0 is

  • y = Ce232a  - xx + a

  • y = Ce23a  - xx + a

  • y = Ce232a  + xx + a

  • y = Ce- 232a  - xx + a


40.

If a = i^ - 2j^5 and b = 2i^ + j^ + 3k^14 are vectors in space, then the value of 2a +b . a × b × a - 2b is

  • 0

  • 1

  • 5

  • 4


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