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


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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


D.

log(x + y + 2) = C + y

Given differential equaton is              dydx = 1x + y + 1or          dxdy = x + y + 1 dxdy - x = y + 1Eq (i) is of the type dxdy + Px = Q,where P and Q are  functions of y or constant termsHere, P = - 1and   Q = y + 1    IF = ePdy = e- 1dy = e- y

Now, general solution is given by        x . IF = IF . Qdy + C1 x . e-y = e-yIy + 1IIdy +C1    xe-y = y + 1e-y- 1 + 1 . e-ydy + C1    xe-y = e-yy + 1 - e-y + C1         x = - y + 1 - 1 + C1ey                   on dividing by e-y x + y + 2 = C1eyOn taking log both sides, we get

      logx + y + 2 = logC1ey logx + y + 2 = logC1 + logey         logmn = logm + logn logx + y + 2 = C + y         put C  = logC1

which is the required solution.


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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


36.

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

  • ylogex + Cx = 1

  • logxe + Cxx = y

  • logCx2 + ex2y = x

  • None of these


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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