Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

41.

The value of 0lnπ2cosex2xex2dx is

  • 1

  • 1 + sin(1)

  • 1 - sin(1)

  • (sin(1) - 1


42.

Distance of the point (2, 3, 4) from the plane 3x - 6 y + 2z + 11 = 0 is

  • 0

  • 1

  • 2

  • 3


43.

The three lines of a triangle are given by (x2 - y2)(2x + 3y - 6) = 0. If the point (- 2, λ) lies inside and (μ, 1) lies outside the triangle, then

  • λ  1, 103, μ  - 3, 5

  • λ  2, 103, μ  - 1, 1

  • λ  - 1, 92, μ  - 2, 103

  • None of the above


44.

If f(x) = 2xdt1 + t4 and g is the inverse of f. Then, the value of g'(0) is

  • 1

  • 17

  • 17

  • None of the above


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45.

The value of i^ . j^ × k^ + j^ . k^ × i^ + k^ . i^ × j^ is

  • 0

  • 1

  • 3

  • - 3


46.

The general solution of the differential equation dydx = ytanx - y2secx is

  • tanx = C + secxy

  • secy = C + tanyx

  • secx = C + tanxy

  • tany = C + secxx


47.

0100ex - xdx is equal to

  • 50(e - 1)

  • 75(e - 1)

  • 90(e - 1)

  • 100(e - 1)


48.

The values of λ, such that (x, y, z) if (0, 0, 0) and i^ + j^ + 3k^x + 3i^ - 3j^ + k^y + - 4i^ + 5j^ are

  • 0, 1

  • - 1, 1

  • - 1, 0

  • - 2, 0


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49.

By Simpson's 13rd rule, the approximate value of the integral 12e- x2dx using four intervals, is

  • 0.377

  • 0.487

  • 0.477

  • 0.387


C.

0.477

Given integral is 12e- x2dxOn dividing interval [1, 2] in four parts, we haveh = 2 - 14 = 14Now, value of f(x) = e- x2 is given below

x 1 54 32 74 e
f(x) 0.6065 0.5352 0.4724 0.4168 0.3679

Now, Simpson's 13 rule is

x0x0 + nhfxdx = h3y0 + yn + 4y1 + y3 + ... + yn - 1+ 2y2 + y4 + ... + yn - 2 12e- x2dx = 1120.6065 + 0.3679 + 40.5352 + 0.4168+ 2(0.4724)                         = 1120.9744 + 3.808 + 0.9448                         = 112 × 5.7272 = 0.477


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50.

A variable plane is at a constant distance p from the origin O and meets the axes at A, B and C. The locus of the centroid of the tetrahedron OABC is

  • 1x2 + 1y2 + 1z2 = 1p2

  • 1x2 + 1y2 + 1z2 = 16p2

  • x2 + y2 + z2 = 16p2

  • x2 + y2 + z2 = p2


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