Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

61.

For any two real numbers a and b, we define a R b if and only if sin2(a) + cos2(b) = 1. The relation R is

  • reflexive but not symmetric

  • symmetric but not transitive

  • transitive but not reflexive

  • an equivalence relation


62.

For the curve x2 + 4xy + 8y = 64 the tangents are parallel to the x-axis only at the points

  • 0, 22 and 0, - 22

  • (8, - 4) and (- 8, 4)

  • 82, - 22 and - 82, 22

  • (9, 0) and (- 8, 0)


63.

If f(x) = x3 - 3x + 2,                      x < 2,x3 - 6x2 + 9x + 2,           x  2

then

  • limx2f(x) does not exist

  • f is not continuous at x = 2

  • f is continuous but not differentiable at x = 2

  • f is continuous and differentiable at x = 2


64.

Let exp (x) denote the exponential function ex. If f (x) = expx1x, x > 0, then the minimum value off in the interval [2, 5] is

  • expe1e

  • exp212

  • exp515

  • exp313


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65.

The minimum value of the function f (x) = 2x - 1 + x - 2 is

  • 0

  • 1

  • 2

  • 3


66.

If P = 2- 2- 4- 1341- 2- 3, then P5 is equal to

  • P

  • 2P

  • - P

  • - 2P


67.

Consider the system of equations x + y + z = 0, αx + βy + γz = 0 and α2x + β2y + γ2z = 0. Then, the system of equations has

  • a unique solution for all values of α, β and γ

  • infinite number of solutions, if any two of α, β, γ are equal.

  • a unique solution, if α, β and γ are distinct

  • more than one, but finite number of solutions depending on values of α, β and γ


68.

The value of the integral

- 11x2013exx2 +cosx + 1exdx is equal to

  • 0

  • 1 - e- 1

  • 2e- 1

  • 21 - e- 1


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69.

The value of I = 0π/4tann + 1xdx + 120π/2tann + 1x2dx is

  • 1n

  • n + 22n + 1

  • 2n - 1n

  • 2n - 33n - 2


A.

1n

Given,

I = 0π/4tann + 1xdx + 120π/2tann - 1x2dxIn second integral, put t = x2 dx = 2dt

 Also, when x = 0 then t = 0,When x = π2 then t = π4Then, I = 0π4tann + 1xdx + 0π/4tann- 1tdt           I = 0π4tann + 1xdx +0π4tann- 1xdx                                        abf(x)dx = 0π4f(y)dy      I = 0π4tann + 1x + tann - 1xdx      I = 0π4tann - 1x . tan2x + 1dx      I= 0π4tann - 1 × sec2xdxPut     t = tanx    dt = sec2xdxAlso, when x = 0, then t = 0when x = π4, then t = 1

I = 01tn - 1dt = tnn01 = 1n


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70.

The value of the integral

12exlogex + x + 1xdx

  • e21 + loge2

  • e2 - e

  • e21 + loge2 - e

  • e2 - e1 + loge2


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