If sin-1x5 + csc-154  = π2, th

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

What must be the matrix X if 2X + 1234 = 3872 ?

  • 132- 1

  • 1- 32- 1

  • 264- 2

  • 2- 64- 2


2.

The value of 111bccaabb + cc + aa + b is

  • 1

  • 0

  • (a - b)(b - c)(c - a)

  • (a + b)(b + c)(c + a)


3.

The value of 441442443445446447449450451 is :

  • 441 × 446 × 4510

  • 0

  • - 1

  • 1


4.

Inverse of the matrix cos2θ- sin2θsin2θcos2θ is

  • cos2θ- sin2θsin2θcos2θ

  • cos2θsin2θsin2θ- cos2θ

  • cos2θsin2θsin2θcos2θ

  • cos2θsin2θ- sin2θcos2θ


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5.

If the distance 's' metres traversed by a particle int seconds is given by s = t3 - 3t2, then the velocity of the particle when the acceleration is zero, in m/s is

  • 3

  • - 2

  • - 3

  • 2


6.

If x = Acos4t + Bsin4t, then d2xdt2 is equal to

  • - 16x

  • 16x

  • x

  • - x


7.

If tangent to the curve x = at2, y = 2at is perpendicular to X - axis, then its point of contact is

  • (a, a)

  • (0, a)

  • (0, 0)

  • (a, 0)


8.

If f(x) = 1 - cosxx, x  0k                , x = 0 is continuous at x = 0, then the value of k is

  • 0

  • 12

  • 14

  • 12


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9.

If cos-1p +cos-1q +cos-1r = π, then p2 + q2 + r2 + 2pqr is equal to

  • 3

  • 1

  • 2

  • - 1


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10.

If sin-1x5 +csc-154  = π2, then x is equal to

  • 1

  • 4

  • 3

  • 5


C.

3

We have, sin-1x5 +csc-154  = π2         sin-1x5 +csc-154 = π2 sin-1x5 × 1 - 452 + 451 - x52 = π2or x5 × 35 + 45 × 25 - x25 = sinπ2 3x + 425 - x2 = 25          425 - x2 = 25 - 3x

On squaring both sides, we get

16(25 - x2) = 625 + 9x2 - 150x

 4500 - 16x2 = 625 + 9x2 - 150x 16x2 + 9x2 - 150x +625 - 400 = 0 25x2 - 150x + 225 = 0         x2 - 6x + 9 = 0or x2 - 3x - 3x + 9 = 0 xx - 3 - 3x - 3 = 0                    x - 32 = 0                           x = 3


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