If cotcos-1x = sectan-1ab2 - a2, then x

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The length of the normal to the curve x = aθ + sinθ, y =  a1 - cosθ at θ = π2 is

  • 2a

  • a2

  • a2

  • 2a


2.

The maximum value of logxx is

  • e

  • 2e

  • 1e

  • 2e


3.

In the interval - π4, π4, the number of real solutions of the equations sinxcosxcosxcosxsinxcosxcosxcosxsinx = 0 is

  • 0

  • 2

  • 1

  • 3


4.

If f(x) = xsin1x, x  0k           , x = 0 is continuous at x = 0, then the value of k will be

  • 1

  • - 1

  • 0

  • None of these


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5.

If cotcos-1x = sectan-1ab2 - a2, then x is equal to

  • b2b2 - a2

  • a2b2 - a2

  • 2b2 - a2a

  • 2b2 - a2b


A.

b2b2 - a2

Given, cotcos-1x = sectan-1ab2 - a2 cotcos-1x1 - x2 = secsec-1bb2 - a2     x1 - x2 = bb2 - a2  x2b2 - a2 = b2 - b2x2 x22b2 - a2 = b2                     x = b2b2 - a2


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6.

Number of solutions ofthe equation

tan-112x + 1 + tan-114x + 1 = tan-12x2 is

  • 1

  • 2

  • 3

  • 4


7.

If f(x) = 4x4x + 2, then f197 + f297 + ... + f9697 is equal to

  • 1

  • 48

  • - 48

  • - 1


8.

If f(x) = xpcos1x, x  00             , x = 0 is differentiable at x = 0, then

  • p < 0

  • 0 < p < 1

  • p = 1

  • p > 1


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9.

If xx21 +x3yy21 + y3zz21 + z3 = 0 and x, y, z are all distinct, then xyz is equal to

  • - 1

  • 1

  • 0

  • 3


10.

If A = 1111, then A100 is equal to

  • 100 A

  • 299 A

  • 2100 A

  • 99 A


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