The quadratic equation whose roots are&n

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 Multiple Choice QuestionsShort Answer Type

231.

The value of 0 . 16log2 . 513 + 132 + 133 + ...  is


 Multiple Choice QuestionsMultiple Choice Questions

232.

limxaa + 2x13 - 3x133a + x13 - 4x13 a  0 = ?

  • 292313

  • 2343

  • 2943

  • 232913


233.

Let f : 0,   0,  be a differentiable function such that f(1) = e and limtx t2f2x - x2f2tt - x. If f(x) = 1, then x is equal to

  • 1e

  • 2e

  • 12e

  • e


234.

limx0xe1 + x2 + x4 - 1/x - 11 + x2 + x4 - 1

  • does not exist

  • is equal to 1

  • is equal to e

  • is equal to 0


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235.

A limx10x - 12tcost2dtx - 1sinx - 1 = ?

  • 1

  • does not exist

  • 12

  •  - 12


236.

If [x] denotes the greatest integer not exceeding x and if the function f defined by

fx =  a + 2cosxx2, x< 0btanπx + 4, x  0

is continuous at x = 0, then the ordered pair(a, b) is equal to

  • (- 2, 1)

  • (- 2, - 1)

  • - 1, 3

  • - 2, 3


237.

If y = 1 +x1 + x21 + x4 . . . 1 + x2n,then dydxx = 0 = ?

  • 0

  • 12

  • 1

  • 2


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238.

The quadratic equation whose roots are l and m,where  l = limθ  0 3sinθ - 4sin2θθ,m = limθ  0 2tanθθ 1 - tan2θ  is

  • x2 + 5x + 6

  • x2 - 5x + 6

  • x2 - 5x - 6

  • x2 + 5x - 6


B.

x2 - 5x + 6

We have,l = limθ03sinθ - 4sin2θθUsing L-hospital rule              = limθ0 3cosθ - 8sinθcosθ1 = 3and    m = limθ02tanθ1 -tan2θ             = limθ0 tan2θθ             = limθ0 tan2θ2θ = 2The required quadratic equation is       x2 - l + mx +lm = 0  x2 - 3 + 2x + 3 . 2 = 0                               x2 - 5x + 6 = 0 


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239.

limx01 + 1 + x2  - 2x - 8 = ?

  • 32

  • 14

  • 124

  • 112


240.

If cos-1x2 - y2x2 + y2 = k a constant, then dydx = ?

  • yx

  • xy

  • x2y2

  • y2x2


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