If limn→∞∑logn + r - logn

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 Multiple Choice QuestionsMultiple Choice Questions

71.

If limx0x1 + acosx - bsinxx3 = 1, then

  • a = - 52, b = - 12

  • a = - 32, b = - 12

  • a = - 32, b = - 52

  • a = - 52, b = - 32


72.

If α and β are the rooots of ax2 + bx + c = 0, then limxα1 - cosax2 + bx + cx - α2 is equal to

  • a22α - β2

  • - a22α - β2

  • 0

  • 1


73.

If sin(x + y) = log(x + y), then dydx is equal to

  • - 1

  • 1

  • 2

  • - 2


74.

If x = acos3t and y = asin3t, then dydxt = π4 is equal to

  • 1

  • - 1

  • 0


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75.

The anti-derivative F of f defined by f(x) = 4x3 - 6x2 + 2x + 5, F(0) = 5, is

  • x4 - 2x3 + x2 + 5x

  • 12x2 - 12x + 2

  • 16x4 - 18x3 + 4x2 + 5x

  • x4 - 2x3 + x2 + 5x + 5


76.

limx0x- 11x

  • 0

  • 1

  • - 1

  • Does not exist


77.

If f(x) = sinxx for x  00         for x = 0

where [x] denotes the greatest integer less than or eual to x, then limx0fx is equal to

  • 1

  • - 1

  • 0

  • Does not exist


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78.

If limnlogn + r - lognn = 2log2 - 12, then limn1nλn + 1λn + 2λ ... n + nλ1n is equal to

  • 4λe

  • 4eλ

  • 4e1λ

  • e4λ


B.

4eλ

Given, limnlogn + r - lognn = 2log2 - 12or limn1nlog1 + rn = 2log2 - 12         ...iLet A = limn1nλn + 1λn + 2λ ... n + nλ1n         = limn1nλ1 + 1nλ1 + 2nλ ... 1 + nnλ1n

On taking log both sides, we get     logA = limn1nlog1 + 1nλ + log1 + 2nλ + ... + log1 + nnλ logA = limn1nr = 1nλlog1 + rn logA = λlimnr = 1n1nlog1 + rn logA = 2λlog2 - 12     from Eq. (i) logA = log4λ - λ logA = log4λ - λloge logA = log4λeλ        A = 4eλ


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79.

limx0sinsinx - sinxax3 + bx5 + c = - 112, then

  • a = 2, b  R, c = 0

  • a = - 2, b  R, c = 0

  • a = 1, b  R, c = 0

  • a = - 1, b  R, c = 0


80.

If f(x) = sinxx, x  00,         x = 0, where [x] denotes the gretest function less than or equal to x, then limx0fx is equal to

  • 1

  • 0

  • - 1

  • Does not exist


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