The value of c in (0, 2) satisfying the mean value theorem for th

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 Multiple Choice QuestionsMultiple Choice Questions

731.

If y = ex . ex2 . ex3 . ... exn ..., for 0 < x < 1, then dydx at x = 12 is

  • e

  • 4e

  • 2e

  • 3e


732.

The derivative of tan-12x1 - x2 with repsect to cos-11 - x2 is

  • 1 - x21 + x2

  • 11 - x2

  • 21 - x21 + x2

  • 21 - x21 + x2


733.

Let f(x) = ex - 12sinxalog1 + x4 for x  0 and f(0) = 12.  If f is continuous at x = 0, then the value of a is equal to

  • 1

  • - 1

  • 2

  • 3


734.

If y = sin-11 - x, then dydx is equal to

  • 11 - x

  • - 121 - x

  • 1x

  • - 12x1 - x


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735.

The derivative of sin-12x1 - x2 with respect to sin-13x - 4x is

  • 23

  • 32

  • 12

  • 1


736.

If fx = x - 2 + x + 1 - x, then f'- 10 is equal to

  • - 3

  • - 2

  • - 1

  • 0


737.

If x = a1 + cosθ, y = aθ + sinθ, then d2ydx2 at θ = π2 is

  • 1a

  • 1a

  • - 1

  • - 2


738.

If y = tan-1cosx1 + sinx, then dydx is equal to

  • 12

  • 2

  • - 2

  • 12


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739.

The value of c in (0, 2) satisfying the mean value theorem for the function f(x) = x(x - 1)2, x [0, 2] is equal to

  • 34

  • 43

  • 13

  • 23


B.

43

Given curve is

f(x) = x(x - 1)2, x  [0,2]

   f'x = x3x - 4 + 1    f'c = c3c - 4 + 1Also, f0 = 0, f2 = 2By mean value theoremf0 - f2 = f'c0 - 2   0 - 2 = c3c - 4 + 1- 2 3c2 - 4c + 1 = 1       c3c - 4 = 0                     c = 0, 43

 


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740.

The value of x in the interval [ 4, 9] at which the function f(x) = x satisfies the mean value theorem is

  • 134

  • 174

  • 214

  • 254


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