The value of cfrom the Lagrange's mean value theorem for which f(

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 Multiple Choice QuestionsMultiple Choice Questions

81.

The derivative of tan-11 + x2 - 1x with respect to tan-12x1 - x21 - 2x2 at x = 0 is

  • 18

  • 14

  • 12

  • 1


82.

Let f(x + y) = f(x) f(y) and f(x) = 1 + sin(2x) g(x) where g(x) is continuous. Then, f'(x) equals

  • f(x) g(0)

  • 2f(x) g(0)

  • 2g(0)

  • None of the above


83.

If fx = ax2 - b,    a  x < 12,               x = 1x + 1,       1 < x  2 Then, the value of the pair (a, b) for which f(x) cannot be continuous at x = 1, is

  • (2, 0)

  • (1, - 1)

  • (4, 2)

  • (1, 1)


84.

Which of the following function is not differentiable at x = 1 ?

  • f(x) = tanx - 1 + x - 1

  • f(x) = sinx - 1 - x - 1

  • f(x) = x2 - 1x - 1x - 2

  • None of the above


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85.

Using Rolle's theorem, the equation a0xn + a1xn - 1 + ... + an = 0 has atleast one root between 0 and 1, if 

  • a0n + a1n - 1 + ... + an - 1 = 0

  • a0n - 1 + a1n - 2 + ... + an - 2 = 0

  • na0 + (n - 1)a1 + ... + an - 1 = 0

  • a0n + 1 + a1n + ... + an = 0


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86.

The value of cfrom the Lagrange's mean value theorem for which f(x) = 25 - x2 in [1, 5], is

  • 5

  • 1

  • 15

  • None of these


C.

15

It is clear that f(x) has a definite and unique value for each x  [1, 5].

Thus, for every point in the interval [1, 5], the value of f(x) exists.

So, f(x) is continuous in the interval [1, 5].

Also, f'(x) = - x25 - x2, hich clearly exists for all x in an open interval (1,5).

Hence, f'(x) is differentiable in (1, 5).

So, there must be a value c  [1, 5] such that

                f'c = f5 - f15 - 1 = 0 - 244                       = 0 - 264 = - 62But           f'(c) = - c25 - c2 - c25 - c2 = - 62 4c2 = 625 - c2 4c2 = 150 - 6c2  10c2 = 150   c2 = 15  c = ± 15     c = 15  1, 5


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87.

Let f'(x), be differentiable a. If f(1) = - 2 and f'(x) 2  x [1, 6], then

  • f(6) < 8

  • f(6)  8

  • f(6)  5

  • f(6)  5


88.

If fx = 1 + sinxasinx,  - π6 < x < 0b,                                  x = 0etan2xtan3x,               0 < x < - π6

then the value of a and b, if f is continuous at x = 0, are respectively

  • 23, 32

  • 23, e23

  • 32, e32

  • None of these


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89.

If f(x) = mx + 1,         x  π2sinx + n,     x > π2 is continuous at x = π2, then

  • m = 1, n = 0

  • m = 2 + 1

  • n = mπ2

  • m = n = π2


90.

If f(x) = x2,         x  02 sinx, x > 0, then x = 0 is

  • point of minima

  • point of maxima

  • point of discontinuity

  • None of the above


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