The value of 'c' of Rolle's theorem for f(x) = ex sin(x) in [0, x

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 Multiple Choice QuestionsMultiple Choice Questions

961.

ddxtan-11 + x2 + 1 - x21 + x2 - 1 - x2 is equal to

  • - x1 - x4

  • x1 - x4

  • - 121 - x4

  • 121 - x4


962.

If y = x2x + 32x + 1, then dydx is equal to

  • y12x + 42x + 3 - 12x + 1

  • y13x + 42x + 3 - 12x + 1

  • y13x + 42x + 3 - 1x + 1

  • None of the above


963.

If y = xy, then x1 - ylogxdydx is equal to

  • x2

  • y2

  • xy2

  • xy


964.

Rolle's theorem is not applicable for the function f(x) = x, where x [- 1, 1] because

  • the function f(x) is not continuous in the interval [- 1, 1]

  • the function f(x) is not differentiable in the interval (- 1, 1)

  • f- 1  f1

  • f- 1 = f1  0


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965.

If the function f(x) = x3 - 6ax2 + 5x satisfies the conditions of Lagrange's Mean Value theorem for the interval [1, 2] and the tangent to the curve y = f(x) at x = 7/4 is parallel to the chord that join the points ofintersection of the curve with the ordinates x = 1 and x = 2 . Then, the value of a is

  • 3516

  • 3548

  • 716

  • 516


966.

If 2a + 3b + 6c = 0, then atleast one root of the equation ax2 + bx + c = 0, lies in the interval

  • (0, 1)

  • (1, 2)

  • (2, 3)

  • None of these


967.

If equation y + y33 + y55 + ... = 2x + x33 + x55 + ... , then value of y is

  • x1 - x2

  • 2x1 - x2

  • 1 - x22x

  • None of these


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968.

The value of 'c' of Rolle's theorem for f(x) = ex sin(x) in [0, x] is given by

  • π4

  • 3π4

  • 5π6

  • π2


B.

3π4

Given, fx = exsinxNow,   f0 = e0sin0 = 0and    fπ = eπsinπ = 0       f0 = fπIt is also continuous in the interval [0, π]and also differentiable (0, π).Now, fx = exsinx   f'x = excosx + exsinxPut  f'x = 0 excosx + exsinx = 0 excosx + sinx = 0 tanx = - 1        x = 3π4        c = 3π4


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969.

- aaxa2 - x2dx is equal to

  • π4

  • π3

  • π8

  • 0


970.

If y = tan-11 - cosx1 + cosx, then dydx will be

  • sinxcosx

  • π2

  • 12

  • 11 + cos2x


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