f(x) and g(x) are differentiable in the interval [0, 1] such that

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381.

f(x) and g(x) are differentiable in the interval [0, 1] such that f(0) = 2, g(0) = 0, f(1) = 6, g(1) = 2, then Rolle's theorem is applicable for which of the following in [0, 1] ?

  • f(x) - g(x)

  • f(x) - 2g(x)

  • f(x) + 3g(x)

  • None of the above


B.

f(x) - 2g(x)

Let ϕx = fx - 2gx, x  0, 1Clearly, ϕ(x) is continuous on [0, 1] and differentiable on (0, 1),as f(x) and g(x) are differentiable on [0, 1].Also, ϕ0 = f0 - 2g0 = 2 - 0 = 2and   ϕ1 = f1 - 2g1 = 6 - 2 × 2 = 2     ϕ0 = ϕ1Thus, ϕ(x) satisfies all the three conditions of Rolle's theorem.Therefore, there exists a point x  (0, 1) such that    ϕ'x = 0  f'x - 2g'x = 0 f'x = 2g'x


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382.

The positive root of equation x2 - 2x - 5 = 0 lies in the interval

  • {0, 1}

  • (1, 2)

  • (2, 3)

  • (3, 4)


383.

One root of the equation x2 - 4x+ 1 = 0 is between 1 and 2. The value ofthis root using Newton-Raphson method will be

  • 1.775

  • 1.850

  • 1.875

  • 1.950


384.

The point/points of discontinuity of the function f(x) = x + 3,     if x  - 3- 2x,         if - 3 < x < 36x + 2,      if x  3 is/are

  • 3, - 3

  • 3

  • - 3

  • None of these


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385.

The value of ddxsintan-1ex at x = 0  is

  • 0

  • - 2

  • - 122

  • - 12


386.

If f(x) = x2 - 10x + 25x2 - 7x + 10 and f is continuous at x = 5, then f(5) is equal to

  • 0

  • 5

  • 10

  • 25


387.

If h(x) = xxx, then at x = 1, h'xhx is equal to

  • h(x)

  • 1hx

  • 1 + loghx

  • - loghx


388.

ddxsin-13x - 4x3 is equal to

  • 34 - x2

  • 31 - x2

  • 14 - x2

  • - 14 - x2


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389.

If f(x) = x2x + a, then f''(a) is equal to

  • 4a

  • 18a

  • 14a

  • 8a


390.

If u = ex2 - y2, then

  • xux = yuy

  • yux = xyy

  • yux + xuy = 0

  • x2uy + y2ux = 0


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