If fx = logex3 - x3 + x13, the

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 Multiple Choice QuestionsMultiple Choice Questions

121.

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


122.

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


123.

If the function f(x) = x,              if x  1cx + k,     if 1 < x < 4- 2x,       if x  4 is continuous everywhere, then the values of c and k are respectively.

  • - 3, - 5

  • - 3, 5

  • - 3, - 4

  • - 3, 4


124.

If y = 5tanx, then dydx at x = π4 is equal to

  • 5log5

  • 10log5

  • 0

  • log52


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125.

If y = sin-1x and z = cos-11 - x2, then dydz is equal to

  • x1 - x2

  • 12

  • - x1 - x2

  • 1


126.

If u = 2(t - sin(t)) and v = 2 (1 - cos(t)), then dvdu at t = 2π3 is equal to

  • 3

  • - 3

  • 23

  • 13


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127.

If fx = logex3 - x3 + x13, then f'(1) is equal to

  • 34

  • 23

  • 13

  • 12


A.

34

Given, f(x) = logex3 - x3 + x13

 fx = logex + 13log3 - x - log3 + xOn differentiating w.r.t. x, we getf'x = 1ex × ex + 13- 13 - x - 13 +x       = 1 + 13- 13 - x - 13 + x       = 1 + 13- 3 - x - 3 + x9 - x2       = 1 + - 69 - x2

At x = 1

f'1 = 1 + 13- 69 - 12       = 1 + 13- 68 = 1 - 14 = 34


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128.

If y = logx2, then dydx at x = e is equal to

  • 2

  • e2

  • e

  • 2e


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129.

If yx = 2x, then dydx is equal to

  • yxlog2y

  • xylog2y

  • yxlogy2

  • xylogy2


130.

If x2 + 2xy + 2y2 = 1, then dydx  at the point where y = 1 is equal to

  • 1

  • 2

  • - 1

  • 0


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