The function f(x) = 2x3 - 15x2 + 36x + 6 is strictly decreasing i

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

If y = f(x2 + 2) and f'(3) = 5, then dydx at x = 1 is

  • 5

  • 25

  • 15

  • 10


2.

Let, f(x) = x2 + bx + 7. If f'(5) = 2f'72, then the value of b is

  • 4

  • 3

  • - 4

  • - 3


3.

If y = sin-12x1 - x2, - 12  x  12, then dydx is equal to

  • x1 - x2

  • 11 - x2

  • 21 - x2

  • 2x1 - x2


4.

A straight line parallel to the line 2x - y + 5 = 0 is also a tangent to the curve y2 = 4x + 5. Then, the point of contact is

  • (2, 1)

  • (- 1, 1)

  • (1, 3)

  • (3, 4)


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5.

The function f(x) = 2x3 - 15x2 + 36x + 6 is strictly decreasing in the interval

  • (2, 3)

  • - , 2

  • (3, 4)

  • - , 3  4, 


A.

(2, 3)

Given, f(x) = 2x3 - 15x2 + 36x + 6

On differentiating both sides w.r.t. x, we get

f'(x) = 6x2 - 30x + 36

For strictly decreasing,

f'(x) < 0

 6x2 - 5x + 6 < 0   x - 3x - 2 < 0                2 < x < 3                  x  2, 3


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6.

The slope of the tangent to the curve y2exy = 9e- 3x2 at (- 1, 3) is

  • - 152

  • - 92

  • 15

  • 152


7.

The radius of a cylinder is increasing at the rate of 5 cm/min so that its volume is constant. When its radius is 5 cm and height is 3 cm, then the rate of decreasing of its height is

  • 6 cm/min

  • 3 cm/min

  • 4 cm/min

  • 5 cm/min


8.

The function fx = 2x2 - 1,     if 1  x  4151 - 30x, if 4 < x  5 is not suitable to apply Rolle's theorem, since

  • f(x) is not continuous on [1, 5]

  • f(1) f(5)

  • f(x) is continuous only at x = 4

  • f(x) is not differentiable at x = 4


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9.

The slope of the normal to the curve y = x21x2 at (- 1, 0) is

  • 14

  • - 14

  • 4

  • - 4


10.

The minimum value of sin(x) + cos(x) is

  • 2

  • - 2

  • 12

  • - 12


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