The points at which the tangent to the curve y = x3 + 5 is perpen

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 Multiple Choice QuestionsMultiple Choice Questions

291.

The function y = 2x3 - 9x2 + 12x - 6 is monotonic decreasing when

  • 1 < x < 2

  • x > 2

  • x < 1

  • None of these


292.

Maximum slope of the curve y = - x3 + 3x2 + 9x - 27 is

  • 0

  • 12

  • 16

  • 32


293.

The points on the curve x2 = 2y which are closest to the point (0, 5) are

  • (2, 2), (- 2, 2)

  • 22, 4, - 22, 4

  • 6, 3, - 63, 3

  • 23, 6, - 23, 6


294.

The point on the curve y = 2x2 - 4x + 5, at which the tangent is parallel to x-axis, will be

  • (1, 3)

  • (- 1, 3)

  • (1, - 3)

  • (- 1, - 3)


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295.

The interval, inwhich the function f(x) = x2e-x is an increasing function, will be

  • - , 

  • (- 2, 0)

  • 2, 

  • (0, 2)


296.

The minimum distance from the point (4, 2) to the parabola y2 = 8x is

  • 2

  • 22

  • 2

  • 32


297.

A particle moves so that the space described in time t is square root of a quadratic function of t. Then,

  • v  1s

  • acceleration  1s3

  • acceleration  s3

  • None of these


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298.

The points at which the tangent to the curve y = x3 + 5 is perpendicular to the line x + 3y = 2 are

  • (6, 1), (- 1, 4)

  • (1, 6), (1, 4)

  • (6, 1), (4, - 1)

  • (1, 6), (- 1, 4)


D.

(1, 6), (- 1, 4)

Given, curve         y = x3 + 5 dydx = 3x2          ...i   Now, equation of tangent to the curve y = x3 + 5and perpendicular to the line x + 3y = 2 is 33x - y + λ = 0     ...iiSo, the slope of Eq. (ii) = 3x2       3 = 3x2 x2 = 1   x = ± 1 y = 1 + 5 = 6 at x = 1 y = - 1 + 5 = 4 at x = - 1Hence, the points are (- 1, 4) and (1, 6). Option (d) (1, 6), (- 1, 4) is correct.


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299.

Let f(x) = a - (x - 3)8/9, then maxima of f(x) is

  • 3

  • a - 3

  • a

  • None of these


300.

If the total cost C(x) in rupees associated with the production of x units of an item is given by C (x) = 3x3 - 2x2 + x + 100. Then, the marginal change in cost, when x = 5, is

  • 200

  • 225

  • 206

  • 226


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