The height of the cone of maximum volume inscribed in a sphere of

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 Multiple Choice QuestionsMultiple Choice Questions

331.

The angle between the curves y = 4x + 4 and y2 = 36(9 - x) is

  • 30°

  • 45°

  • 60°

  • 90°


332.

If m and M respectively denote the minimum and maximum of f(x) = (x - 1)2 + 3 for x [- 3, 1], then the ordered pair (m, M) is equal to

  • (- 3, 19)

  • (3, 19)

  • (- 19, 3)

  • (- 19, - 3)


333.

The length of the subtangent at (2, 2) to thecurve x5 = 2y4 is

  • 52

  • 85

  • 25

  • 58


334.

There is an error of ± 0.04cm in the measurement of the diameter of a sphere. When radius is 10cm, the percentage error in the volume of the sphere is

  • ± 1.2

  • ± 1.0

  • ± 0.8

  • ± 0.6


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335.

The function fx = x3 +ax2 + bx +c, a2  3b has

  • one maximum value

  • one minimum value

  • no extreme value

  • one maximum and one minimum value


336.

The maximum value of log(x)x, 0 < x <  is

  • e

  • 1

  • e - 1


337.

z = tany +ax +y - ax  zxx - a2zyy =?

  • 0

  • 2

  • zx + zy = 0

  • zxzy


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338.

The height of the cone of maximum volume inscribed in a sphere of radius R is

  • R3

  • 2R3

  • 4R3

  • 4R3


C.

4R3

Let the height of the cone = h

and the radius of the cone = r

Given, radius of the sphere = R

Now, In OPB

 R2 = r2 + h - R2 r2 = R2  - h - R2         = R +h - RR -h + R r2 = h2R - hThe volume of the cone isV = 13πr2h V = 13πh2R - hh V = π34Rh2 - 3h3Differentiating with r to hdVdh = 0 π34Rh - 3h2 = 0 h4R - 3h = 0 h = 0, h = 4R3    Not possibleNow, d2Vdh2 = π34R - 6hd2Vdh2at h = 4R3 = π34R - 6 . 4R3                                 = π34R - 8R = - 4π3R  Negativeie, maximumHence, the height og the cone of maximum volume is 4R3

 


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339.

If the distance s travelled by a particle in time t is given by s = t- 2t + 5, then its acceleration is

  • 0

  • 1

  • 2

  • 3


340.

The length of the sub tangent at any point (x1, y1) on the curve y = 5x is

  • 5x1

  • y15x1

  • loge5

  • 1loge5


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