A population p(t) of 1000 bacteria introduced into nutrient mediu

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 Multiple Choice QuestionsMultiple Choice Questions

661.

The slope of the tangent to the curve x = 3t2 + 1, y = t3 - 1 at x = 1 is

  • 12

  • 0

  • - 2


662.

The rate ofchange of the surface area ofthe sphere of radius r when the radius is increasing at the rate of 2 cm/sec is proportional to

  • 1r2

  • 1r

  • r2

  • r


663.

For the curve xy = c2, the subnormal at any point varies as

  • x3

  • x2

  • y3


664.

Let the function f : R R be defined by f(x) = 2x + cos(x), then f

  • has maximum at x = 0

  • has minimum at x = π

  • is an increasing function

  • is a decreasing


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665.

The equation to the tangent to the curve y = be- x/a at the point where it crosses the Y-axis is

  • ax + by = 1

  • xa - yb = 1

  • xa + yb = 1

  • ax - by = 1


666.

The height of the cylinder of maximum volume inscribed in a sphere of radius 'a' is

  • 3a2

  • 2a3

  • a3

  • 2a3


667.

The perimeter of a sector is P. The area of the sectoris maximum when its radius is

  • 1P

  • P2

  • P4

  • P


668.

The function f(x) = x3 - 3x is

  • increasing on - , - 1  [1, ) and decreasing on (- 1, 1)

  • decreasing on - , - 1  [1, ) and increasing on (- 1, 1)

  • increasing on (0, ) and decreasing on (- , 0)

  • decreasing on (0, ) and increasing on (- , 0)


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669.

A population p(t) of 1000 bacteria introduced into nutrient medium grows according to the relation p(t) = 1000 + 1000t100 +t2. The maximum 100+t size of this bacterial population is

  • 1100

  • 1250

  • 1050

  • 5250


C.

1050

 Pt = 1000 + 1000t100 +t2        ...iOn differentiating both sides w.r.t., we getp't = 0 + 100 + t21000 - 1000t2t100 + t22       = 1000100 - t2100 + t22       ...iiPut p'(t) = 0 for maxima or minima 100 - t2 = 0  t = ± 10Now, again differentiating Eq. (ii) w.r.t. x, we getp''t = 1000100 + t22- 2t - 100 - t22100 + t22t100 + t24        = 1000t100 + t2- 2 - 100 - t24100 + t23       = - 1000t600 - 2t2100 + t23

At t = 10, p''t < 0Thus the maximum value isp10 = 1000 + 10000100 + 100 = 1000 + 10000200    using Eq. (i)          = 1000 + 50 = 1050


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670.

If ST and SN are the lengths of the subtangent and the subnormal at the point θ = π2 on the curve x = aθ + sinθ, y = a1 - cosθa  1, then

  • ST = SN

  • ST = 2SN

  • ST2 = aSN3

  • ST3 = aSN


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