The length of the longest size rectangle of maximum area that can

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 Multiple Choice QuestionsMultiple Choice Questions

561.

The sides of an equilateral triangle are increasing at the rate of 2 cm/s. The rate at which the area increases, when the side is 10 cm, is:

  • 3 sq cm/s

  • 10 sq cm/s

  • 103 sq cm/s

  • 103 sq cm/s


562.

The function f(x) = 1 - x3 - x5 is decreasing for :

  • 1  x  5

  • x  1

  • x  1

  • all values of x


563.

If PQ and PR are the two sides of a triangle, then the angle between them which gives maximum area of the triangle, is :

  • π

  • π3

  • π4

  • π2


564.

The function y = a(1 - cos(x)) is maximum when x is equal to

  • π

  • π2

  • - π2

  • π6


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565.

The maximum value of logxx is equal to :

  • 2e

  • 1e

  • e

  • 1


566.

If the volume of a sphere is increasing at a constant rate, then the rate at which its radius is increasing, is

  • a constant

  • proportional to the radius

  • inversely proportional to the radius

  • inversely proportional to the surface area


567.

The length of the subtangent to the curve x2 + xy + y2 = 7 at (1, - 3) is :

  • 3

  • 5

  • 35

  • 15


568.

Twenty two metres are available to fence a flower bed in the form of a circular sector. If the flower bed should have the greatest possible surface area, the radius of the circle must be :

  • 4 m

  • 3 m

  • 6 m

  • 5 m


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569.

The value of x for which the polynomial 2x3 - 9x2 + 12x + 4 is a decreasing function of x, is :

  • - 1 < x < 1

  • 0 < x < 2

  • x > 3

  • 1 < x < 2


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570.

The length of the longest size rectangle of maximum area that can be inscribed in a semicircle of radius 1, so that 2 vertices lie on the diameter, is :

  • 2

  • 2

  • 3

  • 23


A.

2

Let the length and breadth of the rectangle be l and b.

In OAB,      b = sinθand l = 2cosθ A = Area of rectangle        = lb = 2sinθcosθ        = sin2θ

On differentiating w.r.t. θ, we get

dA = 2cos2θ

Put dA = 0 for maxima or minima

 cos2θ = 0          θ = π4

               d2A2 = - 4sin2θd2A2θ = π4 < 0

  Function is maximum at θ = π4 Length of rectangle = 2cosθ = 2cosπ4                                    = 212 = 2


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