Rectangles are inscribed ina circle of radius r. The dimensions o

Subject

Mathematics

Class

JEE Class 12

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

1.

The values of x, y and z for the system of equations x + 2y + 3z = 6, 3x - 2y + z = 2 and 4x + 2y + z = 7 are respectively

  • 1, 1, 1

  • 1, 2, 3

  • 1, 3, 2

  • 2, 3, 1


2.

If the determinant  = 3- 2sin3θ- 78cos2θ- 11142 = 0, then the value of sinθ is

  • 13or 1

  • 12 or 32

  • 0 or 12

  • None of these


3.

The relation R in R defined by R = {(a, b): a  b3), is

  • reflexive

  • symmetric

  • transitive

  • None of these


4.

The value of 2tan-1csctan-1x - tancot-1x is

  • tan-1x

  • tan(x)

  • cot(x)

  • csc-1x


Advertisement
5.

Let f (x + y) = f(x) + f(y) for all x and y. If the function f(x) is continuous at x = 0, then f(x) is continuous

  • only at x = 0

  • at x  R - 0

  • for all x

  • None of these


6.

Let fx = x2sin1x, x  00,             x = 0. Then, f(x) is continuous but not differentiable at x = 0, if

  • n  0, 1

  • n  [1, )

  • n  - , 0

  • n = 0


7.

The altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is

  • r2

  • r3

  • 3r4

  • 4r3


8.

If in a ABCsin3A + sin3B + sin3C = 3sinAsinBsinC, then the value of determinant abcbcacab is equal to

  • 0

  • 1

  • 2

  • 3


Advertisement
9.

Let f(x) = x(x - 1)2, the point at which f(x) assumes maximum and minimum are respectively

  • 13, 1

  • 1, 13

  • 3, 1

  • None of these


Advertisement

10.

Rectangles are inscribed ina circle of radius r. The dimensions of the rectangle which has the maximum area, are

  • r, r

  • 2r, 2r

  • 2r, 2r

  • None of the above


C.

2r, 2r

Let ABCD be the rectangle inscribed in a circle of radius r.

Let  AB = x

and BC = y

Then, x2 + y2 = 4r2

 y = 4r2 - x2      ...iArea of rectangle,         A = xy            = x4r2 - x2Let    u = A2 = x24r2 - x2 dudx = 8r2x - 4x3Put dudx = 0 for maxima or minima     4x2r2 - x2 = 0                        x = 2rAlso,               d2udx2 = 8r2 - 12x2 d2udx2x = 2r = 8r2 - 24r2 < 0 u and A are maximum at x = 2rFrom Eq.(i),        y = 2r Dimensions of the rectangle are 2r, 2r


Advertisement
Advertisement