The three sides of a trapezium are equal, each being 8cm. The area of the trapezium, where it is maximum, is (a)`24sqrt(3)` `c m^2` (b) `48sqrt(3)c m^2` (c)72`sqrt(3)c m^2` (d) none of these

If A(1,p^2),B(0,1) and C(p ,0) are the coordinates of three points, then the value of p for which the area of triangle A B C is the minimum is (a) 1/(sqrt(3)) (b) -1/(sqrt(3)) (c) 1/(sqrt(2)) (d) none of these

If tan^(-1)((a+x)/a)+tan^(-1)((a-x)/a)=pi/6,then x^2= 2sqrt(3)a (b) sqrt(3)a (c) 2sqrt(3)a^2 (d) none of these

A rectangle of the greatest area is inscribed in a trapezium A B C D , one of whose non-parallel sides A B is perpendicular to the base, so that one of the rectangles die lies on the larger base of the trapezium. The base of trapezium are 6cm and 10 cm and A B is 8 cm long. Then the maximum area of the rectangle is 24c m^2 (b) 48c m^2 36c m^2 (d) none of these

The largest area of the trapezium inscribed in a semi-circle or radius R , if the lower base is on the diameter, is (a) (3sqrt(3))/4R^2 (b) (sqrt(3))/2R^2 (3sqrt(3))/8R^2 (d) R^2

Let f: R->R be given by f(x)=x^2-3 . Then, f^(-1) is given by (a)sqrt(x+3) (b) sqrt(x)+3 (c) x+sqrt(3) (d) none of these

The length of the chord of the parabola y^2=x which is bisected at the point (2, 1) is 2sqrt(3) (b) 4sqrt(3) (c) 3sqrt(2) (d) 2sqrt(5)

The solution for x of the equation int_(sqrt(2))^x(dt)/(tsqrt(t^2-1))=pi/2 is pi (b) (sqrt(3))/2 (c) 2sqrt(2) (d) none of these

The volume of the greatest cylinder which can be inscribed in a cone of height 30 cm and semi-vertical angle 30^0 is (a) 4000pi/(sqrt(3)) (b) 4000pi/3c m^3 (c) 4000pi/(sqrt(3)c m^3 (d) none of these

A rectangle is inscribed in an equilateral triangle of side length 2a units. The maximum area of this rectangle can be (a) sqrt(3)a^2 (b) (sqrt(3)a^2)/4 a^2 (d) (sqrt(3)a^2)/2

If the area enclosed between the curves y=kx^2 and x=ky^2 , where kgt0 , is 1 square unit. Then k is: (a) 1/sqrt(3) (b) sqrt(3)/2 (c) 2/sqrt(3) (d) sqrt(3)