The area of the circle `x^2+y^2=16`exterior to the parabola `y^2=6x`is(A) `4/3(4pi-sqrt(3))` (B) `4/3(4pi+sqrt(3))`(C) `4/3(8pi-sqrt(3))` (D) `4/3(8pi+sqrt(3))`

Prove that the area of the region bounded by circle x^2 +y^2 =16 and y^2 = 6x is 4/3 ( 4 pi + sqrt(3)) sq . Units

If 0lt=xlt=pi/3 then range of f(x)=sec(pi/6-x)+sec(pi/6+x) is (4/(sqrt(3)),oo) (b) (4/(sqrt(3)),oo) (0,4/(sqrt(3))) (d) (0,4/(sqrt(3)))

If 0lt=xlt=pi/3 then range of f(x)=sec(pi/6-x)+sec(pi/6+x) is (a) (4/(sqrt(3)),oo) (b) (4/(sqrt(3)),oo) (c) (0,4/(sqrt(3))) (d) (0,4/(sqrt(3)))

If 0lt=xlt=pi/3 then range of f(x)=sec(pi/6-x)+sec(pi/6+x) is (a)(4/(sqrt(3)),oo) (b) (4/(sqrt(3)),oo) (c)(0,4/(sqrt(3))) (d) (0,4/(sqrt(3)))

If the distance between the centers of two circles of unit radii is 1 the common area of the circles is 1) (2 pi)/(3)-sqrt(3) 2) (2 pi)/(3)+sqrt(3) 3) (2 pi)/(3)-(sqrt(3))/(2) 4) (2 pi)/(3)(sqrt(3))/(3)

The area ( in sq.units) of the region {(x,y):y^(2)>=2x and x^(2)+y^(2) =0} is :(A)pi-(4)/(3)(B)pi-(8)/(3)(C)pi-(4sqrt(2))/(3) (D) pi-(2sqrt(2))/(3)

The tangent to the graph of the function y=f(x) at the point with abscissae x=1, x=2, x=3 make angles pi/6,pi/3 and pi/4 respectively. The value of int_1^3f\'(x)f\'\'(x)dx+int_2^3f\'\'(x)dx is (A) (4-3sqrt(3))/3 (B) (4sqrt(3)-1)/(3sqrt(3)) (C) (4-3sqrt(3))/2 (D) (3sqrt(3)-1)/2

The area bounded by the curves x^(2)+y^(2)=1,x^(2)+y^(2)=4 and the pair of lines sqrt(3)x^(2)+sqrt(3)y^(2)=4xy ,in the first quadrant is (1)(pi)/(2)(2)(pi)/(6)(3)(pi)/(4)(4)(pi)/(3)

Circumference of the circle touching the lines 3x+4y-3=0,6x+8y-1=0 is (A) pi (B) (pi)/(3) (C) 6 pi (D) (pi)/(2)

The area bounded by the curves x^2+y^2=1,x^2+y^2=4 and the pair of lines sqrt3 x^2+sqrt3 y^2=4xy , in the first quadrant is (1) pi/2 (2) pi/6 (3) pi/4 (4) pi/3