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)))`

Find the range of f(x)=sec((pi)/(4)cos^(2)x) where -oo

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))

sin(pi-6x)+sqrt3 sin(pi/2+6x)=sqrt3

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

Find the rang of f(x)=log_sqrt5(3+cos((3pi)/4+x)+cos((pi)/4+x)+cos((pi)/4-x)-cos((3pi)/4-x))

If int_(0)^(oo)e^(-x^(2))dx=(sqrt(pi))/(2), then int_(0)^(oo)e^(-ax^(2))dx where a>0 is: (A)(sqrt(pi))/(2) (B) (sqrt(pi))/(2a)(C)2(sqrt(pi))/(a) (D) (1)/(2)(sqrt((pi)/(a)))

Prove that (pi)/(6)

The vertices of a triangle are (0,0),(x,cos x) and (sin^(3)x,0), where0

If f(x)=sqrt(1+cos^(2)(x^(2))), then f'((sqrt(pi))/(2)) is (sqrt(pi))/(6)(b)-sqrt(pi/6)1/sqrt(6)(d)pi/sqrt(6)

If cosxdy/dx-y sin x=6x, (0ltxltx/2)and y(pi/3)=0 , then y(pi/6) is equal to (a) pi^(2) /(2sqrt 3) (b) -pi^(2) /(2sqrt 3) (c) -pi^(2) /(4sqrt 3) (d) (c) -pi^(2) /2