If pi/2 < `alpha` < 3 pi/2 then the modulus argument of `(1+cos 2alpha)+i sin2alpha`

If f(x) is continuous at x = (pi)/2 , where f(x) = (sinx)^(1/(pi-2x)), "for" x != (pi)/2 , then f((pi)/(2)) =

If : f(x) {: ((1-sin x)/(pi-2x)", ... " x != pi//2 ),(=lambda ", ... " x = pi//2):} is continuous at x = pi//2 , then : lambda =

If f(x) is continuous at x=pi/2 , where f(x)=(3^(x- pi/2)-6^(x- pi/2))/(cos x) , for x != pi/2 , then f(pi/2)=

int_0^(pi/2)sin4xcotx dx is equal to -pi/2 (2) 0 (3) pi/2 (4) pi

int_0^pi cos2xlogsinxdx= (A) pi (B) -pi/2 (C) pi/2 (D) none of these

Period of sin^2theta is- pi^2 b. pi/2 c. 2pi d. pi

f(x)=(s i x)/(sqrt(1+tan^2x))+(cos x)/(sqrt(1+cot^2x)) is constant in which of following interval a. \ (0,\ pi/2) b. (pi/2,\ pi) c. (pi,\ ((3pi)/2) d. ((3pi)/2,2pi)

Find the value of expression ((cos pi)/(2)+i sin((pi)/(2)))(cos((pi)/(2^(2)))+i sin((pi)/(2^(2))))......oo

If f(x)=sin x-x, then int_(-2 pi)^(2 pi)|f^(-1)(x)|dx= (A) pi^(2)(B)2 pi^(2)(C)3 pi^(2)(D)4 pi^(2)

The largest interval lying in (-pi/2,pi/2) for which the function [f(x)=4^-x^2+cos^(-1)(x/2-1)+log(cosx)] is defined, is (1) [0,pi] (2) (-pi/2,pi/2) (3) [-pi/4,pi/2) (4) [0,pi/2)