The function `f(x)=sqrt3sin(2x) -cos(2x)+4` is one-one in the interval:
a) `[-pi/2,pi/2]`
b) `[-pi/4,pi/4]`
c) `[-pi/6,pi/3]`
d) `[-pi,pi]`

f(x) = cos x is strictly increasing in the interval : (a) ((pi)/(2),pi) (b) (pi,(3 pi)/(2)) (c) (0,(pi)/(2))

Show that f(x)=tan^(-1)(sin x+cos x) is a decreasing function on the interval on (pi/4,pi/2).

Let the function f'(x)=sin x be one- one and onto. Then a possible domain of f is O [0,2 pi] O [0,pi] O [-(pi)/(2),(pi)/(2)] O [-pi, pi]

Show that f(x)=tan^(-1)(sin x+cos x) is decreasing function on the interval ((pi)/(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)

f(x)=cos2x, interval [-(pi)/(4),(pi)/(4)]

Show that function f(x)=sin(2x+(pi)/(4)) is decreasing on ((3 pi)/(8),(5 pi)/(8))

Range of function f(x)=arccot(2x-x^(2)) is [(pi)/(4),(3 pi)/(3)](b)[0,(pi)/(4)][0,(pi)/(2)] (d) [(pi)/(4),pi]

The maximum value of the function f(x)=sin(x+(pi)/(6))+cos(x+(pi)/(6)) in the interval (0,(pi)/(2)) occurs at (pi)/(12) (b) (pi)/(6)(c)(pi)/(4)(d)(pi)/(3)