The function `f(x)=tan^(-1)(sinx+cosx)` is an increasing function in `(-pi/2,pi/4)` (b) `(0,pi/2)` `(-pi/2,pi/2)` (d) `(pi/4,pi/2)`

Show that the function f given by f(x) = tan^(-1) (sin x+cosx ) , x gt 0 is always an increasing function in (0,(pi)/(4)) .

Range of tan^(-1)((2x)/(1+x^2)) is (a) [-pi/4,pi/4] (b) (-pi/2,pi/2) (c) (-pi/2,pi/4) (d) [pi/4,pi/2]

The range of the function f(x)=int_(-1)^1(sinxdt)/((1-2tcosx+t^2)i s (a) [-pi/2,pi/2] (b) [0,pi] (c) {0,pi} (d) {-pi/2,pi/2}

If f"(x)>0AAx in R ,f'(3)=0,a n df(x)=f("tan"hat2x-2"tan"x+4),0

Let f:(-1,1)rarrB be a function defined by f(x)=tan^(-1)[(2x)/(1-x^2)] . Then f is both one-one and onto when B is the interval. (a) [0,pi/2) (b) (0,pi/2) (c) (-pi/2,pi/2) (d) [-pi/2,pi/2]

If roots of the equation 2x^2-4x+2sintheta-1=0 are of opposite sign, then theta belongs to (pi/6,(5pi)/6) (b) (0,pi/6)uu((5pi)/6,pi) ((13pi)/6,(17pi)/6) (d) (0,pi)

One of the root equation cosx-x+1/2=0 lies in the interval (a) (0,pi/2) (b) (-pi/(2,0)) (c) (pi/2,pi) (d) (pi,(3pi)/2)

Prove that the function f given by f(x) = log sin x is increasing on (0,(pi)/(2)) and decreasing on ((pi)/(2),pi) .

Range of the function f(x)=sqrt(cos^(- 1)(sqrt(log_4x))-pi/2)+sin^(- 1)((1+x^2)/(4x)) is equal to (A) (0,pi/2+sqrt(pi/2)] (B) [pi/2,pi/2+sqrt(pi/2)] (C) [pi/6,pi/4) (D) {pi/6}

Show that the function given by f (x) = sin x is (a) increasing in (0,(pi)/(2)) (b) decreasing in ((pi)/(2) ,pi) (c) neither increasing nor decreasing in (0, pi )