If `f(x) {-x-pi/2,xle-pi/2 and -cosx,-pi/2 < xle0 and x-1,0< x le1 and lnx,x > 1` then-

If f(x) ={{:(-x=(pi)/(2),xle -(pi)/(2)), (- cos x, -(pi)/(2)lt x ,le 0):} ,( x-1, 0 lt x le 1),("in"x, x gt1):}

If F(x) = {(sin{cosx})/(x-pi/2),x!=pi/2 and 1,x=pi/2, where {.} represents the fractional part function, then lim_(xto pi//2)f(x) is

Find alpha and beta, so the function f (x) defined by f(x) =-2 sin x, "for" -pi le x le-pi/2 =alpha sinx+beta"for" -pi/2lt x lt pi/2 =cosx , "for" pi/2 le x le pi, is continuous on [-pi, pi].

tan^(-1)[(cosx)/(1+sinx)] is equal to pi/4-x/2,when (a) x in (-pi/2,(3pi)/2) (b) x in (-pi/2,pi/2) (c) x in (-pi/2,pi) (d) x in (-(3pi)/2,pi/2)

If f(x)={(sin{cosx}/(x-pi/2), x ne pi/2),(1, x=pi/2):} , where {.} denotes the fractional part of x, then f(x) is :