`f(x)={(cosx)/(pi/2-x), x!=pi/2 1, x=pi/2`

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

Let f(x) =(kcosx)/(pi-2x) if x!=pi/2 and f(x)=3 if x=pi/2 then find the value of k if lim_(x->pi/2) f(x)=f(pi/2)

If f(x)={((sin(cosx)-cosx)/((pi-2x)^2) ,, x!=pi/2),(k ,, x=pi/2):} is continuous at x=pi/2, then k is equal to

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 :

f (x) = {(cos x) / ((pi) / (2) -x), x! = (pi) / (2) 1, x = (pi) / (2)

If the function f(x) = (1-sinx)/(pi-2x)^2 , x!=pi/2 is continuous at x=pi/4 , then find f(pi/2) .

If f(x) = {((1-sinx)/(pi-2x), ",", x != pi/2),(lambda, ",", x = pi/2):} , be continuous at x = (pi)/2 , then the value of lambda is

If f(x) = {((1-sinzx)/(pi-2x), ",", x != pi/2),(lambda, ",", x = pi/2):} , be continuous at x = (pi)/2 , then the value of lambda is

f(x)={((5^(cosx)-1)/((pi)/(2)-x)",", x ne (pi)/(2)), (log 5"," , x =(pi)/(2)):} at x =(pi)/(2) is

f(x)={((5^(cosx)-1)/((pi)/(2)-x)",", x ne (pi)/(2)), (log 5"," , x =(pi)/(2)):} at x =(pi)/(2) is