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 0 (b) `1/2` (c) 1 (d) `-1`

f(x)={(sin(cos x)-cos x)/((pi-2x)^(2)),x!=(pi)/(2);k,x=(pi)/(2) is continuous at x=(pi)/(2), then k is equal to 0 (b) (1)/(2)(c)1(d)-1

f(x)={(sin(cos x)-cos x)/((pi-2x)^(3)),quad if x+-(pi)/(2) and k,quad if x=(pi)/(2) is continuous at x=(pi)/(2), then k

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 (a) 0 (b) 1/2 (c) 1 (d) -1

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 (a) 0 (b) 1/2 (c) 1 (d) -1

If the function f(x)={((sqrt(2+cosx)-1)/(pi-x)^2 ,; x != pi), (k, ; x = pi):} is continuous at x = 1, then k equals:

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

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

Find the value of k, the function, f(x)={:{((1-sinx)/((pi-2x)^(2))", "x!=(pi)/(2)),(" k, "x=(pi)/(2)):} is continuous at x=(pi)/(2) .