The value of k so that the function f(x)={kcosxπ−2xx≠π23,x=π2 is continuous at x=π2 is
2
4
6
8
fπ2=limx→π2kcos2xπ-2x 00 form by L'Hospital rule 3= limx→π2-ksin2x-2=k2⇒k=6