f(x) = _{[n][infinity]}(sinx)²ⁿ,

 

f(x) = _{[n][infinity]} (sin²x)ⁿ

 

Now analyze sin²x at x = m(pi)/2    which is 1.

Hence (sin²x)ⁿ at x =m(pi)/2 is 1 as n tends to infinity.

 

As x deviates from m(pi)/2 a little the sin²x < 1

Hence _{[x][infinity ]} (sin²x)ⁿ = 0

 

Since the limits are not equal at x = m(pi)/2

 

Hence the function is discontinuou at x = pi/2, -pi/2, m(pi/2)  m is an integer

 

Hence b,c,d are correct options.