For every integer n, let an and bn be real numbers. Let function f: ℝ → ℝ be given by f(x) = an + sin(πx), for x ∈ [2n, 2n+1) and f(x) = bn + cos(πx), for x ∈ (2n-1, 2n), for all integers n. If f is continuous, then which of the following hold(s) for all n?

Check the continuity at x = 2n
L.H.L = limx→2n-[f(x)] = limh→0[f(2n-h)] = limh→0[bn + cos(π(2n-h))] = limh→0[bn + cos(πh)] = bn + 1
R.H.L = limx→2n+[f(x)] = limh→0[f(2n+h)] = limh→0[an + sin(π(2n+h))] = an
f(2n) = an + sin(2πn) = an
For continuity, limx→2n-f(x) = limx→2n+f(x) = f(2n)
So, an = bn + 1 ⇒ an - bn = 1
Now, check for continuity at x = 2n+1
L.H.L. = limh→0(an + sin(π(2n+1-h))) = an
R.H.L. = limh→0(bn+1 + cos(π(2n+1-h))) = bn+1
f(2n+1) = an
For continuity, an = bn+1 ⇒ an - bn+1 = 0
Both, option B and option D are incorrect. Only an - bn = 1 is correct.