For every integer n, let $a_n$ and $b_n$ be real numbers. Let function $f: R \rightarrow R$ be given by 

$f(x)= \begin{cases}a_n+\sin \pi x & \text { for } x \in[2 n, 2 n+1] \\ b_n+\cos \pi x & \text { for } x \in(2 n-1,2 n)\end{cases}$

for all integers n. If f is continuous, then which of the following does not hold for all n ?

$a_{n-1}-b_{n-1}=0$

For the continuity of f at $x=2 n+1$, we must have

$\lim\limits_{x \rightarrow(2 n+1)^{-}} f(x)=\lim\limits_{x \rightarrow(2 n+1)^{+}} f(x)=f(2 n+1)$

$\Rightarrow \lim\limits_{x \rightarrow(2 n+1)^{-}} a_n+\sin \pi x=\lim\limits_{x \rightarrow(2 n+1)^{+}} b_{n+1}+\cos \pi x =a_n+\sin \pi(2 n+1)$

$\Rightarrow a_n+\sin (2 n+1) \pi =b_{n+1}+\cos (2 n+1) \pi =a_n+\sin (2 n+1) \pi$

$\Rightarrow a_n=b_{n+1}-1 \Rightarrow a_b-b_{n+1}=-1$

Replacing n by n - 1, we get

$a_{n-1}-b_n=-1$

So, options (a) and (d) are correct.

For the continuity of f at x = 2n, we must have

$\lim\limits_{x \rightarrow 2 n^{-}} f(x)=\lim\limits_{x \rightarrow 2 n^{+}} f(x)=f(2 n)$

$\Rightarrow \lim\limits_{x \rightarrow 2 n^{-}} b_n+\cos \pi x=\lim\limits_{x \rightarrow 2 n^{+}} a_n+\sin \pi x=a_n+\sin 2 n \pi$

$\Rightarrow b_n+\cos 2 n \pi=a_n+\sin 2 n \pi $

$\Rightarrow b_n+1=a_n \Rightarrow a_n-b_n=1$

So, option (c) is correct.

Hence, option (b) does not hold.