For every integer \(n\), let \(a_{n}\) and \(b_{n}\) be real numbers. Let function \(f: I R \rightarrow I R\) be given by

\(f(x)=\left\{\begin{array}{lll}

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{array}\right.\)

for all integers \(n\). If \(f\) is continuous, then which of the following hold \((s)\) for all \(n\) ?

Given : \(f(x)=\left\{\begin{array}{ll}a_{n}+\sin \pi x, & x \in[2 n, 2 n+1] \\ b_{n}+\cos \pi x, & x \in(2 n-1,2 n)\end{array}\right.\)

\(\because f\) is continuous for all \(n\)

\(\therefore\) At \(x=2 n, \mathrm{LHL}=\mathrm{RHL}=f(2 n)\)

\(\Rightarrow b_{n}+\cos 2 \pi n=a_{n}+\sin 2 \pi n=a_{n}+\sin 2 \pi n\)

\(\Rightarrow b_{n}+1=a_{n} \Rightarrow a_{n}-b_{n}=1, \therefore\) option (b) is correct.

Also at \(x=2 n+1, \mathrm{LHL}=\mathrm{RHL}=f(2 n+1)\)

\(\Rightarrow \lim _{h \rightarrow 0} a_{n}+\sin \pi(2 n+1-h)\)

\(=\lim _{h \rightarrow 0} b_{n+1}+\cos \pi(2 n+1-h)=a_{n}+\sin (2 n+1) \pi\)

\(\Rightarrow a_{n}=b_{n+1}-1=a_{n} \Rightarrow a_{n}-b_{n+1}=-1\)

\(\therefore\) option (c) is incorrect.

\(\Rightarrow a_{n-1}-b_{n}=-1, \therefore\) option (d) is correct.