For every integer $n$, let $a_{n}$ and $b_{n}$ be real numbers. Let function $f: I R \rightarrow I R$ be…
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 $n$ ?
$a_{n-1}-b_{n-1}=0$
$a_{n}-b_{n}=1$
$a_{n}-b_{n+1}=1$
$a_{n-1}-b_{n}=-1$
Solution
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.