Question

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 hold(s) for all $n$ ?

• A. $a_{n-1}-b_{n-1}=0$
• B. $a_{n}-b_{n}=1$
• C. $a_{n}-b_{n+1}=1$
• D. $a_{n-1}-b_{n}=-1$

Answer 1

Answer: (B), (D)

At $x =2 n$, we have the following:
L.H.L $=\displaystyle\lim _{h \rightarrow 0}\left(b_{n}+\cos \pi(2 n-h)\right)=b_{n}+1$
R.H.L $=\displaystyle\lim _{h \rightarrow 0}\left(a_{n}+\sin \pi(2 n+h)\right)=a_{n}$
Now, $f(2 n)=a_{n} .$
For continuity, $b_{n}+1=a_{n} .$
At $x =2 n +1$, we have
L.H.L. $=\displaystyle\lim _{h \rightarrow 0}\left(a_{n}+\sin \pi(2 n+1-h)\right)=a_{n}$
R.H.L. $=\displaystyle\lim _{h \rightarrow 0}\left(b_{n+1}+\cos (\pi(2 n+1-h))\right)=b_{n+1}-1$
$f (2 n +1)=a_{n}$
For continuity :
$a_{n}=b_{n+1}-1$
$a_{n-1}-b_{n}=-1$