Question

For every integer $n$, let $a_{n}$ and $b_{n}$ be real numbers. Let function $f: IR \rightarrow$ IR be given by $f(x) = \begin{cases} a_n + sin \pi x, & \text{for} x \in [2n, 2n+1] \\[2ex] b_n, & \text{for} x \in (2n-1, 2n) \end{cases}$, 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 $x=2 n$
L.H.L. $=\displaystyle\lim _{h \rightarrow 0}\left(b_{n}+\cos \pi(2 n-h)\right)=b_{n}+1$
R.H.L. $=\displaystyle\lim _{h \rightarrow 0}\left(a_{n}+\sin \pi(2 n+h)\right)=a_{n}$
$f(2 n)= a _{ n }$
For continuity $b_{n}+1=a_{n}$
At $x=2 n+1$
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
$a_{n}=b_{n+1}-1$
$a_{n-1}-b_{n}=-1$