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: (D)

$f(2 n)=a_{n}, f\left(2 n^{+}\right)=a_{n}$
$f\left(2 n^{-}\right)=b_{n}+1$
$\Rightarrow a_{n}-b_{n}=1$
$f (2 n +1)=a_{n}$
$f \left\{(2 n +1)^{-}\right\}=a_{n}$
$f\left\{(2 n+1)^{+}\right\}=b_{n+1}-1$
$\left.\Rightarrow a_{n}=b_{\{} n+1\right\}-1$ or $a_{n}-b_{n+1}=-1$
or $a_{n-1}-b_{n}=-1$