Question

For every integer $n$, let $a_n$ and $b_n$ be real numbers. Let function $f:IR\to IR$ be given by
$f\left(x\right)=\{\begin{array}{c}{a}_n+\sin \pi x,forx\in [2n,2n+1]\\ b_n+\cos \pi x,forx\in (2n-1,2n)\end{array}$$,forallintegersn.$
lf $f$ 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)

The correct options are
B $a_n-b_n=1$
D $a_{n-1}-b_n=-1$

Check the continuity at $x=2n$

$L.H.L=\underset{x\to 2n^{-}}{lim}\left[f\right(x\left)\right]$

$=\underset{h\to 0}{lim}\left[f\right(2n-h\left)\right]$

$=\underset{h\to 0}{lim}[b_n+\cos \pi (2n-h\left)\right]$

$=\underset{h\to 0}{lim}[b_n+\cos \pi h]$

$=b_n+1$

$R.H.L=\underset{x\to 2n^{+}}{lim}\left[f\right(x\left)\right]=\underset{h\to 0}{lim}\left[f\right(2n+h\left)\right]$

$=\underset{h\to 0}{lim}[a_n+\sin \pi (2n+h\left)\right]=a_n$

$f\left(2n\right)=a_n+\sin 2\pi n=a_n$

For continuity, $\underset{x\to 2n^{-}}{lim}f\left(x\right)=\underset{x\to 2n^{+}}{lim}f\left(x\right)=f\left(2n\right)$

So, $a_n=b_n+1\Rightarrow a_n-b_n=1$

Now, check for continuity at $x=2n+1$

$L.H.L.=\underset{h\to 0}{lim}(a_n+\sin \pi (2n+1-h\left)\right)=a_n$

$R.H.L.=\underset{h\to 0}{lim}(b_{n+1}+\cos (\pi (2n+1-h)\left)\right)=b_{n+1}-1$

$f(2n+1)=a_n$.

For continuity, $a_n=b_{n+1}-1\Rightarrow a_{n-1}=b_n-1$

Both, option B and option D are correct.