1)

For every integer n, let  $a_{n}$  and $b_{n}$ be real numbers. Let  function $f:R \rightarrow R$ be given by 

$f(x)=\begin{cases}a_{n}+\sin \pi x & for x\epsilon[2n,2n+1] \\b_{n}+\cos \pi x & for x \epsilon (2n-1,2n)\end{cases}$

for all integers n, 

If f is continuous ,  then which of the following hold9s) 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:

Option B,D

Explanation:

$f(2n)=a_{n},f(2n^{+})=a_{n}$

 $f(2n^{-})=b_{n}+1$

$ \Rightarrow$   $a_{n}-b_{n}=1$

 $f(2n+1)=a_{n}$

 $f((2n+1)^{-})=a_{n}$

    $f((2n+1)^{+})=b_{n+1}-1$

$\Rightarrow$   $a_{n}=b_{n+1}-1$ or $a_{n}-b_{n+1}=-1$

 or     $a_{n-1}-b_{n}=-1$