For every integer n,let a_n and b_n be real n | vedclass

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(A) For $f$ to be continuous at $x = 2n$,the left-hand limit and right-hand limit must be equal to $f(2n)$.$f(2n) = a_n + \sin(2n\pi) = a_n$.$f(2n^+)

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$(B, D)$

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(A) For $f$ to be continuous at $x = 2n$,the left-hand limit and right-hand limit must be equal to $f(2n)$.$f(2n) = a_n + \sin(2n\pi) = a_n$.$f(2n^+) = a_n + \sin(2n\pi) = a_n$.$f(2n^-) = b_n + \cos(2n\pi) = b_n + 1$.Equating $f(2n^+) = f(2n^-)$,we get $a_n = b_n + 1$,which implies $a_n - b_n = 1$. Thus,$(B)$ is correct.For $f$ to be continuous at $x = 2n+1$,the left-hand limit and right-hand limit must be equal to $f(2n+1)$.$f(2n+1) = a_n + \sin((2n+1)\pi) = a_n$.$f((2n+1)^-) = a_n + \sin((2n+1)\pi) = a_n$.$f((2n+1)^+) = b_{n+1} + \cos((2n+1)\pi) = b_{n+1} - 1$.Equating $f((2n+1)^-) = f((2n+1)^+)$,we get $a_n = b_{n+1} - 1$,which implies $a_n - b_{n+1} = -1$. Replacing $n$ with $n-1$,we get $a_{n-1} - b_n = -1$. Thus,$(D)$ is correct.

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