For every integer n, let an and bn be real numbers. Let function f: IR arrow IR be given by f(x) = begincases an + sin π x, textfor x ∈ [2n, 2n+1] [2ex] bn, textfor x ∈ (2n-1, 2n) endcases, for all integers n. If f is continuous, then which of the following hold(s) for all n ?

Question

For every integer n, let an and bn be real numbers. Let function f: IR arrow IR be given by f(x) = begincases an + sin π x, textfor x ∈ [2n, 2n+1] [2ex] bn, textfor x ∈ (2n-1, 2n) endcases, for all integers n. If f is continuous, then which of the following hold(s) for all n ? (A) an-1-bn-1=0 (B) an-bn=1 (C) an-bn+1=1 (D) an-1-bn=-1

Tardigrade Admin · Accepted Answer

At x=2 n L.H.L. = displaystyle lim h arrow 0(bn+ cos π(2 n-h))=bn+1 R.H.L. = displaystyle lim h arrow 0(an+ sin π(2 n+h))=an f(2 n)= a n For continuity bn+1=an At x=2 n+1 L.H.L = displaystyle lim h arrow 0(an+ sin π(2 n+1-h))=an R.H.L = displaystyle lim h arrow 0(bn+1+ cos (π(2 n+1-h)))=bn+1-1 f(2 n+1)=an For continuity an=bn+1-1 an-1-bn=-1

$a_{n - 1} - b_{n - 1} = 0$

$a_{n} - b_{n} = 1$

$a_{n} - b_{n + 1} = 1$

$a_{n - 1} - b_{n} = - 1$

Q. For every integer n, let an​ and bn​ be real numbers. Let function f:IR→ IR be given by f(x)=⎩⎨⎧​an​+sinπx,bn​,​forx∈[2n,2n+1]forx∈(2n−1,2n)​, for all integers n. If f is continuous, then which of the following hold(s) for all n ?

an−1​−bn−1​=0

an​−bn​=1

an​−bn+1​=1

an−1​−bn​=−1

$a_{n - 1} - b_{n - 1} = 0$

$a_{n} - b_{n} = 1$

$a_{n} - b_{n + 1} = 1$

$a_{n - 1} - b_{n} = - 1$