For every integer n, let an and bn be real numbers. Let function f: R arrow R be given by f(x)= begincasesan+ sin π x text for x ∈[2 n, 2 n+1] bn+ cos π x text for x ∈(2 n-1,2 n) 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: R arrow R be given by f(x)= begincasesan+ sin π x text for x ∈[2 n, 2 n+1] bn+ cos π x text for x ∈(2 n-1,2 n) 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, we have the following: 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 Now, f(2 n)=an . For continuity, bn+1=an . At x =2 n +1, we have 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

Q. For every integer $n$ , let $a_{n}$ and $b_{n}$ be real numbers. Let function
$f : R \to R$ be given by
$f (x) = {a_{n} + sin π x b_{n} + cos π x for x \in [2 n, 2 n + 1] for x \in (2 n - 1, 2 n)$ ,
for all integers $n$ . If $f$ is continuous, then which of the following hold(s) for all $n$ ?

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

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

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

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

Q. For every integer n, let an​ and bn​ be real numbers. Let function f:R→R be given by f(x)={an​+sinπxbn​+cosπx​ for x∈[2n,2n+1] for x∈(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

Q. For every integer $n$ , let $a_{n}$ and $b_{n}$ be real numbers. Let function
$f : R \to R$ be given by
$f (x) = {a_{n} + sin π x b_{n} + cos π x for x \in [2 n, 2 n + 1] for x \in (2 n - 1, 2 n)$ ,
for all integers $n$ . If $f$ is continuous, then which of the following hold(s) for all $n$ ?

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

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

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

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