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

f(2 n)=an, f(2 n+)=an f(2 n-)=bn+1 ⇒ an-bn=1 f (2 n +1)=an f (2 n +1)- =an f (2 n+1)+ =bn+1-1 .⇒ an=b n+1 -1 or an-bn+1=-1 or 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
$f (x) = {a_{n} + sin π x, for x \in [2 n, 2 n + 1] b_{n} + cos π x, for x \in (2 n - 1, 2 n)$
for all integers $n$ .
If $f$ is continuous, then which of the following hold(s) for all $n$ ?

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

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

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

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

$f (2 n) = a_{n}, f (2 n^{+}) = a_{n}$
$f (2 n^{-}) = b_{n} + 1$
$\Rightarrow a_{n} - b_{n} = 1$
$f (2 n + 1) = a_{n}$
$f {(2 n + 1)^{-}} = a_{n}$
$f {(2 n + 1)^{+}} = b_{n + 1} - 1$
$\Rightarrow a_{n} = b_{{} n + 1} - 1$ or $a_{n} - b_{n + 1} = - 1$
or $a_{n - 1} - b_{n} = - 1$

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