For every integer $n$, let $a_n$ and $b_n$ be real numbers. Let function $f: \mathbb{R} \to \mathbb{R}$ be given by $f(x)=\begin{cases} a_n+\sin \pi x, & \text{for } x \in [2n,2n+1] \\ b_n+\cos \pi x, & \text{for } x \in (2n-1,2n) \end{cases}$ for all integers $n$. If $f$ is continuous, then which of the following hold(s) for all $n$ ?

For every integer n, let aₙ and bₙ be real numbers. Let function f: R → R be given by f(x)= aₙ+sin π x, for x ∈ [2n,2n+1] bₙ+cos π x, for x ∈ (2n-1,2n) for all integers n. If f is continuous, then which of the following hold(s) for all n ? $f(x) = \begin{cases} a_n+\sin\pi x, & x \in [2n, 2n+1] \\ b_n+\cos\pi x, & x \in (2n-1, 2n) \end{cases}$ $f \text{ is continuous on } \mathbb{R}$ $\text{To find: which relations hold for all } n$ $1.\ \text{find the break points (integers)}$ $2.\ \text{impose left value} = \text{right value there}$ $3.\ \text{compare with the four given relations}$ $\text{Break points: } x = 2n \text{ and } x = 2n+1$ $x = 2n:\quad \text{left} = b_n + \cos(2n\pi) = b_n + 1$ $\text{value} = a_n + \sin(2n\pi) = a_n$ $a_n = b_n + 1 \Rightarrow a_n - b_n = 1 \Rightarrow (2) \text{ holds}$ $x = 2n+1:\quad \text{value} = a_n + \sin(2n+1)\pi = a_n$ $\text{right} = b_{n+1} + \cos(2n+1)\pi = b_{n+1} - 1$ $\Rightarrow a_n - b_{n+1} = -1$ $(3)\ a_n - b_{n+1} = 1 \text{ is FALSE (we got } -1)$ $n \to n-1:\quad a_{n-1} - b_n = -1 \Rightarrow (4) \text{ holds}$ $\text{From (2): } a_{n-1} - b_{n-1} = 1 \neq 0 \Rightarrow (1) \text{ is false}$ $b_n = 2n,\ a_n = 2n+1:\ a_n - b_n = 1,\ a_{n-1} - b_n = -1\ \checkmark$ $\boxed{\text{Answer: options (2) and (4)}}$