## Task 4 - 221234

If $c$ is a positive real number, then $f$ denotes the function defined for all real $x \neq 0$ by $f(x)=\sin \frac{c}{x}$.

Given is now an arbitrary natural number $m>1$.

a) Determine (in dependence on $m$) all positive real numbers $c$ for which the function $f$ has exactly $m$ zeros in the interval $10 \leq x \leq 20$, including the numbers 10 and 20 themselves.

b) For each $c$ found in a), prove that $f$ has only finitely many zeros in the interval $20 \leq x < \infty$.

Furthermore, determine (in dependence on $m$ and for each $c$ found for the respective $m$) the largest zero of $f$.
a) For every positive real $c$, it holds: A positive number $x$ is a zero of $f$ if and only if there exists an integer $k$ such that

$$
\frac{c}{x}=k \pi
$$

Now, (1) is equivalent to

$$
x=\frac{c}{k \pi}
$$

Therefore, an integer $k$ leads to a zero $x$ with $10 \leq x \leq 20$ if and only if it satisfies $10 \leq \frac{c}{k \pi} \leq 20$. This is equivalent to

$$
\frac{c}{20 \pi} \leq k \leq \frac{c}{10 \pi}
$$

In particular, an integer $k_{1}$ or $k_{2}$ leads to 10 or 20 as a zero via (2) if and only if

$$
k_{1}=\frac{c}{10 \pi} \quad ; \quad k_{2}=\frac{c}{20 \pi}
$$

holds. For a positive real $c$, these two numbers are integers if and only if $\frac{c}{20 \pi}$ is an integer (4), and

$$
\frac{c}{10 \pi}=\frac{c}{20 \pi}+m-1
$$

holds. If (4) and (5) are satisfied for a positive real $c$, then $2 c=c+20(m-1) \pi$, so $c=20(m-1) \pi$.

Conversely, for this $c$, $\frac{c}{20 \pi}=m-1$ is indeed an integer, so (4) is satisfied, and (5) holds. Therefore, exactly the number $c=20(m-1) \pi$ has the property required in a).
b) For it, it further holds: An integer $k$ leads to a zero $x \geq 20$ via (2) if and only if $\frac{c}{k \pi} \geq 20$. This is equivalent to

$$
0 < k \leq \frac{c}{20 \pi}
$$

For any two zeros $x=\frac{c}{k \pi}, x^{\prime}=\frac{c}{k^{\prime} \pi}$ corresponding to a positive real $c$, $x < x^{\prime}$ holds if and only if $k > k^{\prime}$. Therefore, the largest zero of $f$ belongs to the smallest value of $k$ satisfying (6), i.e., to the value $k=1$. Thus, the largest zero sought in b) is

$$
x=\frac{c}{\pi}=20(m-1)
$$

Adapted from $[5]$