\section*{Problem 6A - 161236A}

For each real number \(x\) with \(0 < x \leq \frac{\pi}{2}\), in a plane right-angled Cartesian coordinate system, a circle is drawn through the points \(P(0 ; 1)\), \(Q(x ; \cos x)\), and \(R(-x ; \cos (-x))\).

a) Provide a function \(f\) such that for each of these numbers \(x\), the mentioned circle has the radius \(r = f(x)\).

b) Calculate \(\lim _{x \rightarrow \infty} f(x)\), if this limit exists.

c) Determine the range of the function \(f\) with the set of all numbers \(x\) for which \(0 < x \leq \frac{\pi}{2}\) holds, as the domain.
}

With \(\cos (-x)=\cos (x)\), the following sketch is obtained:

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-2017.jpg?height=454&width=580&top_left_y=1252&top_left_x=732)

It holds that

\[
\begin{gathered}
(r-(1-\cos x))^{2}+x^{2}=r^{2} \\
r^{2}-2 r(1-\cos x)+(1-\cos x)^{2}+x^{2}=r^{2} \\
2 r(1-\cos x)=(1-\cos x)^{2}+x^{2} \\
r=\frac{1}{2}\left(1-\cos x+\frac{x^{2}}{1-\cos x}\right)
\end{gathered}
\]

Thus, the function sought in part a) of the problem is

\[
f(x)=\frac{1}{2}\left(1-\cos x+\frac{x^{2}}{1-\cos x}\right)
\]