## Task B-4.1.

Given is the function

$$
f: \mathbf{R} \rightarrow \mathbf{R}, \quad f(x)=\frac{1}{2}\left(a^{x}+a^{-x}\right)
$$

where $a$ is a positive real number different from 1. What is $f(p+t)+f(p-t)$ if $f(t)=20$ and $f(p)=25 ?$
## Solution.

By direct calculation, we obtain the following sequence of equalities:

$$
\begin{array}{cc}
f(p+t)+f(p-t)=\frac{1}{2}\left(a^{p+t}+a^{-p-t}\right)+\frac{1}{2}\left(a^{p-t}+a^{-p+t}\right)= & 1 \text { point } \\
=\frac{1}{2}\left(a^{p+t}+a^{-p-t}+a^{p-t}+a^{-p+t}\right)=\frac{1}{2}\left(a^{p}\left(a^{t}+a^{-t}\right)+a^{-p}\left(a^{-t}+a^{t}\right)\right)= & 2 \text { points } \\
=\frac{1}{2}\left(a^{t}+a^{-t}\right) \cdot\left(a^{p}+a^{-p}\right)= & 1 \text { point } \\
\frac{1}{2} \cdot 2 f(t) \cdot 2 f(p)=2 \cdot 20 \cdot 25=1000 . & 2 \text { points }
\end{array}
$$