Determine the minimum value of the function $k \sqrt{a^{2}+x^{2}}-x \quad(k>1)$ without using derivatives.
I. Solution: It is sufficient to examine the expression for positive values, because if $xk \sqrt{a^{2}+x^{2}}-|x|=$ $k \sqrt{a^{2}+|x|^{2}}-|x|$, this is the value of the expression at $|x|$. Let $x>0$ and denote the smallest value of the expression $k \sqrt{a^{2}+x^{2}}-x$ by $l$, which is certainly positive. Then

$$
k \sqrt{a^{2}+x^{2}} \geq x+l>0
$$

and thus

$$
\begin{gathered}
k^{2} a^{2}+k^{2} x^{2} \geq x^{2}+2 l x+l^{2} \\
\left(k^{2}-1\right) x^{2}-2 l x+k^{2} a^{2}-l^{2} \geq 0
\end{gathered}
$$

and since $k^{2}-1>0$, we can divide by $k^{2}-1$

$$
x^{2}-2 \frac{l}{k^{2}-1} x+\frac{a^{2} k^{2}-l^{2}}{k^{2}-1}=\left(x-\frac{l}{k^{2}-1}\right)^{2}+\frac{\left(a^{2} k^{2}-l^{2}\right)\left(k^{2}-1\right)-l^{2}}{\left(k^{2}-1\right)^{2}} \geq 0
$$

Since here $l$ is the smallest value achieved by the function, this can only be true if

$$
\left(a^{2} k^{2}-l^{2}\right)\left(k^{2}-1\right)-l^{2}=k^{2}\left(a^{2} k^{2}-l^{2}-a^{2}\right)=0
$$

that is

$$
\begin{aligned}
l^{2} & =a^{2}\left(k^{2}-1\right) \\
l & =a \sqrt{k^{2}-1}
\end{aligned}
$$

and at the point where the function reaches this minimum, the expression (1) is also 0. There is only one such value of $x$:

$$
x=\frac{l}{k^{2}-1}=\frac{a}{\sqrt{k^{2}-1}}
$$

II. Solution: The problem can also be solved geometrically. It is sufficient to focus on positive $x$ values. In this case, $\sqrt{a^{2}+x^{2}}$ represents the hypotenuse of a right-angled triangle whose legs are of lengths $a$ and $x$. Let the angle opposite the $x$ side be $\alpha$, then $x=\frac{a}{\tan \alpha}$ and $\sqrt{a^{2}+x^{2}}=\frac{a}{\sin \alpha}$. It will be useful to find a trigonometric expression for $k$ as well. Since $k>1$, there is an acute angle $\varphi$ such that $k=\frac{1}{\cos \varphi}$ since $k \neq 1, \varphi>0$, then the function can be rewritten as:

$$
f(x)=\frac{1}{\cos \varphi} \cdot \frac{a}{\sin \alpha}-\frac{a}{\tan \alpha}=a\left(\frac{1}{\cos \varphi} \cdot \frac{1}{\sin \alpha}-\frac{\cos \alpha}{\sin \alpha}\right)=a\left(\frac{1-\cos \alpha \cos \varphi}{\sin \alpha \cos \varphi}\right)
$$

Subtract and add $a \cdot \tan \varphi$ to the function:

$$
\begin{gathered}
f(x)=a\left(\frac{1-\cos \alpha \cos \varphi}{\sin \alpha \cos \varphi}-\frac{\sin \varphi}{\cos \varphi}+\tan \varphi\right)= \\
=a\left(\frac{1-(\cos \alpha \cos \varphi+\sin \alpha \sin \varphi)}{\sin \alpha \cos \varphi}+\tan \varphi\right)=a\left(\frac{1-\cos (\alpha-\varphi)}{\sin \alpha \cos \varphi}+\tan \varphi\right)
\end{gathered}
$$

The value of the function will be minimal when the first term is 0, that is, when $\cos (\alpha-\varphi)=1$ and thus $\alpha=\varphi$, so $x=\frac{a}{\tan \varphi}=$ $\frac{a}{\sqrt{\frac{1}{\cos ^{2} \varphi}-1}}=\frac{a}{\sqrt{k^{2}-1}}$

and the minimum value is

$$
a \tan \varphi=a \sqrt{k^{2}-1}
$$