91Ó°ÊÓ

To find the range of \(f(x)=\min \cdot\\{\tan x, \cot x\\}\), we should compare the values of \(\tan x\) and \(\cot x\). Notice that: - If \(0 < x < \frac{\pi}{2}\), then \(\tan x > \cot x\) and the minimum value will be \(f(x) = \cot x\). - If \(-\frac{\pi}{2} < x < 0\), then \(\tan x < \cot x\) and the minimum value will be \(f(x) = \tan x\). Thus, if \(-\frac{\pi}{2} < x < \frac{\pi}{2}\), the function \(f(x)\) takes its minimum value from the functions \(\tan x\) over the interval \((-\frac{\pi}{2},0)\) and from \(\cot x\) over the interval \((0,\frac{\pi}{2})\). Since \(\tan x\) approaches \(+\infty\) as \(x\) approaches \(\frac{\pi}{2}\), and \(\cot x\) approaches \(+\infty\) as \(x\) approaches \(0^+\), we deduce that the range of \(f(x)\) is \((-\infty, +\infty)\). Therefore, statement (a) is false.

Considering the initial analysis of the range of the function, we can conclude that \(f(x)\) will inherit its periodic behavior from the functions \(\tan x\) and \(\cot x\). Since both \(\tan x\) and \(\cot x\) have a period of \(\pi\), the function \(f(x)\) will also have a period of \(\pi\). Therefore, statement (b) is true.

The points of discontinuity for the function \(f(x)\) will be the points where neither \(\tan x\) nor \(\cot x\) are defined. \(\tan x\) has vertical asymptotes and is undefined for \(x = \pm \frac{\pi}{2} + k\pi\) for integer values of \(k\). Similarly, \(\cot x\) has vertical asymptotes and is undefined for \(x = k\pi\) for integer values of \(k\). Therefore, the points of discontinuity for the function \(f(x)\) will be the points that satisfy both conditions, that is, \(0, \pm \frac{\pi}{2}, \pm \pi, \pm \frac{3\pi}{2}, \ldots\). Statement (c) is true.

To find the points of non-differentiability of \(f(x)\), we need to determine where the first derivative of \(f(x)\) does not exist. Since \(\tan x\) and \(\cot x\) have a complicated relationship, we can instead analyze their individual points of non-differentiability and deduce the non-differentiability for \(f(x)\). Recall that \(\tan x\) and \(\cot x\) are discontinuous at the same points, and these discontinuities must also be points of non-differentiability. Additionally, we need to check for any points where there is a "cusp" in the graph - these are points where the first derivative exists, but there is a sudden change in slope, causing non-differentiability. For \(-\frac{\pi}{2} < x < 0\), we are looking at the function \(\tan x\). The slope of \(\tan x\) continuously increases from \(-\infty\) to 1 as \(x\) ranges from \(-\frac{\pi}{2}\) to 0. For \(0 < x < \frac{\pi}{2}\), we are looking at the function \(\cot x\). The slope of \(\cot x\) continuously decreases from 1 to \(-\infty\) as \(x\) ranges from 0 to \(\frac{\pi}{2}\). Since the slope of \(\tan x\) at 0 is equal to the slope of \(\cot x\) also at 0, there are no additional "cusps" between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). As a result, the points of non-differentiability are the discontinuities: \(0, \pm \frac{\pi}{2}, \pm \pi, \pm \frac{3\pi}{2}, \ldots\). Statement (d) is false. To summarize, statements (b) and (c) are true while statements (a) and (d) are false.