Chapter 4.6, Problem 9E

To determine

To sketch: the period of the given function.

Explanation of Solution

Given information:

Given function is

  $y=\sec 2x$ .

  

Calculation:

The vertical asymptotes for $y=\sec x$ are at $x=-\left(\frac{\pi }{2}\right)$ and $x=\frac{\pi }{2}$ and its period is $2\pi$ .

So for the given equation, let us we find out the values of the two consecutive asymptotes.

On comparing we get,

  $\begin{array}{c}2x=-\frac{\pi }{2}\\ x=-\frac{\pi }{4}\end{array}$

Also,

  $\begin{array}{c}2x=\frac{\pi }{2}\\ x=\frac{\pi }{4}\end{array}$

Therefore, the two consecutive asymptotes occur at $x=-\left(\frac{\pi }{4}\right)$ and $x=\frac{\pi }{4}$

Next, we calculate the values of $y$ foe a few values of $x$ between these two asymptotes.

For example,

Substituting $x=-\frac{\pi }{8}$

We get,

  $\begin{array}{c}y=\sec \left(2\right)\left(-\frac{\pi }{8}\right)\\ y=\sec \left(-\frac{\pi }{4}\right)\\ y=\sqrt{2}\end{array}$

Similarly, we calculate other values of $y$ and write them down in tabular form.

$x$	$-\left(\frac{\pi }{4}\right)$	$-\left(\frac{\pi }{8}\right)$	0	$\left(\frac{\pi }{8}\right)$	$\left(\frac{\pi }{4}\right)$
$y$	Not defined	$\sqrt{2}$	$1$	$\sqrt{2}$	Not defined

Now using these values, let us plot the graph from $-\pi$ to $\pi$

  

The above graph represents the equation $y=\sec 2x$ resembles the graph given in option (e).

Thus, the correct option is (e) and the period of the given function is $\pi$ .