Chapter 13.1, Problem 39PPSE

Solution

To find: The graph of the function which doesn’t represent a periodic function shown below and choose the correct option.

  

The graph of the function $\text{II}\text{.}$ doesn’t represent a periodic function. So, option $\left(B\right)\text{ II only}$ is correct option.

Given:

The graph of the functions is shown below.

  

Concept used:

A periodic function is a function that repeats itself a pattern of outputs ( $y-$ values) at regular intervals where one complete pattern represents a cycle. The period of the function is the horizontal length- the distance along the $x-$ axis of one complete cycle.

Calculation:

Check the period of the graphs of the functions with the help of concept stated above.

1. From the graph of the function, it can be interpreted that the function repeats itself for $y-$ values between $x=-4\text{ to }x=1$ and $x=1\text{ to }x=6$ . Therefore, the given graph represents a periodic function whose period is the horizontal length- the distance along the $x-$ axis of one complete cycle whose period $\left(T_I\right)$ is evaluated as:

  $\begin{array}{c}{T}_I=\left(6\right)-\left(1\right)\\ =5\mathrm{....}\left(\text{Periodic of period }5\right)\end{array}$

2. From the graph of the function, it can be interpreted that the graph of the function doesn’t repeat itself for $y-$ values for any value of $\left(x\right)$ in any interval. Therefore, the given graph doesn’t represent a periodic function. Therefore, the period of the function is not defined. …. $\left(1\right)$
3. From the graph of the function, it can be interpreted that the function repeats itself for $y-$ values between $x=0\text{ to }x=2$ and $x=2\text{ to }x=4$ . Therefore, the given graph represents a periodic function whose period is the horizontal length- the distance along the $x-$ axis of one complete cycle whose period $\left(T_{III}\right)$ is evaluated as:

  $\begin{array}{c}{T}_{III}=\left(2\right)-\left(0\right)\\ =2\mathrm{....}\left(\text{Periodic of period }2\right)\end{array}$

Conclusion:

From the equation $\left(1\right)$ , the graph of the function $\text{II}\text{.}$ doesn’t represent a periodic function. So, option $\left(B\right)\text{ II only}$ is correct option.