Question

Cosine function whose period is 5 is
$A.\cos \left( {\dfrac{{7\pi x}}{5}} \right) \\
  B.\cos \left( {\dfrac{{2\pi x}}{5}} \right) \\
  C.\cos \left( {\dfrac{{3\pi x}}{5}} \right) \\
  D.\cos \left( {\dfrac{{4\pi x}}{5}} \right) \\$

Answer 1

Hint: A function that repeats itself after every constant interval of time is known as the
periodic function, and the concept of modelling the periodic function using trigonometric identities are known as modelling of periodic behaviour functions.
In this question, we need to find the cosine function whose time period is 5 for which we need to equate the terms associated with the cosine with the standard function as $\cos \theta = \cos \left( {\dfrac{{2\pi \omega }}{T}} \right)$ where, T is the time period of the corresponding cosine function.

Complete step by step answer:
Substitute T=5 in the standard cosine function as: $\cos \left( {\dfrac{{2\pi \omega }}{5}} \right)$
If we see the options, there we can find that instead of $\omega $, $x$ has been used which hardly puts any effect on the time period and consequently on our answer.
So, we can say that the cosine function having a time period of 5 is given as $\cos \left( {\dfrac{{2\pi x}}{5}} \right)$.

Hence the option B is correct.

Note: One complete pattern is known as a cycle. The horizontal length of the cycle is known as the period of the function. Trigonometric identities should be well known by the candidates to solve this type of question. Alternatively, we can also go through options eliminations as well such that:
Option A: Equating the given function with the standard function, we get:
$\cos \left( {\dfrac{{2\pi x}}{T}} \right) = \cos \left( {\dfrac{{7\pi x}}{5}} \right) \\
\dfrac{{2\pi x}}{T} = \dfrac{{7\pi x}}{5} \\
T = \dfrac{{10}}{7}$
Option B: Equating the given function with the standard function, we get:
$ \cos \left( {\dfrac{{2\pi x}}{T}} \right) = \cos \left( {\dfrac{{2\pi x}}{5}} \right) \\
\dfrac{{2\pi x}}{T} = \dfrac{{2\pi x}}{5} \\
T = \dfrac{{10}}{2} = 5 \\$
Option C: Equating the given function with the standard function, we get:
$\cos \left( {\dfrac{{2\pi x}}{T}} \right) = \cos \left( {\dfrac{{3\pi x}}{5}} \right) \\
\dfrac{{2\pi x}}{T} = \dfrac{{3\pi x}}{5} \\
T = \dfrac{{10}}{3} \\$
Option D: Equating the given function with the standard function, we get:
$\cos \left( {\dfrac{{2\pi x}}{T}} \right) = \cos \left( {\dfrac{{4\pi x}}{5}} \right) \\
\dfrac{{2\pi x}}{T} = \dfrac{{4\pi x}}{5} \\
T = \dfrac{{10}}{4} = \dfrac{5}{2} \\$