The period of $f(x) = 3\sin \left( {\dfrac{{\pi x}}{3}} \right) + 4\cos \left( {\dfrac{{\pi x}}{4}} \right)$ is
A. $6$
B. $24$
C. $8$
D. $2\pi $
Answer
257.7k+ views
Hint: We can observe the trigonometric function's modulus in this question. The sine and cotangent functions' modulus. We will therefore arrive at our final outcome by proceeding in this way while keeping the modulus function in mind.
Complete step by step solution:
In the question, the equation is given by:
$f(x) = 3\sin \left( {\dfrac{{\pi x}}{3}} \right) + 4\cos \left( {\dfrac{{\pi x}}{4}} \right)$
To calculate the period of the first term, as we know that the period of the sine function is $2\pi $, then we have:
$Period\,of\,\sin \left( {x \times \dfrac{\pi }{3}} \right) = 2\pi \div \dfrac{\pi }{3} \\$
$\Rightarrow Period\,of\,\sin \left( {\dfrac{{\pi x}}{3}} \right) = 2\pi \times \dfrac{3}{\pi } \\$
$\Rightarrow Period\,of\,\sin \left( {\dfrac{{\pi x}}{3}} \right) = 6 \\$
Now, to calculate the period of the second term, as we know that the period of the cosine function is $2\pi $, then:
$Period\,of\,\cos \left( {x \times \dfrac{\pi }{4}} \right) = 2\pi \div \dfrac{\pi }{4} \\$
$\Rightarrow Period\,of\,\cos \left( {\dfrac{{\pi x}}{4}} \right) = 2\pi \times \dfrac{4}{\pi } \\$
$\Rightarrow Period\,of\,\cos \left( {\dfrac{{\pi x}}{4}} \right) = 8 \\$
For the given function, we have to find period of $f(x)$ which is equal to $LCM$ of period of$\sin \left( {\dfrac{{\pi x}}{3}} \right)$and $\cos \left( {\dfrac{{\pi x}}{4}} \right)$, then we obtain:
$Period\,of\,f(x) = LCM\left( {\sin \left( {\dfrac{{\pi x}}{3}} \right),\cos \left( {\dfrac{{\pi x}}{4}} \right)} \right) \\$
$\Rightarrow Period\,of\,f(x) = LCM(6,8) \\$
The lowest common factor of $(6,8)$ is $24$, so:
$Period\,of\,f(x) = 24$
Thus, the correct option is :(B) $24$
Note: It should be noted that the graph of the function makes complete cycles between $0$and $2$ and each function has the period, $p = 2\pi /s$, according to the procedure for determining the period of a sine function $f(x) = \sin (xs)$. The trigonometric function is periodic is provided here, and it is as follows: - The sine function Cosine function for $2\pi $: Tangent function for$\pi $: Cosecant function over time: Secant function for $2\pi $; cotangent function for $\pi $.
Complete step by step solution:
In the question, the equation is given by:
$f(x) = 3\sin \left( {\dfrac{{\pi x}}{3}} \right) + 4\cos \left( {\dfrac{{\pi x}}{4}} \right)$
To calculate the period of the first term, as we know that the period of the sine function is $2\pi $, then we have:
$Period\,of\,\sin \left( {x \times \dfrac{\pi }{3}} \right) = 2\pi \div \dfrac{\pi }{3} \\$
$\Rightarrow Period\,of\,\sin \left( {\dfrac{{\pi x}}{3}} \right) = 2\pi \times \dfrac{3}{\pi } \\$
$\Rightarrow Period\,of\,\sin \left( {\dfrac{{\pi x}}{3}} \right) = 6 \\$
Now, to calculate the period of the second term, as we know that the period of the cosine function is $2\pi $, then:
$Period\,of\,\cos \left( {x \times \dfrac{\pi }{4}} \right) = 2\pi \div \dfrac{\pi }{4} \\$
$\Rightarrow Period\,of\,\cos \left( {\dfrac{{\pi x}}{4}} \right) = 2\pi \times \dfrac{4}{\pi } \\$
$\Rightarrow Period\,of\,\cos \left( {\dfrac{{\pi x}}{4}} \right) = 8 \\$
For the given function, we have to find period of $f(x)$ which is equal to $LCM$ of period of$\sin \left( {\dfrac{{\pi x}}{3}} \right)$and $\cos \left( {\dfrac{{\pi x}}{4}} \right)$, then we obtain:
$Period\,of\,f(x) = LCM\left( {\sin \left( {\dfrac{{\pi x}}{3}} \right),\cos \left( {\dfrac{{\pi x}}{4}} \right)} \right) \\$
$\Rightarrow Period\,of\,f(x) = LCM(6,8) \\$
The lowest common factor of $(6,8)$ is $24$, so:
$Period\,of\,f(x) = 24$
Thus, the correct option is :(B) $24$
Note: It should be noted that the graph of the function makes complete cycles between $0$and $2$ and each function has the period, $p = 2\pi /s$, according to the procedure for determining the period of a sine function $f(x) = \sin (xs)$. The trigonometric function is periodic is provided here, and it is as follows: - The sine function Cosine function for $2\pi $: Tangent function for$\pi $: Cosecant function over time: Secant function for $2\pi $; cotangent function for $\pi $.
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