
The period of the function \[f(x) = {{\mathop{\rm \csc}\nolimits} ^2}3x + \cot 4x\] is:
A. \[\dfrac{\pi }{3}\]
B. \[\dfrac{\pi}{3}\]
C. \[\dfrac{\pi }{6}\]
D. \[\pi \]
Answer
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Hint: We will use the concept of periodic function to solve the problem which states that a period is the amount of time between two waves, whereas a periodic function is a function whose values recur at regular intervals or periods. A function that repeats its values after a specific interval is referred to as a periodic function. The periodic function is described in this manner. Then we will find the period of \[{{\mathop{\rm \csc}\nolimits} ^2}x\] and with its help we will find the period of \[{{\mathop{\rm \csc}\nolimits} ^2}3x\]. Similarly, we will find the period of \[\cot x\] and with its help we will find the period of \[\cot 4x\]. Then we will find the LCM of the numerators and HCF of the denominators of the period of \[{{\mathop{\rm \csc}\nolimits} ^2}3x\] and \[\cot 4x\]. Then we will divide the LCM of the numerator with the HCF of the denominator to get the answer.
Formula used:
The period of any function can be determined as:
The period of \[\cot x\] is \[\pi\].
Then the period of \[\cot mx\] is \[\dfrac{\pi}{m}\].
The period of \[\csc x\] is \[\pi\].
Then the period of \[\csc mx\] is \[\dfrac{\pi}{m}\].
Complete step by step solution:
Given the function \[f(x) = {{\mathop{\rm \csc}\nolimits} ^2}3x + \cot 4x\].
To find the period of the function \[f(x)\], we will find the period of \[\csc^{2} 3x\] and \[\cot 4x\].
We know that, the period of \[\csc mx\] is \[\dfrac{\pi}{m}\].
If \[m= 3\], then the period of \[\csc 3x\] is \[\dfrac{\pi}{3}\].
So the period of \[\csc^{2} 3x\] is \[\dfrac{\pi}{3}\].
Again we know that, the period of \[\cot x\] is \[\pi\].
If \[m= 4\], then the period of \[\cot 4x\] is \[\dfrac{\pi}{4}\].
Now we will divide the LCM of the numerators of \[(\dfrac{\pi }{3},\dfrac{\pi }{4})\] with the HCF of the denominators of \[(\dfrac{\pi }{3},\dfrac{\pi }{4})\]:
The HCF of 3 and 4 is 1.
The LCM of \[\pi\] and \[\pi\] is \[\pi\].
The LCM of \[\dfrac{\pi }{3}\] and \[\dfrac{\pi }{4}\] is \[\dfrac{\text{LCM of } \pi \text{ and } \pi}{\text{HCF of } 3 \text{ and } 4}\].
\[ = \dfrac{\pi}{1}\]
\[ = \pi \]
Hence option D is correct.
Note: Students can make mistakes in calculating the LCM & HCF as they might calculate the LCM & HCF of whole radians in spite of calculating the LCM of numerators and HCF of denominators.
Formula used:
The period of any function can be determined as:
The period of \[\cot x\] is \[\pi\].
Then the period of \[\cot mx\] is \[\dfrac{\pi}{m}\].
The period of \[\csc x\] is \[\pi\].
Then the period of \[\csc mx\] is \[\dfrac{\pi}{m}\].
Complete step by step solution:
Given the function \[f(x) = {{\mathop{\rm \csc}\nolimits} ^2}3x + \cot 4x\].
To find the period of the function \[f(x)\], we will find the period of \[\csc^{2} 3x\] and \[\cot 4x\].
We know that, the period of \[\csc mx\] is \[\dfrac{\pi}{m}\].
If \[m= 3\], then the period of \[\csc 3x\] is \[\dfrac{\pi}{3}\].
So the period of \[\csc^{2} 3x\] is \[\dfrac{\pi}{3}\].
Again we know that, the period of \[\cot x\] is \[\pi\].
If \[m= 4\], then the period of \[\cot 4x\] is \[\dfrac{\pi}{4}\].
Now we will divide the LCM of the numerators of \[(\dfrac{\pi }{3},\dfrac{\pi }{4})\] with the HCF of the denominators of \[(\dfrac{\pi }{3},\dfrac{\pi }{4})\]:
The HCF of 3 and 4 is 1.
The LCM of \[\pi\] and \[\pi\] is \[\pi\].
The LCM of \[\dfrac{\pi }{3}\] and \[\dfrac{\pi }{4}\] is \[\dfrac{\text{LCM of } \pi \text{ and } \pi}{\text{HCF of } 3 \text{ and } 4}\].
\[ = \dfrac{\pi}{1}\]
\[ = \pi \]
Hence option D is correct.
Note: Students can make mistakes in calculating the LCM & HCF as they might calculate the LCM & HCF of whole radians in spite of calculating the LCM of numerators and HCF of denominators.
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