Question

The period of the function \[f\left( x \right) = \cos e{c^2}\left( {3x} \right) + \cot \left( {4x} \right)\] is
\[\left( a \right)\dfrac{\pi }{3}\]
\[\left( b \right)\dfrac{\pi }{4}\]
\[\left( c \right)\dfrac{\pi }{6}\]
\[\left( d \right)\pi \]

Answer 1

Hint: We solve this problem by finding the period of \[\cos e{c^2}\left( {3x} \right)\] and \[\cot \left( {4x} \right)\]then by taking the least common multiple of their respective periods. Period of \[\cos ec\left( x \right)\]and \[\cot \left( x \right)\]should be kept in mind while doing the simplifications. We will first find the periods of the functions individually and then find their LCM.

Complete step-by-step answer:
A function is periodic if and only if \[f\left( {x + T} \right) = f\left( x \right)\]; where T is the time period of the function, that is, a constant.
The given function \[f\left( x \right) = \cos e{c^2}\left( {3x} \right) + \cot \left( {4x} \right)\] is periodic if
\[f\left( {x + T} \right) = \cos e{c^2}\left( {3x + T} \right) + \cot \left( {4x + T} \right) = f\left( x \right)\]
We know that the period of\[\cot \left( x \right)\]is \[\pi \].
If Period of \[f\left( x \right) = T\], then period of \[f\left( {ax} \right) = \dfrac{T}{a}\]
This implies that period of \[\cot \left( {4x} \right)\] is \[\dfrac{\pi }{4}\]
Similarly, \[\cos ec\left( x \right)\] is \[2\pi \].
In general, the odd powers of \[\cos ec\left( x \right)\] have a period of \[2\pi \] and the even ones have a period of \[\pi \],
This implies that \[\cos e{c^2}\left( x \right)\] has a period of \[\pi \].
If Period of \[f\left( x \right) = T\], then period of \[f\left( {ax} \right) = \dfrac{T}{a}\]
This implies that the period of \[\cos e{c^2}\left( {3x} \right)\] is \[\dfrac{\pi }{3}\].
Now, \[f\left( x \right) = \cos e{c^2}\left( {3x} \right) + \cot \left( {4x} \right)\] has a period of least common multiple of \[\dfrac{\pi }{4}\] and \[\dfrac{\pi }{3}\]
L.C.M of a fraction = L.C.M of numerator/H.C.F of denominator
So,
L.C.M \[\left[ {\dfrac{\pi }{3}and\dfrac{\pi }{4}\;} \right]\] = \[\dfrac{{LCM\left[ {\pi ,\pi } \right]}}{{HCF\left[ {3,4} \right]}} = \dfrac{\pi }{1} = \pi \]
Therefore, the period of \[f\left( x \right)\] is \[\pi \].
Hence, option (d) is the correct answer.
So, the correct answer is “Option d”.

Note: Such questions require grip over the trigonometrical concepts. One must know the period of the trigonometric functions and should remember that if Period of \[f\left( x \right) = T\], then period of \[f\left( {ax} \right) = \dfrac{T}{a}\]and this should also be kept in mind that L.C.M of a fraction = L.C.M of numerator/H.C.F of denominator. One should be careful while doing the calculations. Care should be taken while handling the calculations.