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.