Question

If the period of the function f(x)=\dfrac{\sin \left\\{ \left( \sin (nx) \right) \right\\}}{\tan \left( \dfrac{x}{n} \right)},n\in N , is $6\pi$ , then $n$ is equal to
(a) 3
(b) 2
(c) 1
(d) None of these

Answer 1

Use the period of $\sin x$ and $\tan x.$ So if there is function $f(ax)$ then its period is $\dfrac{T}{a}$ ,
For a function $f\left( \dfrac{x}{a} \right)$ the period is $aT$ . Use these properties. So find the period of f(x)=\dfrac{\sin \left\\{ \left( \sin (nx) \right) \right\\}}{\tan \left( \dfrac{x}{n} \right)} and put it equal to $6\pi$ you will get the value of $n$ .

Complete step-by-step answer:
In question it is given that the period of function f(x)=\dfrac{\sin \left\\{ \sin (nx) \right\\}}{\tan \left( \dfrac{x}{n} \right)} is $6\pi$ .
So we have to find the value of $n$ .
Now first period of function,
The period of a periodic function is the interval between two “matching” points on the graph. In other words, it's the distance along the $x$ -axis that the function has to travel before it starts to repeat its pattern. The basic sine and cosine functions have a period of $2\pi$ , while tangent has a period of $\pi$ .
The time interval between two waves is known as a Period whereas a function that repeats its values at regular intervals or periods is known as a Periodic Function. Or you can say, Periodic function is a function that repeats its values after every particular interval.
The period of the function is this particular interval mentioned above.
A function $f$ will be periodic with period $m$ , so if we have
$f(a+m)=f(a),$ for every $m>0$
It Shows that the function $f(a)$ possesses the same values after an interval of $m$ . One can say that after every interval of $m$ the function $f$ repeats all its values.
Now we know if there is a function $f(x)$ which has a period function of $T$ ,
$f(x+T)=f(x)$ where $T$ is period,
and $f(x+b)$ has period $T$ since we are adding in $x$ that will shift $x$ -axis .
e.g. $\sin (x+5)$ has period $\pi$ .
similar to this
$f(ax+b)=f\left( a(x+\dfrac{b}{a}) \right)=f(az)$ where $z=\left( x+\dfrac{b}{a} \right)$
if $a=1$ ,
then $f(az)=f(z)=\left( x+b \right)$ so the period is $T$ ,
but now we have $x=az$ so the period is $Ta$ .
So if $f(x)$ has period $T$ ,
So we get,
So if there is function $f(ax)$ then its period is $\dfrac{T}{a}$ ,
For function $f\left( \dfrac{x}{a} \right)$ the period is $aT$ ,
So we know the period function of $\sin \theta$ ,
So $\sin \theta$ has period of $2\pi$ ,
So we are given $\sin (nx)$ ,
So the period function for $\sin (\sin (nx))$ is $\dfrac{2\pi }{n}$ ,
And we know period of $\tan x$ ,
So period of $\tan x$ is $\pi$ ,
So we get the period of $\tan \left( \dfrac{x}{n} \right)$ as $\pi n$ ,
So if there is function $\dfrac{p(x)}{g(x)}$ , and if period of $p(x)$ is $\dfrac{a}{b}$ and the period of $g(x)$ is $\dfrac{c}{d}$ ,
So we write period of $\dfrac{p(x)}{g(x)}$ as ratio of L.C.M of $(a,c)$ to H.C.F of $(b,d)$ ,
So now applying the above to the problem, we get,
So we want to find the \dfrac{\sin \left\\{ \sin (nx) \right\\}}{\tan \left( \dfrac{x}{n} \right)} ,
So let p(x)=\sin \left\\{ \sin (nx) \right\\} and $g(x)=\tan \left( \dfrac{x}{n} \right)$ ,
So period of $p(x)\to \dfrac{2\pi }{n}$ and $g(x)\to \dfrac{n\pi }{1}$ ,
So overall period of f(x)=\dfrac{\sin \left\\{ \sin (nx) \right\\}}{\tan \left( \dfrac{x}{n} \right)}, We get,
So period of $f(x)=\dfrac{LCMof(2\pi ,n\pi )}{HCFof(n,1)}$
So period of $f(x)=6\pi$ ………….. (Given in question)
So we get,
$6\pi =\dfrac{2n\pi }{1}$
So simplifying we get the value of $n$ as $3$ .
Hence, option (a) is correct.

Note: So you should know that the period of all trigonometric identities such as $\sin x,\tan x$ etc.
You should know how we have to calculate the period of function. Don’t jumble yourself while taking LCM and HCF. So if there is function $\dfrac{p(x)}{g(x)}$ , and if period of $p(x)$ is $\dfrac{a}{b}$ and the period of $g(x)$ is $\dfrac{c}{d}$ , So we write period of $\dfrac{p(x)}{g(x)}$ as ratio of L.C.M of $(a,c)$ to H.C.F of $(b,d)$ .