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  • Let <img alt="f(x)=\frac{\sin 2 n x}{1+\cos ^2 n x}, n \in N" src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x%29%3D%5Cfrac%7B%5Csin%202%20n%20x%7D%7B1&plus;%5Ccos%20%5E2%20n%20x

Let f(x)=\frac{\sin 2 n x}{1+\cos ^2 n x}, n \in N has \frac{\pi }{6} as its fundamental period, then n is equal to 

 

Option: 1

2

 


Option: 2

4


Option: 3

6

 


Option: 4

8


Answers (1)

best_answer

\begin{aligned} & \because f(x)=\frac{\sin 2 n x}{1+\left(\frac{1+\cos 2 n x}{2}\right)}=\frac{2 \sin 2 n x}{3+\cos 2 n x} \\ & \therefore \quad T_1=\text { period of } \sin 2 n x=\frac{2 \pi}{2 n}=\frac{\pi}{n} \end{aligned}

and  T_2=\text { period of } \cos 2 n x=\frac{2 \pi}{2 n}=\frac{\pi}{n}

\begin{aligned} \therefore \text { Period of } f(x) & =\operatorname{LCM}\left\{T_1, T_2\right\} \\ \\& =\operatorname{LCM}\left\{\frac{\pi}{n}, \frac{\pi}{n}\right\} \\ \\\therefore & =\frac{\pi}{n}=\frac{\pi}{6} \text { (given) } \\ \\\therefore & n=6 \end{aligned}

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seema garhwal

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