The period of the function f(x)=sinx+tanx2+sinx22+tanx23+...+sinx2n−1+tanx2n

f(x)=(sinx+tanx2)+(sinx22+tanx23)+...+(sinx2n−1+tanx2n)=f1(x)f2(x)+...+fn−1(x),f1(x)=sinx+tanx2,f2(x)=sinx22+tanx23,...,fn−1(x)=sinx2n−1+tanx2h,.., f1(x)=sinxtanx2⇒periodof.f1(x)=2πf2=(sinx22+tanx23⇒f2(x)=23π....................................................................fn−1(x)sinx2n−1+tanπ2n ⇒fn−1(x)=LCMof(2π12n−1andππ2π)=2nπf(x)=LCM(2π,23π,....,2nπ)=2nπ

The period of the function f(x)=sinx+tanx2+sinx22+tanx23+...+sinx2n−1+tanx2n

f(x)=(sinx+tanx2)+(sinx22+tanx23)+...+(sinx2n−1+tanx2n)=f1(x)f2(x)+...+fn−1(x),f1(x)=sinx+tanx2,f2(x)=sinx22+tanx23,...,fn−1(x)=sinx2n−1+tanx2h,.., f1(x)=sinxtanx2⇒periodof.f1(x)=2πf2=(sinx22+tanx23⇒f2(x)=23π....................................................................fn−1(x)sinx2n−1+tanπ2n ⇒fn−1(x)=LCMof(2π12n−1andππ2π)=2nπf(x)=LCM(2π,23π,....,2nπ)=2nπ

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The period of the function $f (x) = \sin x + \tan \frac{x}{2} + \sin \frac{x}{2^{2}} + \tan \frac{x}{2^{3}} + ... + \sin \frac{x}{2^{n - 1}} + \tan \frac{x}{2^{n}}$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$2^{n} π$

$2^{n - 1} π$

$2^{n + 1} π$

$2 π$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (x) = (\sin x + \tan \frac{x}{2}) + (\sin \frac{x}{2^{2}} + \tan \frac{x}{2^{3}}) + ... + (\sin \frac{x}{2^{n - 1}} + \tan \frac{x}{2^{n}}) \\ = f 1 (x) f 2 (x) + ... + f n - 1 (x), \\ f 1 (x) = \sin x + \tan \frac{x}{2}, f 2 (x) = \sin \frac{x}{2^{2}} + \tan \frac{x}{2^{3}}, ..., f_{n - 1} (x) = \sin \frac{x}{2^{n - 1}} + \tan \frac{x}{2^{h}}, .., \end{array} \begin{array}{l} f 1 (x) = \sin x \tan \frac{x}{2} \Rightarrow p e r i o d o f . f 1 (x) = 2 π \\ f 2 = (\sin \frac{x}{2^{2}} + \tan \frac{x}{2^{3}} \Rightarrow f 2 (x) = 2^{3} π \\ .................................................................... \\ f n - 1 (x) \sin \frac{x}{2^{n - 1}} + \tan \frac{π}{2^{n}} \end{array} \begin{array}{l} \Rightarrow f n - 1 (x) = L C M o f (\frac{2 π}{\frac{1}{2 n - 1}} a n d \frac{π}{\frac{π}{2 π}}) = 2^{n} π \\ f (x) = L C M (2 π, 2^{3} π, ...., 2^{n} π) = 2^{n} π \end{array}$