1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let f(x+p)=1+[2−3f(x)+3(f(x))2−(f(x))3]1/3,∀x∈R, where p>0. Then, f(x) is periodic with period

A
p
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2p
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4p
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2pWe have, f(x+p)=1+[2−3f(x)+3{f(x)}2−{f(x)}2]1/3⇒f(x+p)=1+[1+{1−f(x)}3]1/3⇒f(x+p)−1=[1−{f(x)−1}3]1/3⇒g(x+p)=[1−{g(x)}3]1/3 ...(1)where g(x)=f(x)−1 and g(x+p)=f(x+p)−1⇒g(x+2p)=[1−{g(x+p)}3]1/3 ...(2)⇒g(x+2p)=[1−{{1−{g(x)}3}1/3}3]1/3From (1) and (2), we have ⇒g(x+2p)=[1−{1−{g(x)}3}]1/3⇒g(x+2p)=[1−1+{g(x)}3]1/3⇒g(x+2p)=[{g(x)}3]1/3⇒g(x+2p)=g(x)which shows f(x+2p)−1=f(x)−1f(x+2p)=f(x)Hence, f(x) is periodic with period 2p.

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Why Do We Need to Manage Our Resources?
MATHEMATICS
Watch in App
Join BYJU'S Learning Program