Functions Question 164
The period of $ f(x)=x^2-
[x] $ , if it is periodic, is [AMU 2000]
Options:
A) $ f(x) $ is not periodic
B) $ \frac{1}{2} $
1
2
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ f(x) $ be periodic with period T. Then, $ f(x+T)=f(x) $ for all $ x\in R $
Þ $ x+T-[x+T]=x-[x] $ , for all $ x\in R $
Þ $ x+T-x=[x+T]-[x] $
Þ $ \lfloor x+T \rfloor - \lfloor x \rfloor = T $ for all $ x\in \mathbb{R} $
Þ $ =\underset{x\to 0}{\mathop{\lim }}\frac{{e^{\alpha x}}-1}{x}-\underset{x\to 0}{\mathop{\lim }}\frac{{e^{\beta x}}-1}{x} $ The smallest value of T satisfying $ f(x+T)=f(x) $ for all $ x\in R $ is 1. Hence $ f(x)=x-\lfloor x \rfloor $ has period 1.
BETA
