A) \[f(x)\] is not periodic
B) \[\frac{1}{2}\]
C) 1
D) 2
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 \[[x+T]-[x]=T\] for all \[x\in R\] 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 \[f(x)=x-[x]\] has period 1.
You need to login to perform this action.
You will be redirected in
3 sec
