Let A=x R:[x+3]+[x+4] 3, B=x R:3^x( _r=1^310^r)^x-3

Question

Let $A={x\in R:[x+3]+[x+4]\le 3},$

$B={x\in R:{3}^{x}{(\sum _{r=1}^{\infty}\frac{3}{{10}^{r}})}^{x-3}<{3}^{-3x}},$ where [t] denotes greatest integer function. Then,

Accepted Answer

We have,

$\begin{gathered} A={x\in R:[x+3]+[x+4]\le 3} \ [x+3]+[x+4]\le 3 \ \Rightarrow [x]+3+[x]+4\le 3 \ (\because [x+n]=[x]+n,n\in I) \ \Rightarrow 2[x]+4\le 0\Rightarrow [x]\le -2 \ \Rightarrow x\in (-\infty,-1) \ A\equiv (-\infty,-1)...........(i) \end{gathered}$

Also,

$B=x\in R:{3}^{x}{(\sum _{r=1}^{\infty}\frac{3}{{10}^{r}})}^{x-3}3\Rightarrow x<-1   \Rightarrow \& x\in (-\infty,-1)   \Rightarrow \& B\equiv (-\infty,-1)...........(ii)$

From equations (i) and (ii), we get

$A=B$

Solutions · 1