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

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

We have, <br/><br/>$$ \begin{aligned} & A=\{x \in R:[x+3]+[x+4] \leq 3\} \\\\ & \text { Here, }[x+3]+[x+4] \leq 3 \\\\ & \Rightarrow [x]+3+[x]+4 \leq 3 \\\\ & (\because[x+n]=[x]+n, n \in I) \\\\ & \Rightarrow 2[x]+4 \leq 0 \Rightarrow[x] \leq-2 \\\\ & \Rightarrow x \in(-\infty,-1) \\\\ & A \equiv(-\infty,-1) ...........(i) \end{aligned} $$ <br/><br/>Also, <br/><br/>$$ B=\left\{x \in R: 3^x\left(\sum\limits_{r=1}^{\infty} \frac{3}{10^r}\right)^{x-3}<3^{-3 x}\right\} $$ <br/><br/>Here, <br/><br/>$$ \begin{aligned} & 3^x\left(\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+\ldots\right)^{x-3}<3^{-3 x} \\\\ \Rightarrow & 3^x\left(\frac{\frac{3}{10}}{1-\frac{1}{10}}\right)^{x-3}<3^{-3 x} \\\\ \Rightarrow & 3^x\left(\frac{1}{3}\right)^{x-3}<3^{-3 x} \end{aligned} $$ <br/><br/>$$ \begin{array}{ll} \Rightarrow & 3^{x-x+3}<3^{-3 x} \Rightarrow 3^3<3^{-3 x} \\\\ \Rightarrow & -3 x>3 \Rightarrow x<-1 \\\\ \Rightarrow & x \in(-\infty,-1) \\\\ \Rightarrow & B \equiv(-\infty,-1) ...........(ii) \end{array} $$ <br/><br/>From equations (i) and (ii), we get <br/><br/>$A=B$