Question

Let $A=\{x \in R :[x+3]+[x+4] \leq 3\}$ and $B=\left\{x \in R : 3^x\left(\displaystyle\sum_{r=1}^{\infty} \frac{3}{10^r}\right)^{x-3}< 3^{-3 x}\right\}$ then

• A. $A=B$
• B. $A \underset{\neq}{\subset} B$
• C. $B \underset{\neq}{\subset} A$
• D. $A \cap B=\phi$

Answer 1

Answer: (A)

Solution: Let $x \in A,[x+3]+[x+4] \leq 3$
$\Rightarrow {[x]+3+[x]+4 \leq 3} $
$\Rightarrow 2[x] \leq-4 \Rightarrow[x] \leq-2$
$\Rightarrow x \in(-\infty,-1) $
$ A=(-\infty,-1)$
If $x \in B$ then $3^x 3^{x-3}\left(\displaystyle\sum_{r=1}^{\infty} \frac{1}{10^r}\right)^{x-3}<3^{-3 x}$
$\Rightarrow 3^{2 x-3}\left(\frac{1 / 10}{1-1 / 10}\right)^{x-3}<3^{-3 x} $
$\Rightarrow 3^{2 x-3}\left(3^{-2}\right)^{x-3}<3^{-3 x}$
$\Rightarrow 3^3<3^{-3 x} \Rightarrow 3<-3 x $
$ \text { so } x \in(-\infty,-1)$
Hence $B=(-\infty,-1)$. Thus $A=B$.