Question

Let $A=\{x∈R:[x+3]+[x+4]≤3\}$ and $B=âx∈R:3^x{\left(\underset{r=1}{\overset{∞}{∑}}\frac{3}{10^r}\right)}^{x−3}<3^{−3x}â$ then

• A. $A=B$
• B. $A\underset{î€=}{⊂}B$
• C. $B\underset{î€=}{⊂}A$
• D. $Aâˆ\copyright B=Ï•$

Answer 1

Answer: (A)

Solution: Let $x∈A,[x+3]+[x+4]≤3$
$⇒[x]+3+[x]+4≤3$
$⇒2[x]≤−4⇒[x]≤−2$
$⇒x∈(−∞,−1)$
$A=(−∞,−1)$
If $x∈B$ then $3^x{3}^{x−3}{\left(\underset{r=1}{\overset{∞}{∑}}\frac{1}{10^r}\right)}^{x−3}<3^{−3x}$
$⇒3^{2x−3}{\left(\frac{1/10}{1−1/10}\right)}^{x−3}<3^{−3x}$
$⇒3^{2x−3}{\left(3^{−2}\right)}^{x−3}<3^{−3x}$
$⇒3^3<3^{−3x}⇒3<−3x$
$\text{ so }x∈(−∞,−1)$
Hence $B=(−∞,−1)$. Thus $A=B$.