Let A={x∈R:[x+3]+[x+4]≤3}; B = \left\{x \in \mathbb{R} : 3^x \left( \sum_{r=3}^{\infty} \frac{3}{10^r} \right) < 3^{-3x} \right\}, where [t] denotes greatest integer function. Then
Explanation
A={x∈R:[x+3]+[x+4]≤3}
⇒2[x]+7≤3⇒2[x]≤−4
\Rightarrow [x] \le -2 \Rightarrow x < -1
B = \left\{x \in \mathbb{R} : 3^x \left( \sum_{r=1}^{\infty} \left( \frac{3}{10^r} \right)^{x-3} \right) < 3^{-3x} \right\}
\Rightarrow 3^{2x-3} \left( \frac{\frac{1}{10}}{1 - \frac{1}{10}} \right)^{x-3} < 3^{-3x}
\Rightarrow 3^{6-2x} < 3^{3-5x}
\Rightarrow 6 - 2x < 3 - 5x \Rightarrow x < -1
Therefore, A=B
Explanation
A={x∈R:[x+3]+[x+4]≤3}
⇒2[x]+7≤3⇒2[x]≤−4
\Rightarrow [x] \le -2 \Rightarrow x < -1
B = \left\{x \in \mathbb{R} : 3^x \left( \sum_{r=1}^{\infty} \left( \frac{3}{10^r} \right)^{x-3} \right) < 3^{-3x} \right\}
\Rightarrow 3^{2x-3} \left( \frac{\frac{1}{10}}{1 - \frac{1}{10}} \right)^{x-3} < 3^{-3x}
\Rightarrow 3^{6-2x} < 3^{3-5x}
\Rightarrow 6 - 2x < 3 - 5x \Rightarrow x < -1
Therefore, A=B
