Let A = x ∈ ℝ: [x + 3] + [x + 4] ≤ 3 , B = \ x ∈ ℝ: 3 ^ x ( ∑ _ r =...

Given,

A = { x \in \mathbb{R}: \left[x + 3\right] + \left[x + 4\right] \le 3 } \text{\text{ and } } B = \left\{x \in \mathbb{R}: {3}^{x}{\left(\mathop{∑}_{r = 1}^{\infty}\dfrac{3}{{10}^{r}}\right)}^{x - 3}< {3}^{- 3x}\right\},$</p><p>Now solving$A$we get,</p><p>\left[x + 3\right] + \left[x + 4\right] \le 3$

$⇒ \left[x\right] + 3 + \left[x\right] + 4 \le 3$

$⇒ 2\left[x\right] \le - 4$

$⇒ \left[x\right] \le - 2$

$⇒ x < - 1 ⇒ A = \left(- \infty , - 1\right)$

Now solving $B$ we get,

${3}^{x}{\left(\mathop{∑}_{r = 1}^{\infty}\dfrac{3}{{10}^{r}}\right)}^{x - 3} < {3}^{- 3x}$

Now using infinite $G . P$ formula ${S}_{\infty} = \dfrac{a}{1 - r}$ we get,

$⇒ {3}^{x}{\left(\dfrac{3 \cdot \dfrac{1}{10}}{1 - \dfrac{1}{10}}\right)}^{x - 3} < {3}^{- 3x}$

$⇒ {3}^{x}{\left(\dfrac{1}{3}\right)}^{x - 3} < {3}^{- 3x}$

$⇒ {3}^{x - x + 3 + 3x} < 1$

$⇒ {3}^{3\left(x + 1\right)} < 1$

$⇒ 3\left(x + 1\right) < 0$

$⇒ x < - 1 ⇒ B = \left(- \infty , - 1\right)$

⇒ A = B