Let A = { x : [ x + 3 ] + [ x + 4 ] 3 } , B = { x : 3 x ( r = 1 3 10 r ) x - 3 < 3 - 3 x } , where [t] denotes greatest integer function. Then, [2023]

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Solution

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

$B = {x \in ℝ : 3^{x} {(\sum_{r = 1}^{\infty} \frac{3}{10^{r}})}^{x - 3} < 3^{- 3 x}},$

where [t] denotes greatest integer function. Then, [2023]

1 $A \cap B = ϕ$

2 $B \subset C, A \neq B$

3 $A \subset B, A \neq B$

4 $A = B$

Ans.
(4)

We have, $[x + 3] + [x + 4] \leq 3$

$\Rightarrow [x] + 3 + [x] + 4 \leq 3 \Rightarrow 2 [x] \leq - 4 \Rightarrow [x] \leq - 2$

$\Rightarrow x \in (- \infty, - 1)$

Now,
$3^{x} (\sum_{r = 1}^{\infty} {(\frac{3}{10^{r}}))}^{x - 3} < 3^{- 3 x} \Rightarrow 3^{x} {(\frac{1}{3})}^{x - 3} < 3^{- 3 x}$

$\Rightarrow 3^{x} \cdot 3^{3 - x} < 3^{- 3 x} \Rightarrow 3^{3} < 3^{- 3 x} \Rightarrow 3 < - 3 x \Rightarrow x < - 1$

$\Rightarrow x \in (- \infty, - 1)$ $∴ Both sets A and B are equal.$

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