Let $\mathrm{A}=\{\mathrm{x} \in \mathbb{R}:[\mathrm{x}+3]+[\mathrm{x}+4] \leq 3\}$, $B=\left\{x \in \mathbb{R}: 3^{x}\left(\sum_{r=1}^{\infty} \frac{3}{10^{r}}\right)^{x-3}<3^{-3 x}\right\}$, where $[t]$ denotes greatest integer function. Then,

(A) $\mathrm{A} \cap \mathrm{B}=\phi$

(C) $\mathrm{B} \subset \mathrm{C}, \mathrm{A} \neq \mathrm{B}$

(D) $\mathrm{A} \subset \mathrm{B}, \mathrm{A} \neq \mathrm{B}$

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Sets Relations and Functions

Let

A = {x \in R : [x + 3] + [x + 4] \leq 3}

B = {x \in R : 3^{x} (\sum_{r = 1}^{\infty} \frac{3}{1 0 ^{r}})^{x - 3} < 3^{- 3 x}}

, where

[t]

denotes greatest integer function. Then,

Answer: (B)

For any real

x

[x + 3] + [x + 4] = 2 [x] + 7

, so the condition

[x + 3] + [x + 4] \leq 3

reduces to

[x] \leq - 2

, giving

A = (- \infty, - 1)

. The geometric series

\sum_{r = 1}^{\infty} \frac{3}{1 0 ^{r}} = \frac{1}{3}

, so the inequality defining

B

becomes

3^{x} (\frac{1}{3})^{x - 3} < 3^{- 3 x}

, that is

3^{3} < 3^{- 3 x}

, hence

- 3 x > 3

and

x < - 1

, so

B = (- \infty, - 1)

. Therefore

A = B

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Single correct option