Let A={x∈R:[x+3]+[x+4]≤3} and B=x∈R:3x∑r=1∞ 310rx−3<3−3x then

Let A={x∈R:[x+3]+[x+4]≤3} and B=x∈R:3x∑r=1∞ 310rx−3

A
$A = B$
B
$A ⫋ B$
C
$B ⊊ A$
D
$A \cap B = ϕ$

Solution:

Let $x \in A, [x + 3] + [x + 4] \leq 3$

$\begin{array}{cl} \Rightarrow & [x] + 3 + [x] + 4 \leq 3 \\ \Rightarrow & 2 [x] \leq - 4 \Rightarrow [x] \leq - 2 \\ \Rightarrow & x \in (- \infty, - 1) \\ A = (- \infty, - 1) \end{array}$

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

$\begin{array}{ll} \Rightarrow 3^{2 x - 3} {(\frac{1 / 10}{1 - 1 / 10})}^{x - 3} < 3^{- 3 x} \\ \Rightarrow 3^{2 x - 3} {(3^{- 2})}^{x - 3} < 3^{- 3 x} \\ \Rightarrow 3^{3} < 3^{- 3 x} \Rightarrow 3 < - 3 x \\ so x \in (- \infty, - 1) \end{array}$

Hence $B = (- \infty, - 1)$

Thus $A = B$ .

Let $A = {x \in R : [x + 3] + [x + 4] \leq 3}$ and $B = \{x \in R : 3^{x} {(\sum_{r = 1}^{\infty} \frac{3}{10^{r}})}^{x - 3} < 3^{- 3 x}\}$ then

Solution:

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