Let [A = x R:[x + 3] + [x + 4] 3 ,]

Question

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

Accepted Answer

We have,

$$
\begin{aligned}
& A=\{x \in R:[x+3]+[x+4] \leq 3\} \\
& \text { Here, }[x+3]+[x+4] \leq 3 \\
& \Rightarrow [x]+3+[x]+4 \leq 3 \\
& (\because[x+n]=[x]+n, n \in I) \\
& \Rightarrow 2[x]+4 \leq 0 \Rightarrow[x] \leq-2 \\
& \Rightarrow  x \in(-\infty,-1) \\
& A \equiv(-\infty,-1) ...........(i)
\end{aligned}
$$

Also,

$$
B=\left\{x \in R: 3^x\left(\sum\limits_{r=1}^{\infty} \frac{3}{10^r}\right)^{x-3}3 \Rightarrow x<-1 \\
\Rightarrow \& x \in(-\infty,-1) \\
\Rightarrow \& B \equiv(-\infty,-1) ...........(ii)

$$

From equations (i) and (ii), we get

$A=B$

Solutions · 1