Let A = { x in R : [x + 3] + [x + 4] leq 3 } | vedclass

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(B) For set $A$: $[x + 3] + [x + 4] \leq 3 \implies [x] + 3 + [x] + 4 \leq 3$.$2[x] + 7 \leq 3 \implies 2[x] \leq -4 \implies [x] \leq -2$.Since $[x]

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$A = B$

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(B) For set $A$: $[x + 3] + [x + 4] \leq 3 \implies [x] + 3 + [x] + 4 \leq 3$.$2[x] + 7 \leq 3 \implies 2[x] \leq -4 \implies [x] \leq -2$.Since $[x] \leq -2$,we have $x For set $B$: The sum is a geometric series $\sum_{n=1}^{\infty} \frac{3}{10^n} = 3 \left( \frac{1/10}{1 - 1/10} \right) = 3 \left( \frac{1/10}{9/10} \right) = 3 \left( \frac{1}{9} \right) = \frac{1}{3} = 3^{-1}$.The inequality becomes $3^x (3^{-1})^{x-3} $3^x \cdot 3^{-x+3} $27 Comparing exponents: $3 Thus,$B = (-\infty, -1)$.Since $A = (-\infty, -1)$ and $B = (-\infty, -1)$,we conclude $A = B$.

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

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