Key Concepts and Formulas

• Greatest Integer Function: $[x+n] = [x]+n$ where $n$ is an integer; if $[x] = k$, then $k \le x < k+1$.
• Infinite Geometric Series: $a + ar + ar^2 + \dots = \frac{a}{1-r}$ if $|r| < 1$.
• Exponential Inequalities: If $b > 1$, $b^p < b^q$ is equivalent to $p < q$.

Step-by-Step Solution

Step 1: Analyzing Set A

We are given $A = \{ x \in R:[x + 3] + [x + 4] \le 3\}$. Our goal is to determine the range of real numbers $x$ that satisfy this inequality.

• Applying the property $[x+n] = [x]+n$, we have: $[x+3] = [x] + 3$ and $[x+4] = [x] + 4$.
• Substituting into the given inequality: $[x] + 3 + [x] + 4 \le 3$
• Simplifying the inequality: $2[x] + 7 \le 3$
• Isolating the greatest integer term: $2[x] \le -4$
• Solving for $[x]$: $[x] \le -2$
• Interpreting the result: This means $x$ must be less than $-1$. If $[x] = -2$, then $-2 \le x < -1$. If $[x] = -3$, then $-3 \le x < -2$, and so on. The union of all such intervals gives $x < -1$.
• Expressing Set A: Therefore, $A = (-\infty, -1)$.

Step 2: Analyzing Set B

We are given $B = \left\{ {x \in R:{3^x}{{\left( {\sum\limits_{r = 1}^\infty {{3 \over {{{10}^r}}}} } \right)}^{x - 3}} < {3^{ - 3x}}} \right\}$. We need to determine the range of real numbers $x$ that satisfy this exponential inequality.

• Evaluating the infinite geometric series: The series $\sum_{r=1}^{\infty} \frac{3}{10^r} = \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \dots$ is a geometric series with first term $a = \frac{3}{10}$ and common ratio $r = \frac{1}{10}$. Since $|r| < 1$, the sum is $S = \frac{a}{1-r} = \frac{\frac{3}{10}}{1-\frac{1}{10}} = \frac{\frac{3}{10}}{\frac{9}{10}} = \frac{3}{9} = \frac{1}{3}$.
• Substituting the sum into the inequality: $3^x \left(\frac{1}{3}\right)^{x-3} < 3^{-3x}$
• Simplifying using exponent rules: $3^x \cdot (3^{-1})^{x-3} < 3^{-3x}$ $3^x \cdot 3^{-x+3} < 3^{-3x}$ $3^{x - x + 3} < 3^{-3x}$ $3^3 < 3^{-3x}$
• Comparing exponents (since the base is greater than 1): $3 < -3x$
• Solving for $x$: $x < -1$
• Expressing Set B: Therefore, $B = (-\infty, -1)$.

Step 3: Comparing Sets A and B

We found $A = (-\infty, -1)$ and $B = (-\infty, -1)$. Therefore, $A = B$.

Step 4: Evaluating the Options

Since $A = B$, we can evaluate the given options:

• (A) $B \subset C,A \ne B$: This is false because $A = B$.
• (B) $A \subset B,A \ne B$: This is false because $A = B$.
• (C) $A = B$: This is true since we found $A = B$.
• (D) $A \cap B = \phi$: This is false because $A \cap B = A = B = (-\infty, -1)$, which is not an empty set.

Common Mistakes & Tips

• Be careful when applying the greatest integer function property; remember that $[x] \le k$ implies $x < k+1$.
• Remember to flip the inequality sign when dividing by a negative number.
• When dealing with geometric series, make sure $|r| < 1$ before applying the formula.

Summary

We analyzed the given sets $A$ and $B$ by simplifying the expressions and solving the resulting inequalities. By applying the properties of the greatest integer function and the formula for the sum of an infinite geometric series, we determined that both sets are equal to $(-\infty, -1)$. Therefore, the correct option is (C).

Final Answer

The final answer is \boxed{A = B}, which corresponds to option (C).