Question

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

• A. \(A \subset B, A \ne B\)
• B. \(A \cap B = \phi\)
• C. \(A = B\)
• D. \(B \subset C, A \ne B\)

Answer 1

The Correct Option is C

To solve this problem, we need to determine the set \(A\) and the set \(B\), and then check their relationship.

Step 1: Determine the Set \(A\)

The set \(A\) is defined as:

\(A = \{x \in \mathbb{R}: |x+3| + |x+4| \le 3\}\)

We need to solve the inequality \(|x+3| + |x+4| \le 3\). Consider different cases based on the expressions inside the absolute values:

1. Case 1: \(x \ge -3\)
   • Here, \(|x+3| = x + 3\) and \(|x+4| = x + 4\)
   • The inequality becomes \((x + 3) + (x + 4) \le 3 \Rightarrow 2x + 7 \le 3\)
   • Solving, we get \(2x \le -4 \Rightarrow x \le -2\)
2. Case 2: \(x < -4\)
   • Here, \(|x+3| = -(x + 3) = -x - 3\) and \(|x+4| = -(x + 4) = -x - 4\)
   • The inequality becomes \(-x - 3 - x - 4 \le 3 \Rightarrow -2x - 7 \le 3\)
   • Solving, we get \(-2x \le 10 \Rightarrow x \ge -5\)
3. Case 3: \(-4 \le x < -3\)
   • Here, \(|x+3| = -(x + 3)\) and \(|x+4| = x + 4\)
   • The inequality becomes \(-(x + 3) + (x + 4) \le 3 \Rightarrow 1 \le 3\), which is always true.

Combining results from above, the solution is \(x \in [-5, -2]\).

Step 2: Determine the Set \(B\)

The set \(B\) is defined as:

\(B = \left\{ x \in \mathbb{R} : 3 \cdot \sum_{r=1}^{\infty} \frac{3^{x-3}}{10^r} < 3^{-3x} \right\}\)

We simplify the expression:

1. The sum inside is a geometric series with the first term \(\frac{3^{x-3}}{10}\) and common ratio \(\frac{1}{10}\).
2. The sum evaluates to: \(\sum_{r=1}^{\infty} \frac{3^{x-3}}{10^r} = \frac{3^{x-3} / 10}{1 - 1/10} = \frac{3^{x-3}}{9}\)
3. Substitute back, the inequality becomes: \(3 \left(\frac{3^{x-3}}{9}\right) < 3^{-3x} \Rightarrow 3^{x-3} < 3^{-3x}\)
4. This simplifies to: \(3^{x-3+ 3x} < 1 \Rightarrow 3^{4x-3} < 1\)
5. The exponential inequality simplifies to: \(4x - 3 < 0 \Rightarrow x < \frac{3}{4}\)

Ensuring the greatest integer condition, we have \(x \leq -2\) since it must be an integer (as \([x]\) implies integer).

Conclusion

Comparing sets \(A\) and \(B\):

• Set \(A = [-5, -2]\)
• Set \(B\) includes integers up to \(-2\) and also makes the inequality true.

Therefore, \(A = B\) is the correct relationship.