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

First, let's find the set A. \[ |x+3| + |x+4| \le 3. \] If \(x \ge -3\), then: \[ x+3 + x+4 \le 3 \implies 2x+7 \le 3 \implies 2x \le -4 \implies x \le -2. \] So, \(-3 \le x \le -2\). 

If \(-4 \le x < -3\), then: \[ -(x+3) + x+4 \le 3 \implies -x-3 + x+4 \le 3 \implies 1 \le 3, \] which is always true. So, \(-4 \le x < -3\).

If \(x < -4\), then: \[ -(x+3) - (x+4) \le 3 \implies -2x - 7 \le 3 \implies -2x \le 10 \implies x \ge -5. \] So, \(-5 \le x < -4\).

Combining these cases, we get \(A = [-5, -2]\).

Now, let's analyze set B: \[ 3 \cdot \sum_{r=1}^{\infty} \frac{3^{x-3}}{10^r} < 3^{-[x]}. \] The summation is a geometric series: \[ \sum_{r=1}^{\infty} \frac{1}{10^r} = \frac{1/10}{1 - 1/10} = \frac{1/10}{9/10} = \frac{1}{9}. \] So, \[ 3 \cdot 3^{x-3} \cdot \frac{1}{9} < 3^{-[x]} \implies 3^{x-2} \cdot \frac{1}{3} < 3^{-[x]} \implies 3^{x-3} < 3^{-[x]}. \] Since the base is 3 (greater than 1), we compare the exponents: \[ x - 3 < -[x] \implies x + [x] < 3. \] If we let \(x = n + \epsilon\), where \(n\) is an integer and \(0 \le \epsilon < 1\), then: \[ n + \epsilon + n < 3 \implies 2n + \epsilon < 3. \] If \(n = -2\), then: \[ -4 + \epsilon < 3 \implies \epsilon < 7, \] which is always true since \(0 \leq \epsilon < 1\).

So, \(x < -1\).

Since \([x]\) is an integer, the range of \(x\) will be \([-5, -2]\).

✅ Comparing A and B: We have \(A = B = [-5, -2].\)