Question

If [x+6]+[x+3] ≤ 7 and let call its solution as set A and set B is the solution of inequality 3^5x-8 < 3^-3x.

• A. B ⊂ A, A≠B
• B. A ⊂ B, A≠B
• C. A∩B = ϕ
• D. A∪B = ϕ

Answer 1

The Correct Option is B

To solve the given problem, we need to find the solutions for each inequality separately and then compare their solution sets. 

Step 1: Solve the inequality [x+6]+[x+3] ≤ 7

• Since [.] denotes the greatest integer function or floor function, we have:
• \(\lfloor x+6 \rfloor + \lfloor x+3 \rfloor \leq 7\)
• Let's consider the expression \(\lfloor x+6 \rfloor + \lfloor x+3 \rfloor\):
• Assume \(a = \lfloor x+6 \rfloor\) and \(b = \lfloor x+3 \rfloor\).
• Then \(a \leq x+6 < a+1\) and \(b \leq x+3 < b+1\).
• Given \(a + b \leq 7\), we solve for \(x\).
• Since \(b \leq a - 3\), substitute to get \(a + (a - 3) \leq 7\), simplifying gives \(a \leq 5\).
• For \(a \leq 5\), \(a - 3 \leq 5\) gives \(a - 3 \geq 2\) thus \(b = a - 3\).
• From these, the valid range of \(x\) is \(2 \leq x < 3\).

The solution set A is \(x \in [2, 3)\).

Step 2: Solve the inequality \(3^{5x-8} < 3^{-3x}\)

• Rewriting the inequality using exponential properties:
• \(3^{5x-8} < 3^{-3x}\) implies \(5x - 8 < -3x\).
• Solve for \(x\),
• Adding \(3x\) to both sides: \(5x + 3x < 8\) simplifies to \(8x < 8\).
• Dividing through by 8 gives \(x < 1\).

The solution set B is \(x \in (-\infty, 1)\).

Comparison of sets A and B

• A = \([2, 3)\)
• B = \((- \infty, 1)\)
• It is clear that every element of A fits within the boundary conditions of B even they don't overlap.

Thus, set A is a subset of set B but they are not equal.

Conclusion

Therefore, the correct answer is A ⊂ B, A≠B.