<p>Which of the following is/are true about the root/s of inequality \(\log_{|x|}|x^2+x+1| \leq 1\)</p>
Step-by-Step Solution
Key Concept: Convert the logarithmic inequality using the definition log_b(a) ≤ 1 means a ≤ b when b > 1, and a ≥ b when 0 < b < 1. Also recognize that |x| as a base requires |x| > 0 and |x| ≠ 1, meaning x ≠ 0, ±1.
<p><strong>Step 1:</strong> Identify constraints. Base |x| requires: |x| > 0, |x| ≠ 1, so x ≠ 0, ±1</p><p><strong>Step 2:</strong> Note that x² + x + 1 = (x + 1/2)² + 3/4 > 0 always, so |x² + x + 1| = x² + x + 1</p><p><strong>Step 3:</strong> Rewrite inequality: log_{|x|}(x² + x + 1) ≤ 1</p><p><strong>Case 1: |x| > 1</strong> (i.e., x < -1 or x > 1)</p><p>Then: x² + x + 1 ≤ |x|</p><p>• If x > 1: x² + x + 1 ≤ x has no solutions (LHS > RHS)</p><p>• If x < -1: x² + x + 1 ≤ -x gives x² + 2x + 1 ≤ 0, so (x+1)² ≤ 0, which gives x = -1 (excluded)</p><p><strong>Case 2: 0 < |x| < 1</strong> (i.e., -1 < x < 0 or 0 < x < 1)</p><p>Then: x² + x + 1 ≥ |x| (inequality reverses)</p><p>• If 0 < x < 1: x² + x + 1 ≥ x is always true (since x² + 1 ≥ 0)</p><p>• If -1 < x < 0: x² + x + 1 ≥ -x gives x² + 2x + 1 ≥ 0, so (x+1)² ≥ 0, always true</p><p><strong>Step 4:</strong> Solution set: (-1, 0) ∪ (0, 1)</p><p>∴ Answer: AD (verify which statements match this solution set)
Correct Answer: AD