Basic Mathematics & Logarithm
Logarithmic inequations
Grade 11
Question:
<p>If <i>x</i> and <i>C</i> are real, then the inequation \(\log_2 x + \log_x 2 + 2\cos C ≥ 0\)</p>
<p>(a) has no solution</p>
<p>(b) has exactly two solutions</p>
<p>(c) is satisfied for any real <i>C</i> and any real <i>x</i> in (0, 1)</p>
<p>(d) is satisfied for any real <i>C</i> and any real <i>x</i> in (1, ∞)</p>
Step-by-Step Solution
Key Concept: Use AM-GM inequality to bound the sum of logarithmic terms and analyze when the inequality holds.
<p>Let $y = \log_2 x$. Then $\log_x 2 = \frac{1}{y}$. The inequation becomes $y + \frac{1}{y} + 2\cos C ≥ 0$. By AM-GM, $y + \frac{1}{y} ≥ 2$ when $y > 0$ (i.e., $x > 1$), and $y + \frac{1}{y} ≤ -2$ when $y < 0$ (i.e., $0 < x < 1$). Since $−1 ≤ \cos C ≤ 1$, we have $2\cos C ≥ -2$. For $x > 1$: $y + \frac{1}{y} ≥ 2$ and $2\cos C ≥ -2$, so the inequality holds for all <i>C</i>. Therefore, the answer is (d).</p>
Correct Answer: D