Basic Mathematics & Logarithm
Inequalities involving means
Grade 11
Question:
<p>The least value of \(6\tan^2\phi + 54\cot^2\phi + 18\) is<br>(I) 54 when A.M. \(\geq\) G.M. is applicable for \(6\tan^2\phi\), \(54\cot^2\phi\) and 18 is added further<br>(II) 54 when A.M. \(\geq\) G.M. is applicable for \(6\tan^2\phi\), \(54\cot^2\phi\), and 18 is added further<br>(III) 78 when \(\tan^2\phi = \cot^2\phi\)</p>
<p>(1) (I) is correct, (II) is false</p>
<p>(2) (I) and (II) are correct</p>
<p>(3) (III) is correct</p>
<p>(4) none of the above are correct</p>
Step-by-Step Solution
Key Concept: Apply AM-GM inequality directly to the two variable terms: the minimum of (6tan²φ + 54cot²φ) occurs when 6tan²φ = 54cot²φ, then add the constant 18. This gives 2√(6·54) + 18 = 2√324 + 18 = 36 + 18 = 54.
<p><strong>Step 1:</strong> Identify the expression: f(φ) = 6tan²φ + 54cot²φ + 18</p><p><strong>Step 2:</strong> Apply AM-GM inequality only to variable terms (6tan²φ and 54cot²φ):<br/>√(6tan²φ · 54cot²φ) ≤ (6tan²φ + 54cot²φ)/2</p><p><strong>Step 3:</strong> Simplify the geometric mean:<br/>√(6 · 54 · tan²φ · cot²φ) = √(324 · 1) = 18<br/>So: 6tan²φ + 54cot²φ ≥ 2(18) = 36</p><p><strong>Step 4:</strong> Therefore: f(φ) = (6tan²φ + 54cot²φ) + 18 ≥ 36 + 18 = 54</p><p><strong>Step 5:</strong> Equality holds when 6tan²φ = 54cot²φ, giving tan⁴φ = 9, or tan²φ = 3 (valid condition).</p><p>∴ Answer: B (The minimum value is 54)</p>
Correct Answer: B