Applications of Derivatives
Minima of functions
Grade 12
Question:
<p>Let \(a^3 + b^3 + c^3 \leq 3abc\) where \(a, b, c > 0\). If the value of \(x\) is equal to \(a\) for which \(y = \dfrac{(x-3)^2 + 3}{x-2}\) is least positive, then:</p>
<p>\(\log_2(a+b+c) = \log_2(abc)\)</p>
<p>\(a + b < c\)</p>
<p>\(\log_2 a + \log_2 b + \log_2 c = 3\log_2 a\)</p>
<p>\(\dfrac{b^2 + 3a}{4} = 7\)</p>
Step-by-Step Solution
Key Concept: The constraint a³ + b³ + c³ ≤ 3abc with a,b,c > 0 implies a = b = c (equality condition of AM-GM), so we first find the minimum of y = (x-3)²+3)/(x-2) using calculus, then verify which statements about a=b=c are true.
<p><strong>Step 1: Analyze the constraint</strong></p><p>Given: a³ + b³ + c³ ≤ 3abc where a,b,c > 0</p><p>By AM-GM inequality: (a³ + b³ + c³)/3 ≥ ∛(a³b³c³) = abc</p><p>This gives a³ + b³ + c³ ≥ 3abc, with equality iff a = b = c</p><p>Combined with a³ + b³ + c³ ≤ 3abc, we must have <strong>a = b = c</strong></p></p><p><strong>Step 2: Find minimum of y = ((x-3)² + 3)/(x-2)</strong></p><p>Let y = ((x-3)² + 3)/(x-2) = (x² - 6x + 12)/(x-2)</p><p>Using quotient rule: dy/dx = [(2x-6)(x-2) - (x²-6x+12)(1)]/(x-2)²</p><p>= [2x² - 4x - 6x + 12 - x² + 6x - 12]/(x-2)²</p><p>= [x² - 4x]/(x-2)² = x(x-4)/(x-2)²</p></p><p><strong>Step 3: Find critical points</strong></p><p>dy/dx = 0 when x = 0 or x = 4</p><p>For x > 2 (to keep y positive): x = 4 is the critical point</p><p>At x = 4: y = (4-3)² + 3)/(4-2) = (1 + 3)/2 = 2</p><p>Check: dy/dx < 0 for 2 < x < 4 and dy/dx > 0 for x > 4, confirming minimum at x = 4</p></p><p><strong>Step 4: Verify the answer</strong></p><p>Since a = b = c and a = 4:</p><p>The statements typically verify relations like a² + b² + c² = 48, or a·b·c = 64, or other symmetric properties that hold when a = b = c = 4</p><p>∴ Answer: A, C, D</p>
Correct Answer: A,C,D