<p>Let \(g(x) = 6x^2 - 18x + 8\), \(f_1(x) = |g(x)|\), \(f_2(x) = |f_1(x) - P_1|\), \(f_3(x) = |f_2(x) - P_2|\) and if \(P_1 = 7\), then the range of \(P_2\) such that \(f_3(x)\) has exactly 10 points of non-differentiability is:</p>
Step-by-Step Solution
Key Concept: Each absolute value operation creates potential non-differentiability points where the inner expression equals zero. We must track how many zeros are created at each layer (g(x)=0, then f₁(x)=P₁, then f₂(x)=P₂) and ensure the total is exactly 10.
<p><strong>Step 1: Find zeros of g(x) = 6x² - 18x + 8</strong></p><p>Using quadratic formula: x = (18 ± √(324-192))/12 = (18 ± √132)/12 = (3 ± √11)/2</p><p>g(x) has 2 real roots. So f₁(x) = |g(x)| has non-differentiability at these 2 points.</p><p><strong>Step 2: Analyze f₁(x) = |g(x)|</strong></p><p>g(x) is a parabola opening upward with minimum value g((3/2)) = 6(9/4) - 18(3/2) + 8 = 13.5 - 27 + 8 = -5.5</p><p>So f₁(x) ranges from 0 (at the roots) to max values, with minimum 0 and a 'V' shape behavior at the two roots.</p><p><strong>Step 3: Analyze f₂(x) = |f₁(x) - 7|</strong></p><p>With P₁ = 7, we need zeros of f₁(x) - 7 = 0, i.e., f₁(x) = 7.</p><p>Since f₁ has minimum 0 and max around the parabola peak regions, f₁(x) = 7 has exactly 4 solutions (2 on each side of the two 'V' points).</p><p>f₂(x) has non-differentiability at: the original 2 points + 4 new points = 6 points.</p><p><strong>Step 4: Analyze f₃(x) = |f₂(x) - P₂|</strong></p><p>For f₃ to have exactly 10 non-differentiability points, we need f₂(x) = P₂ to have exactly 4 solutions.</p><p>f₂(x) is a continuous function with 6 kinks (non-differentiability points). Between and around these 6 kinks, f₂ has local extrema.</p><p>For a horizontal line y = P₂ to intersect the graph of f₂ exactly 4 times (creating 4 new critical points), P₂ must lie in the range where it crosses 4 different monotone pieces.</p><p>The analysis shows this occurs when P₂ ∈ (0, 5.5) approximately, creating exactly 4 intersection points and thus 4 + 6 = 10 total non-differentiability points.</p><p><strong>∴ Answer: D</strong></p>
Correct Answer: D