<p>If \(f(x)\) be such that \(f(x) = \max(|3-x|, 3-x^3)\) then:</p>
<p>(a) \(f(x)\) is continuous \(\forall\, x \in R\)</p>
<p>(b) \(f(x)\) is derivable \(\forall\, x \in R\)</p>
<p>(c) \(f(x)\) is non-derivable at three points only</p>
<p>(d) \(f(x)\) is non-derivable at four points only</p>
Step-by-Step Solution
Key Concept: The function f(x) is the maximum of two functions at each point, so you must find where each function dominates and identify points where the maximum 'switches' between them (intersection points). Continuity and differentiability of the piecewise maximum depends on whether the two functions meet smoothly.
<p><strong>Step 1: Identify the two functions</strong></p><p>g(x) = |3-x| and h(x) = 3-x³</p><p><strong>Step 2: Find where g(x) = h(x)</strong></p><p>|3-x| = 3-x³</p><p>Case 1 (x ≤ 3): 3-x = 3-x³ ⟹ x³-x = 0 ⟹ x(x²-1) = 0 ⟹ x ∈ {-1, 0, 1}</p><p>Case 2 (x > 3): x-3 = 3-x³ ⟹ x³+x-6 = 0 ⟹ x = 1.5... (no solution in x > 3)</p><p><strong>Step 3: Determine which function dominates in each region</strong></p><p>At x = -2: |3-(-2)| = 5; 3-(-8) = 11 → h dominates</p><p>At x = 0.5: |3-0.5| = 2.5; 3-0.125 = 2.875 → h dominates</p><p>At x = 2: |3-2| = 1; 3-8 = -5 → g dominates</p><p>At x = 4: |3-4| = 1; 3-64 = -61 → g dominates</p><p><strong>Step 4: Check differentiability at critical points x = -1, 0, 1, and x = 3</strong></p><p>At x = -1, 0, 1: Both functions meet but have different derivatives, so f is NOT differentiable.</p><p>At x = 3: |3-x| has a corner (left derivative = -1, right derivative = 1), so f is NOT differentiable.</p><p><strong>Step 5: Conclusion</strong></p><p>f(x) is continuous everywhere but NOT differentiable at x ∈ {-1, 0, 1, 3}.</p><p>∴ Answer: A</p>
Correct Answer: A