Limits, Continuity & Differentiability
Continuity and Differentiability
Grade 12
<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 expressions at each point; identify where each expression dominates by finding intersection points, then analyze continuity and differentiability at these critical points.
<p><strong>Step 1:</strong> Find intersection points where |3-x| = 3-x³</p><p>Case 1: If x ≤ 3, then |3-x| = 3-x, so: 3-x = 3-x³ ⟹ x³ = x ⟹ x ∈ {-1, 0, 1}</p><p>Case 2: If x > 3, then |3-x| = x-3, so: x-3 = 3-x³ ⟹ x³+x-6 = 0 ⟹ x = 2 (rejected since x > 3)</p><p><strong>Step 2:</strong> Analyze behavior in intervals:</p><p>• For x < -1: |3-x| = 3-x dominates (grows faster)</p><p>• For -1 < x < 1: 3-x³ dominates (higher values)</p><p>• For x > 1: |3-x| dominates</p><p><strong>Step 3:</strong> Check differentiability at x = -1, 0, 1</p><p>At x = -1: Left derivative = -1, Right derivative = 3x²|_{x=-1} = 3 ⟹ NOT differentiable</p><p>At x = 0: Both expressions equal 3, but derivatives differ (left: -1, right: 0) ⟹ NOT differentiable</p><p>At x = 1: Left derivative = 3, Right derivative = -1 ⟹ NOT differentiable</p><p><strong>Step 4:</strong> f(x) is continuous everywhere but NOT differentiable at x ∈ {-1, 0, 1}</p><p>∴ Answer: A (f is continuous but has non-differentiable points)</p>
Correct Answer: A