<p><strong>Paragraph for Question nos. 628 and 629</strong><br>Let \(f\) be a differentiable function satisfying<br>\(\sqrt[3]{f(x+y)} = \sqrt[3]{f(x)} + \sqrt[3]{f(y)} + 1 \ \forall\ x, y \in R\) and \(f'(0) = 3\).<br><br>If \(h(x) = f(x) - x^3\), then number of point(s) where \(y = h(|x|)\) is non-derivable is(are):</p>
Step-by-Step Solution
Key Concept: Substitute u = ∛f(x) to convert the functional equation into u(x+y) = u(x) + u(y) + 1, then solve for u to find f(x) = (x+1)³ - 1. Use f'(0) = 3 to verify, then analyze h(x) = f(x) - x³ = 3x² + 3x and h(|x|) for non-derivability points.
<p><strong>Step 1:</strong> Let u(x) = ∛f(x). Then the functional equation becomes:<br/>u(x+y) = u(x) + u(y) + 1</p><p><strong>Step 2:</strong> Let v(x) = u(x) + c for some constant c. We get:<br/>v(x+y) + c = v(x) + c + v(y) + c + 1<br/>v(x+y) = v(x) + v(y) + c + 1<br/>Setting c = -1: v(x+y) = v(x) + v(y), so v(x) = ax for some constant a.</p><p><strong>Step 3:</strong> Thus u(x) = ax - 1, which means ∛f(x) = ax - 1<br/>Therefore f(x) = (ax - 1)³</p><p><strong>Step 4:</strong> Use f'(0) = 3:<br/>f'(x) = 3a(ax - 1)²<br/>f'(0) = 3a(1) = 3 ⟹ a = 1<br/>So f(x) = (x - 1)³ = x³ - 3x² + 3x - 1</p><p><strong>Step 5:</strong> Find h(x) = f(x) - x³ = -3x² + 3x - 1<br/>h'(x) = -6x + 3, which is smooth everywhere</p><p><strong>Step 6:</strong> Consider h(|x|) = -3x² + 3|x| - 1<br/>For x > 0: h(x) = -3x² + 3x - 1, h'(x⁺) = -6x + 3<br/>For x < 0: h(-x) = -3x² - 3x - 1, h'(x⁻) = -6x - 3</p><p><strong>Step 7:</strong> At x = 0:<br/>Right derivative: lim(h→0⁺) = 3<br/>Left derivative: lim(h→0⁻) = -3<br/>Since left and right derivatives are unequal, y = h(|x|) is non-derivable at x = 0.</p><p>∴ Answer: <strong>B</strong> (1 point of non-derivability)</p>
Correct Answer: B