Limits, Continuity & Differentiability
Differentiability at a point
Grade 12
Question:
<p>Given \(f(t) = (|\lambda|e^{|t|} - \mu)\sin(2|t|)\). If \(f(t)\) is differentiable at \(t = 0\), then the set \(S\) of all possible values of \((\lambda, \mu)\) is a subset of:</p>
<p>\(R \times [0, \infty)\)</p>
<p>\([0, \infty) \times R\)</p>
<p>\(R \times R\)</p>
<p>\([0,\infty) \times [0,\infty)\)</p>
Step-by-Step Solution
Key Concept: For f(t) to be differentiable at t=0, it must be continuous at t=0 AND the left and right derivatives must be equal. Since f(t) involves |t| and sin(2|t|), we need to check the derivative from both sides using the property that f(0)=0 (from the sin term) and match the derivative conditions.
<p><strong>Step 1:</strong> Check continuity at t=0.</p><p>f(0) = (|λ|e⁰ - μ)sin(0) = 0</p><p>lim(t→0) f(t) = lim(t→0) (|λ|e^|t| - μ)sin(2|t|) = (|λ| - μ)·0 = 0</p><p>So f is continuous at t=0 for all λ, μ.</p><p><strong>Step 2:</strong> Find right derivative at t=0.</p><p>For t > 0: f(t) = (|λ|e^t - μ)sin(2t)</p><p>f'(t) = |λ|e^t·sin(2t) + (|λ|e^t - μ)·2cos(2t)</p><p>f'(0⁺) = |λ|·0 + (|λ| - μ)·2 = 2(|λ| - μ)</p><p><strong>Step 3:</strong> Find left derivative at t=0.</p><p>For t < 0: f(t) = (|λ|e^(-t) - μ)sin(-2t) = -(|λ|e^(-t) - μ)sin(2t)</p><p>f'(t) = -[-|λ|e^(-t)·sin(2t) + (|λ|e^(-t) - μ)·2cos(2t)]</p><p>f'(0⁻) = -[0 + (|λ| - μ)·2] = -2(|λ| - μ)</p><p><strong>Step 4:</strong> For differentiability, f'(0⁺) = f'(0⁻).</p><p>2(|λ| - μ) = -2(|λ| - μ)</p><p>4(|λ| - μ) = 0</p><p>∴ |λ| = μ</p><p><strong>Conclusion:</strong> The set S = {(λ, μ) : μ = |λ|}, which represents all points on the V-shaped curve where μ equals the absolute value of λ.</p>
Correct Answer: A