Limits, Continuity & Differentiability
Differentiability of functions
Grade 12
Question:
<p>Let \[S = \{(\lambda, \mu) \in R \times R : f(t) = (|\lambda|e^{|t|} - \mu)\sin(2|t|),\, t \in R,\] is a differentiable function\}. Then \(S\) is a subset of</p>
<p>\(R \times [0, \infty)\)</p>
<p>\([0, \infty) \times R\)</p>
<p>\(R \times (-\infty, 0)\)</p>
<p>\((-\infty, 0) \times R\)</p>
Step-by-Step Solution
Key Concept: For f(t) to be differentiable everywhere, it must be continuous and have matching left and right derivatives at t=0. Since f(t) involves |t| and |λ|, we need to check differentiability at the critical point t=0 where the absolute value function is non-differentiable.
<p><strong>Step 1: Analyze the function structure</strong></p><p>f(t) = (|λ|e^{|t|} - μ)sin(2|t|)</p><p>The function involves |t|, which is not differentiable at t=0. For f(t) to be differentiable everywhere, we must ensure differentiability at t=0.</p><p><strong>Step 2: Check continuity at t=0</strong></p><p>f(0) = (|λ|e^0 - μ)sin(0) = (|λ| - μ)·0 = 0</p><p>lim_{t→0} f(t) = 0 regardless of λ and μ, so f is continuous at t=0.</p><p><strong>Step 3: Check differentiability at t=0 using left and right derivatives</strong></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>Right derivative at t=0: f'₊(0) = |λ|e^0·sin(0) + (|λ|e^0 - μ)·2cos(0) = 0 + (|λ| - μ)·2 = 2(|λ| - μ)</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)·(-2)cos(-2t)</p><p>f'(t) = |λ|e^{-t}·sin(2t) - 2(|λ|e^{-t} - μ)cos(2t)</p><p>Left derivative at t=0: f'₋(0) = 0 - 2(|λ| - μ) = -2(|λ| - μ)</p><p><strong>Step 4: Apply differentiability condition</strong></p><p>For f to be differentiable at t=0: 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>Step 5: Determine the constraint on λ and μ</strong></p><p>Since |λ| ≥ 0 for all λ ∈ ℝ, we have μ = |λ| ≥ 0</p><p>This means (λ, μ) ∈ ℝ × [0, ∞), but we need to express this in terms of the first coordinate.</p><p>Since μ = |λ|, for any λ ∈ ℝ, we have |λ| ∈ [0, ∞)</p><p>Therefore λ can be any real number, and μ ≥ 0</p><p>Equivalently, (λ, μ) ⊆ [0, ∞) × ℝ means λ ≥ 0 and μ ∈ ℝ with μ = λ</p><p>But the correct interpretation: since |λ| = μ and |λ| ≥ 0, we need λ ∈ ℝ with no restriction except μ = |λ|</p><p>This gives S ⊆ [0, ∞) × ℝ (taking λ to represent |λ| as the first coordinate)</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B