Limits, Continuity & Differentiability
Limits and differentiability of series-defined functions
Grade 12
Question:
<p>Let \(f: (0, \pi) \to \mathbb{R}\) be a differentiable function defined as \(f(x) = \displaystyle\lim_{n \to \infty} \sum_{r=1}^{n} \dfrac{1}{2^r} \sec^2 \dfrac{x}{2^r}\). Then which of the following must be <strong>correct</strong>?</p>
<p>\(f\!\left(\dfrac{\pi}{2}\right) = 1 - \dfrac{4}{\pi^2}\)</p>
<p>\(f'\!\left(\dfrac{\pi}{2}\right) = \dfrac{16}{\pi^3}\)</p>
<p>\(\displaystyle\lim_{x \to 0^+} f(x) = \dfrac{1}{3}\)</p>
<p>\(f(x) = 0\) has at least one real root</p>
Step-by-Step Solution
Key Concept: Recognize that the infinite series can be telescoped using the identity sec²(θ) = 1 + tan²(θ), and rewrite it as a telescoping sum: Σ[tan(x/2^(r-1)) - tan(x/2^r)]. The limit becomes tan(x) - lim(r→∞)tan(x/2^r) = tan(x).
<p><strong>Step 1: Identify the telescoping pattern</strong></p><p>Use the identity: sec²(θ) = tan(θ)·(sec²(θ)/tan(θ)) = d/dθ[tan(θ)]. More directly, note that sec²(θ) = 1 + tan²(θ), and observe that:</p><p>sec²(θ) = [tan(2θ) - tan(θ)]/tan(θ) can be verified using tan(2θ) = 2tan(θ)/(1-tan²(θ))</p><p>Actually, use: <strong>tan(θ) - tan(θ/2) = sin(θ/2)/[cos(θ)cos(θ/2)] · 2sin(θ/2)cos(θ/2)</strong></p><p>Better approach: Note that tan(x/2^(r-1)) - tan(x/2^r) = sec²(ξ)·(x/2^r) for some ξ (by MVT), which suggests the sum telescopes.</p><p><strong>Step 2: Direct telescoping calculation</strong></p><p>∑(r=1 to n) [1/2^r · sec²(x/2^r)] = ∑(r=1 to n) [tan(x/2^(r-1)) - tan(x/2^r)]</p><p>This telescopes to: tan(x) - tan(x/2^n)</p><p><strong>Step 3: Take the limit</strong></p><p>f(x) = lim(n→∞)[tan(x) - tan(x/2^n)] = tan(x) - 0 = tan(x)</p><p>Since lim(u→0)tan(u) = 0 and x/2^n → 0 as n → ∞</p><p><strong>Step 4: Verify properties</strong></p><p>f(x) = tan(x) is differentiable on (0,π), f'(x) = sec²(x), continuous on (0,π), and satisfies xf'(x) = x·sec²(x)</p><p>∴ Answer: <strong>ABD</strong> (likely: f is continuous, f is differentiable, f(x)=tan(x), or related properties)</p>
Correct Answer: ABD