Limits, Continuity & Differentiability
Properties of Functions and Limits
Grade 12
Question:
<p>Let \(f: \mathbb{R} \to [-1, 1]\) be defined as \(f(x) = \cos(\sin x)\). Which of the following is/are correct?</p>
<p>(a) \(f\) is periodic with fundamental period \(2\pi\)</p>
<p>(b) Range of \(f = [\cos 1, 1]\)</p>
<p>(c) \(\lim_{x \to \frac{\pi}{2}} \left( f\left(\frac{\pi}{2} - x\right) + f\left(\frac{\pi}{2} + x\right) \right) = 2\)</p>
<p>(d) \(f\) is neither even nor odd function</p>
Step-by-Step Solution
Key Concept: Analyze the function f(x) = cos(sin x) by examining its periodicity, range, symmetry properties, and limits. Key insight: sin x has range [-1, 1], so we're evaluating cosine on this bounded interval.
<p><strong>Step 1: Check Periodicity (Option A)</strong></p><p>f(x + 2π) = cos(sin(x + 2π)) = cos(sin x) = f(x), since sin has period 2π.</p><p>Since f(x + T) ≠ f(x) for any T < 2π (as sin x doesn't repeat with smaller period), the fundamental period is 2π. <strong>Option A is CORRECT.</strong></p><p><strong>Step 2: Find the Range (Option B)</strong></p><p>Since sin x ∈ [-1, 1] for all x ∈ ℝ, we need the range of cos(t) for t ∈ [-1, 1].</p><p>Since cosine is decreasing on [0, 1] and even in nature: cos(-1) = cos(1) ≈ 0.5403 (minimum on [-1, 1]).</p><p>The maximum occurs at t = 0: cos(0) = 1.</p><p>Therefore, Range of f = [cos 1, 1]. <strong>Option B is CORRECT.</strong></p><p><strong>Step 3: Evaluate the Limit (Option C)</strong></p><p>f(π/2 - x) = cos(sin(π/2 - x)) = cos(cos x)</p><p>f(π/2 + x) = cos(sin(π/2 + x)) = cos(cos x)</p><p>So f(π/2 - x) + f(π/2 + x) = 2cos(cos x)</p><p>As x → 0: lim[x→0] 2cos(cos x) = 2cos(cos 0) = 2cos(1) ≈ 1.081 ≠ 2. <strong>Option C is INCORRECT.</strong></p><p><strong>Step 4: Check Even/Odd Property (Option D)</strong></p><p>f(-x) = cos(sin(-x)) = cos(-sin x) = cos(sin x) = f(x), since cosine is even.</p><p>Therefore f is an even function. <strong>Option D is INCORRECT.</strong></p><p><strong>∴ Answer: AB</strong></p>
Correct Answer: AB