Sets, Relations & Functions
Composition of Functions
Grade 11
Question:
<p>Let \(f(x) = \log_e(\sin x)\), \((0 < x < \pi)\) and \(g(x) = \sin^{-1}(e^{-x})\), \((x \geq 0)\). If \(\alpha\) is a positive real number such that \(a = (f \circ g)'(\alpha)\) and \(b = (f \circ g)(\alpha)\), then</p>
<p>\(a\alpha^2 + b\alpha + a = 0\)</p>
<p>\(a\alpha^2 - b\alpha - a = 1\)</p>
<p>\(a\alpha^2 - b\alpha - a = 0\)</p>
<p>\(a\alpha^2 + b\alpha - a = -2\alpha^2\)</p>
Step-by-Step Solution
Key Concept: For f to be invertible, it must be strictly monotonic on its domain. The sign of f'(x) = cos(x)/sin(x) = cot(x) determines monotonicity, which changes at x = π/2 in the interval (0, π).
<p><strong>Step 1:</strong> For f(x) = log_e(sin x) to be invertible, it must be strictly monotonic (either strictly increasing or strictly decreasing) on its domain.</p><p><strong>Step 2:</strong> Find f'(x): f'(x) = (1/sin x) · cos x = cot(x)</p><p><strong>Step 3:</strong> Analyze monotonicity:</p><ul><li>For x ∈ (0, π/2): sin x > 0 and cos x > 0, so cot(x) > 0 ⟹ f is strictly increasing</li><li>For x ∈ (π/2, π): sin x > 0 and cos x < 0, so cot(x) < 0 ⟹ f is strictly decreasing</li><li>At x = π/2: cot(x) = 0 (critical point where monotonicity changes)</li></ul><p><strong>Step 4:</strong> Since f'(x) changes sign at x = π/2, the function is not monotonic on the entire interval (0, π). For invertibility, we must restrict to either (0, π/2) or (π/2, π) where f is strictly monotonic.</p><p><strong>Step 5:</strong> The standard choice is (0, π/2) where f is strictly increasing, making it invertible.</p><p>∴ Answer: A</p>
Correct Answer: A