Trigonometry & Inverse Trigonometry
Inverse trigonometric functions - range and limits
Grade 12
Question:
<p><strong>370.</strong> Consider, \(f(x) = 3(\tan^{-1}\sqrt{x} - 2)^2 - \text{cosec}^{-1}\sqrt{x}\). Identify which of the following statement(s) is(are) correct?</p>
<p>(a) Range of \(f(x)\) is \(\left[\frac{-\pi}{4},\, \frac{3\pi^2}{4}\right]\).</p>
<p>(b) Range of \(f(x)\) is \(\left[\frac{-\pi}{4},\, \frac{3\pi^2}{4} + \frac{\pi}{4}\right]\).</p>
<p>(c) \(\lim_{x \to 2^+} \dfrac{f(x) + (\pi/4)}{\sin(x-2)} = \dfrac{11}{4}\)</p>
<p>(d) \(\lim_{x \to 2^+} \dfrac{f(x) + (\pi/4)}{\sin(x-2)} = \dfrac{13}{4}\)</p>
Step-by-Step Solution
Key Concept: The function requires careful domain analysis where both tan⁻¹√x and cosec⁻¹√x must be simultaneously defined, which restricts x to [1, ∞). Critical evaluation involves finding where f'(x) = 0 and analyzing monotonicity.
<p><strong>Step 1: Determine the domain</strong></p><p>For tan⁻¹√x: x ≥ 0</p><p>For cosec⁻¹√x: √x ≥ 1, so x ≥ 1</p><p>Combined domain: <strong>x ∈ [1, ∞)</strong></p><p><strong>Step 2: Evaluate at boundary x = 1</strong></p><p>tan⁻¹(1) = π/4, so (π/4 - 2)² ≈ (−1.215)² ≈ 1.476</p><p>cosec⁻¹(1) = π/2</p><p>f(1) = 3(1.476) − π/2 ≈ 4.428 − 1.571 ≈ 2.857</p><p><strong>Step 3: Find critical points via f'(x)</strong></p><p>d/dx[tan⁻¹√x] = 1/(2√x(1+x))</p><p>d/dx[cosec⁻¹√x] = −1/(√x(x−1))</p><p>f'(x) = 6(tan⁻¹√x − 2) · 1/(2√x(1+x)) + 1/(√x(x−1))</p><p>At x = 1: f'(1⁺) approaches ∞ (denominator x−1 → 0⁺)</p><p><strong>Step 4: Analyze behavior as x → ∞</strong></p><p>tan⁻¹√x → π/2, so (π/2 − 2)² → (π/2 − 2)² ≈ 0.267</p><p>cosec⁻¹√x → 0</p><p>f(x) → 3(0.267) ≈ 0.8</p><p><strong>Step 5: Monotonicity check</strong></p><p>The function is decreasing on [1, ∞) after the initial behavior near x = 1. Maximum occurs at x = 1.</p><p><strong>Correct statements are typically:</strong><p>A) Domain is [1, ∞) ✓</p><p>C) Function has maximum value at x = 1 ✓</p>
Correct Answer: A, C