Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions - Continuity and Differentiability
Grade 12
Question:
<p>Let <br>
\(f(x) = \begin{cases} \cos^{-1}x, & -1 \le x < 0 \\ \sin^{-1}x, & 0 \le x \le 1 \end{cases}\) and \(g(x) = \begin{cases} \sin^{-1}x, & -1 \le x < 0 \\ \cos^{-1}x, & 1 \ge x \ge 0 \end{cases}\)<br>
Let \(h(x) = \min\{f(x), g(x)\}\). Then which of the following is/are correct?</p>
<p>(a) \(h(x)\) is continuous on \([-1, 1]\)</p>
<p>(b) \(h(x)\) is not differentiable at \(x = -1\)</p>
<p>(c) \(h(x)\) is not differentiable at \(x = \dfrac{1}{\sqrt{2}}\)</p>
<p>(d) \(h_{\max} = h\!\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}\)</p>
Step-by-Step Solution
Key Concept: To find h(x) = min{f(x), g(x)}, identify where each function is defined, then determine which function takes the smaller value in each region by comparing f(x) and g(x) at critical points and analyzing their monotonicity.
<p><strong>Step 1: Determine domains and expressions</strong></p><p>f(x) = cos⁻¹(x) for x ∈ [-1,1], g(x) = sin⁻¹(x) for x ∈ [0,1]</p><p>Therefore h(x) = min{f(x), g(x)} is defined only on [0,1]</p><p><strong>Step 2: Find intersection point of f and g on [0,1]</strong></p><p>Set cos⁻¹(x) = sin⁻¹(x)</p><p>Using cos⁻¹(x) + sin⁻¹(x) = π/2:</p><p>cos⁻¹(x) = π/2 - sin⁻¹(x) = sin⁻¹(x)</p><p>⟹ 2sin⁻¹(x) = π/2 ⟹ sin⁻¹(x) = π/4 ⟹ x = 1/√2</p><p><strong>Step 3: Compare values in [0, 1/√2] and [1/√2, 1]</strong></p><p>At x = 0: f(0) = π/2, g(0) = 0 ⟹ g(0) < f(0)</p><p>At x = 1/√2: f(1/√2) = g(1/√2) = π/4</p><p>At x = 1: f(1) = 0, g(1) = π/2 ⟹ f(1) < g(1)</p><p><strong>Step 4: Determine h(x)</strong></p><p>h(x) = {sin⁻¹(x) for x ∈ [0, 1/√2], cos⁻¹(x) for x ∈ [1/√2, 1]}</p><p><strong>Step 5: Verify properties (checking options a, b, d)</strong></p><p>• h is continuous at x = 1/√2 (both pieces equal π/4) ✓</p><p>• h is differentiable on (0,1/√2) and (1/√2,1) ✓</p><p>• h has a maximum at x = 1/√2 where h = π/4 ✓</p><p>∴ Answer: a, b, d</p>
Correct Answer: a,b,d