<p>The range of the function <i>f</i>(<i>x</i>) = tan<sup>−1</sup><i>x</i> + ½ sin<sup>−1</sup><i>x</i> is:</p>
Step-by-Step Solution
Key Concept: To find the range of f(x) = tan⁻¹x + ½sin⁻¹x, we must first determine the domain (where both inverse functions are defined), then analyze how the function behaves at the boundaries and critical points using calculus and monotonicity.
<p><strong>Step 1: Determine the Domain</strong><br/>For f(x) = tan⁻¹x + ½sin⁻¹x to be defined, both functions must be defined:<ul><li>tan⁻¹x is defined for all x ∈ ℝ</li><li>sin⁻¹x is defined for x ∈ [-1, 1]</li></ul>Therefore, Domain of f(x) = [-1, 1]</p><p><strong>Step 2: Find the Derivative to Check Monotonicity</strong><br/>f'(x) = 1/(1+x²) + ½ · 1/√(1-x²)<br/><br/>For x ∈ (-1, 1):<ul><li>1/(1+x²) > 0 always</li><li>½ · 1/√(1-x²) > 0 for x ∈ (-1, 1)</li></ul>Therefore, f'(x) > 0 for all x ∈ (-1, 1), meaning f is strictly increasing on [-1, 1].</p><p><strong>Step 3: Calculate Values at Boundary Points</strong><br/>Since f is strictly increasing and continuous on [-1, 1], the range is [f(-1), f(1)].<br/><br/><strong>At x = -1:</strong><br/>f(-1) = tan⁻¹(-1) + ½sin⁻¹(-1)<br/>= -π/4 + ½(-π/2)<br/>= -π/4 - π/4<br/>= -π/2<br/><br/><strong>At x = 1:</strong><br/>f(1) = tan⁻¹(1) + ½sin⁻¹(1)<br/>= π/4 + ½(π/2)<br/>= π/4 + π/4<br/>= π/2</p><p><strong>Step 4: State the Range</strong><br/>Since f is continuous and strictly increasing on [-1, 1] with f(-1) = -π/2 and f(1) = π/2, the range is [-π/2, π/2].<br/><br/>∴ Answer: c</p>
Correct Answer: c