Relations & Functions
One-one and onto functions
Grade None
Question:
<p>Let \(f: (-1, 1) \to B\), be a function defined by \(f(x) = \tan^{-1}\dfrac{2x}{1-x^2}\), then \(f\) is both one-one and onto when \(B\) is the interval</p>
<p>\(\left(0, \dfrac{\pi}{2}\right)\)</p>
<p>\(\left[0, \dfrac{\pi}{2}\right)\)</p>
<p>\(\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\)</p>
<p>\(\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) = tan⁻¹(2x/(1-x²)) relates to the double angle formula for tangent: tan(2θ) = 2tan(θ)/(1-tan²(θ)). If x = tan(θ), then f(x) = tan⁻¹(tan(2θ)) = 2θ, so f(x) = 2tan⁻¹(x). The range depends on the domain constraint for the inverse function to be valid.
<p><strong>Step 1: Substitute x = tan(θ)</strong></p><p>If x = tan(θ), then f(x) = tan⁻¹(2tan(θ)/(1 - tan²(θ))) = tan⁻¹(tan(2θ)) = 2θ = 2tan⁻¹(x)</p><p><strong>Step 2: Determine the domain constraint on θ</strong></p><p>Since x ∈ (-1, 1), we have θ = tan⁻¹(x) ∈ (-π/4, π/4)</p><p><strong>Step 3: Find the range of f</strong></p><p>Therefore, f(x) = 2tan⁻¹(x) ∈ 2(-π/4, π/4) = (-π/2, π/2)</p><p><strong>Step 4: Verify one-one and onto</strong></p><p>f(x) = 2tan⁻¹(x) is strictly increasing on (-1, 1) (since derivative is positive), making it one-one. For onto, B must equal the range (-π/2, π/2).</p><p>∴ Answer: B = (-π/2, π/2), which is option D</p>
Correct Answer: D