Trigonometry & Inverse Trigonometry
Inverse trigonometric functions - range
Grade 12
Question:
<p>If \(f(x) = \sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right) - 2\tan^{-1}x\) and \(g(x) = \sin^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right) + 4\tan^{-1}x\), then range of \((f(x) - g(x))\) for \(x \in (-\infty, -1]\) is:</p>
<p>\(\left[0,\, \dfrac{3\pi}{2}\right)\)</p>
<p>\(\left[\dfrac{-3\pi}{2},\, \pi\right)\)</p>
<p>\(\left[-\pi,\, \dfrac{-\pi}{2}\right)\)</p>
<p>\(\left[\pi,\, \dfrac{7\pi}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: Recognize that $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x$ and $\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \pi - 2\tan^{-1}x$ for $x \leq -1$ (using the substitution $x = \tan\theta$ and careful handling of branch cuts). This transforms the problem into finding the range of a simple expression.
<p><strong>Step 1:</strong> Use the substitution $x = \tan\theta$. For $x \in (-\infty, -1]$, we have $\theta \in (-\pi/2, -\pi/4]$.</p><p><strong>Step 2:</strong> Apply the identity: $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x$ (valid here).</p><p><strong>Step 3:</strong> For $x \leq -1$: $\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \pi - 2\tan^{-1}x$ (since $\frac{1-x^2}{1+x^2} = \cos(2\theta)$ and we're in the region where the inverse must be adjusted).</p><p><strong>Step 4:</strong> Calculate $f(x) = 2\tan^{-1}x - 2\tan^{-1}x = 0$.</p><p><strong>Step 5:</strong> Calculate $g(x) = \pi - 2\tan^{-1}x + 4\tan^{-1}x = \pi + 2\tan^{-1}x$.</p><p><strong>Step 6:</strong> Thus $f(x) - g(x) = 0 - (\pi + 2\tan^{-1}x) = -\pi - 2\tan^{-1}x$.</p><p><strong>Step 7:</strong> For $x \in (-\infty, -1]$, $\tan^{-1}x \in (-\pi/2, -\pi/4]$, so $2\tan^{-1}x \in (-\pi, -\pi/2]$.</p><p><strong>Step 8:</strong> Therefore $-\pi - 2\tan^{-1}x \in [-\pi/2, 0)$.</p><p>∴ Answer: B</p>
Correct Answer: B