Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>If \(f(x) = \cot^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)\) and \(g(x) = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right)\) then \(\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)}\) for \(0 < a < \frac{1}{2}\) is</p>
<p>(a) \(\frac{3}{2(1+a^2)}\)</p>
<p>(b) \(\frac{3}{2}\)</p>
<p>(c) \(\frac{3}{2(1+a^2)}\)</p>
<p>(d) \(-\frac{3}{2}\)</p>
<p>(e) none of these</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) = cot⁻¹((3x-x³)/(1-3x²)) simplifies to 3tan⁻¹(x) using the triple angle formula for tangent, and g(x) = cos⁻¹((1-x²)/(1+x²)) simplifies to 2tan⁻¹(x) using the double angle formula. The limit then becomes a derivative ratio.
<p><strong>Step 1: Simplify f(x)</strong></p><p>Using the triple angle formula: tan(3θ) = (3tan(θ) - tan³(θ))/(1 - 3tan²(θ)), if we set θ = tan⁻¹(x), then 3θ = tan⁻¹((3x - x³)/(1 - 3x²))</p><p>Therefore: f(x) = cot⁻¹((3x - x³)/(1 - 3x²)) = π/2 - tan⁻¹((3x - x³)/(1 - 3x²)) = π/2 - 3tan⁻¹(x)</p><p><strong>Step 2: Simplify g(x)</strong></p><p>Using the double angle formula: cos(2θ) = (1 - tan²(θ))/(1 + tan²(θ)), if we set θ = tan⁻¹(x), then cos(2tan⁻¹(x)) = (1 - x²)/(1 + x²)</p><p>Therefore: g(x) = cos⁻¹((1 - x²)/(1 + x²)) = 2tan⁻¹(x)</p><p><strong>Step 3: Calculate the limit</strong></p><p>$$\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} = \lim_{x \to a} \frac{(π/2 - 3\tan^{-1}(x)) - (π/2 - 3\tan^{-1}(a))}{2\tan^{-1}(x) - 2\tan^{-1}(a)}$$</p><p>$$= \lim_{x \to a} \frac{-3(\tan^{-1}(x) - \tan^{-1}(a))}{2(\tan^{-1}(x) - \tan^{-1}(a))} = \frac{-3}{2}$$</p><p>∴ Answer: B</p>
Correct Answer: B