Q.1 (iii) If tan⁻¹(2x) + tan⁻¹(3x) = π/4 , then x = _____. A. -1 B. 1/6 C. 1/3 D. 3/2

Answer: B. 1/6 Answer: (b) 1/6 tan⁻¹2x + 3x/1 - 6x² = π/4 gives 5x/1 - 6x² = 1 , so 6x² + 5x - 1 = 0 , i.e. (6x - 1)(x + 1) = 0 . x = -1 makes both angles negative, so their sum cannot be π/4 ; hence x = 1/6 .

If \(\tan^{-1}(2x) + \tan^{-1}(3x) = \dfrac{\pi}{4}\), then \(x =\) _____.

Q.1 (iii) If \(\tan^{-1}(2x) + \tan^{-1}(3x) = \dfrac{\pi}{4}\), then \(x =\) _____.

A. \(-1\)

B. \(\dfrac{1}{6}\)

C. \(\dfrac{1}{3}\)

D. \(\dfrac{3}{2}\)

Doubtly Asked 28 Oct 2026 Board exam · March 2026 hsc 12th Save WhatsApp Report

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Answer: B. \(\dfrac{1}{6}\)

Answer: (b) \(\dfrac{1}{6}\)

\(\tan^{-1}\dfrac{2x + 3x}{1 - 6x^2} = \dfrac{\pi}{4}\) gives \(\dfrac{5x}{1 - 6x^2} = 1\), so \(6x^2 + 5x - 1 = 0\), i.e. \((6x - 1)(x + 1) = 0\). \(x = -1\) makes both angles negative, so their sum cannot be \(\pi/4\); hence \(x = \dfrac{1}{6}\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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If \(\tan^{-1}(2x) + \tan^{-1}(3x) = \dfrac{\pi}{4}\), then \(x =\) _____.

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