<p>Roots \(r,s,t\) of \(x(x-2)(3x-7)=2\) are real and positive. \(\tan^{-1}r+\tan^{-1}s+\tan^{-1}t=\)</p>
Step-by-Step Solution
<div class="solution"><p><strong>Step 1:</strong> Expand: \(3x^3-13x^2+14x-2=0\).</p><p><strong>Step 2:</strong> Vieta's: \(S_1=13/3,\;S_2=14/3,\;S_3=2/3\).</p><p><strong>Step 3:</strong> <span class="math-block">\[\tan(\tan^{-1}r+\tan^{-1}s+\tan^{-1}t)=\frac{S_1-S_3}{1-S_2}=\frac{13/3-2/3}{1-14/3}=\frac{11/3}{-11/3}=-1\]</p><p><strong>Step 4:</strong> Since \(r,s,t>0\), each \(\tan^{-1}\in(0,\pi/2)\), so sum \(\in(0,3\pi/2)\). Only \(3\pi/4\) gives \(\tan=-1\) there.</p><p><strong>Answer: (B) \(3\pi/4\)</strong></p><div class="trap-box"><strong>Trap:</strong> tan = -1 also at -\pi/4, but sum of positive arc-tangents must be positive.<div class="key-concept"><strong>Key Concept:</strong> Vieta + tan⁻^1 addition formula + quadrant check for the sum
Correct Answer: 2