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Black Advanced Problems in Mathematics
Exam
Black Advanced Problems in Mathematics
Subject
Mathematics
Chapter
Sequences and Series

Black Advanced Problems in Mathematics · Mathematics · Sequences and Series

If ((π)/(12) - x) , \ tan(π)/(12), ((π)/(12) + x) in order are...

If tan⁡\tan\left(\frac{\pi}{12} -x\right), $tanπ12,\$tan\frac{\pi}{12},\tan (π12+x)\left(\frac{\pi}{12} + x\right) in order are three consecutive terms of a G.P. then sum of all the solutions in [0,314][0, 314] is kπk\pi. The value of kk is:

Options

  1. A49504950
  2. B50505050
  3. C25252525
  4. D50105010

Correct answer

B. 50505050

Step-by-step solution

Since they are in G.P., we have: tan⁡2π12=tan⁡(π12−x)tan⁡(π12+x)tan⁡2π12=tan⁡2π12−tan⁡2x1−tan⁡2π12tan⁡2x\tan^2\frac{\pi}{12} = \tan\left(\frac{\pi}{12} - x\right)\tan\left(\frac{\pi}{12} + x\right)\tan^2\frac{\pi}{12} = \frac{\tan^2\frac{\pi}{12} - \tan^2 x}{1 - \tan^2\frac{\pi}{12}\tan^2 x} Let t=tan⁡2π12t = \tan^2\frac{\pi}{12} and y=tan⁡2xy = \tan^2 x. t=t−y1−ty  ⟹  t−t2y=t−y  ⟹  y(1−t2)=0t = \frac{t - y}{1 - ty} \implies t - t^2y = t - y \implies y(1 - t^2) = 0. So tan⁡2x(1−tan⁡4π12)=0\tan^2 x (1 - \tan^4\frac{\pi}{12}) = 0. Since tan⁡π12≠1\tan\frac{\pi}{12} \ne 1, we must have tan⁡2x=0  ⟹  tan⁡x=0\tan^2 x = 0 \implies \tan x = 0. The solutions are x=nπx = n\pi, where n∈Zn \in \mathbb{Z}. In the interval [0,314][0, 314], since π≈3.14159\pi \approx 3.14159, 100π≈314.159100\pi \approx 314.159, so the solutions are x=0,π,2π,…,99πx = 0, \pi, 2\pi, \dots, 99\pi. The sum is 0+π+2π+⋯+99π=99×1002π=4950π0 + \pi + 2\pi + \dots + 99\pi = \frac{99 \times 100}{2} \pi = 4950\pi. Hence, k=4950k = 4950.
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