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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

The value of& x satisfying the equation | x x| + 2 + ^ 2 x + ^ 2 x...

The value of& xx satisfying the equation \left|\sin x \cos x\right| + \sqrt{2 + \tan^{2} x + \cot^{2} x} = \sqrt{3}$&

Options

  1. Abelongs to [0,π3]\left[0, \dfrac{\pi}{3}\right]
  2. Bbelongs to (π3,π2)\left(\dfrac{\pi}{3}, \dfrac{\pi}{2}\right)
  3. Cbelongs to [3π4,π)\left[\dfrac{3\pi}{4}, \pi\right)
  4. Ddoes not exist

Correct answer

D. does not exist

Step-by-step solution

∣sin⁡xcos⁡x∣+2+tan⁡2x+cot⁡2x=3\left|\sin x \cos x\right| + \sqrt{2 + \tan^{2} x + \cot^{2} x} = \sqrt{3}
⇒∣sin⁡xcos⁡x∣+(tanx+cotx)2=3⇒ \left|\sin x \cos x\right| + \sqrt{{\left(tan x + cot x\right)}^{2}} = \sqrt{3}
⇒∣sin⁡xcos⁡x∣+∣tan⁡x+cotx∣=3⇒ \left|\sin x \cos x\right| + \left|\tan x + cot x\right| = \sqrt{3}
⇒12∣sin⁡2x∣+∣tan⁡x+cotx∣=3⇒ \dfrac{1}{2}\left|\sin 2x\right| + \left|\tan x + cot x\right| = \sqrt{3}
⇒∣tanx1+tan⁡2x∣+∣1+tan⁡2xtanx∣=3⇒ \left|\dfrac{tan x}{1 + \tan^{2} x}\right| + \left|\dfrac{1 + \tan^{2} x}{tan x}\right| = \sqrt{3}
Let t=∣tanx1+tan⁡2x∣t = \left|\dfrac{tan x}{1 + \tan^{2} x}\right|
t+1t=3⇒t2−3t+1=0⇒D<0t + \dfrac{1}{t} = \sqrt{3} ⇒ t^{2} - \sqrt{3}t + 1 = 0 ⇒ D < 0
So, no real values of tt
Hence, xx does not exists.
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