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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

If 2y = ( ^ -1 ( 3 cos x + sin x cos x - 3 sin x ))^ 2 x ∈ (0,...

If 2y=(cot⁡−1(3cosx+sinxcosx−3sinx))2∀x∈(0,π2),2y = \left(\cot^{-1} \left(\dfrac{\sqrt{3}cos x + sin x}{cos x - \sqrt{3}sin x}\right)\right)^{2} \forall x \in \left(0, \dfrac{\pi}{2}\right), then dydx\dfrac{dy}{dx} is equal to

Options

  1. Aπ6−x\dfrac{\pi}{6} - x
  2. B2x−π32x - \dfrac{\pi}{3}
  3. Cx−π6x - \dfrac{\pi}{6}
  4. Dπ3−x\dfrac{\pi}{3} - x

Correct answer

C. x−π6x - \dfrac{\pi}{6}

Step-by-step solution

2y=(cot⁡−13cos⁡x+sin⁡xcosx−3sinx)22y = \left(\cot^{- 1} \dfrac{\sqrt{3}\cos x + \sin x}{cos x - \sqrt{3}sin x}\right)2
=(cot⁡−1(3+tan⁡x1−3tan⁡x))2= \left(\cot^{- 1} \left(\dfrac{\sqrt{3} + \tan x}{1 - \sqrt{3}\tan x}\right)\right)2
=(cot⁡−1tan⁡(π3+x))2= {\left(\cot^{- 1} \tan \left(\dfrac{\pi}{3} + x\right)\right)}^{2}
=(cot⁡−1cot⁡(π2−(π3+x)))2= {\left(\cot^{- 1} \cot \left(\dfrac{\pi}{2} - \left(\dfrac{\pi}{3} + x\right)\right)\right)}^{2}
2y=(π6−x)2y = { \left(\dfrac{\pi}{6} -x\right)}^{2}, x \in \left(0,0,\dfrac{\pi}{6}\right)
=(π+π6−x)2,x∈(π6,π2)= { \left(\pi + \dfrac{\pi}{6} - x\right)}^{2}, x \in \left(\dfrac{\pi}{6}, \dfrac{\pi}{2}\right)
⇒ \dfrac{2dy}{dx} = 2\left(\dfrac{\pi}{6} - x\right) . \left(-1\right) ⇒ dydx=\dfrac{dy}{dx} =x - π6,\dfrac{\pi}{6},x \in \left(0,0,\dfrac{\pi}{6}\right)
\dfrac{dy}{dx} =x−x -\dfrac{7\pi}{6},x∈(π6,π2)x \in \left(\dfrac{\pi}{6}, \dfrac{\pi}{2}\right)
x−π6x - \dfrac{\pi}{6} is matching with one of the given options.
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