Verified Solution • 2025–26 Session moderate

QJEEM19Ma55f3: JEE Main 2019 Mathematics - Class 12 JEEM19Ma55f3 Solution: If , x then is equal to:

If 2y=(cot⁡−1(3cos⁡x+sin⁡xcos⁡x−3sin⁡x))22y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2} ,

x ∈\in (0,π2)\left( {0,{\pi \over 2}} \right) then dydxdy \over dx is equal to:

Multiple Choice Options

  • A 2x−π32x - {\pi \over 3}
  • B π6−x{\pi \over 6} - x
  • C π3−x{\pi \over 3} - x
  • D x−π6x - {\pi \over 6}

Step-by-Step Verified Solution

2y=(cot⁡−1(3cos⁡x+sin⁡xcos⁡x−3sin⁡x))22y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}

⇒\Rightarrow 2y = (cot⁡−1(3+tan⁡x1−3tan⁡x))2{\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 + \tan x} \over {1 - \sqrt 3 \tan x}}} \right)} \right)^2}

⇒\Rightarrow 2y = (cot⁡−1(tan⁡π3+tan⁡x1−tan⁡π3tan⁡x))2{\left( {{{\cot }^{ - 1}}\left( {{{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}} \right)} \right)^2}

⇒\Rightarrow 2y = (cot⁡−1tan⁡(π3+x))2{\left( {{{\cot }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}

⇒\Rightarrow 2y = (π2−tan⁡−1tan⁡(π3+x))2{\left( {{\pi \over 2} - {{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}

As x ∈\in (0,π2)\left( {0,{\pi \over 2}} \right) then

tan⁡−1tan⁡(π3+x){{{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} = (π3+x){\left( {{\pi \over 3} + x} \right)}

⇒\Rightarrow 2y = (π2−(π3+x))2{\left( {{\pi \over 2} - \left( {{\pi \over 3} + x} \right)} \right)^2}

⇒\Rightarrow 2y = (π6−x)2{\left( {{\pi \over 6} - x} \right)^2}

∴\therefore 2dydx=2(π6−x)(−1)2{{dy} \over {dx}} = 2\left( {{\pi \over 6} - x} \right)\left( { - 1} \right)

⇒\Rightarrow dydx=(x−π6){{dy} \over {dx}} = \left( {x - {\pi \over 6}} \right)

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