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QJEEM21Ma153d: JEE Main 2021 Mathematics - Class 11 JEEM21Ma153d Solution: If satisfies the equation t 2 - 9t + 8 = 0, then the value of…

If e(cos⁡2x+cos⁡4x+cos⁡6x+...∞)log⁡e2{e^{\left( {{{\cos }^2}x + {{\cos }^4}x + {{\cos }^6}x + ...\infty } \right){{\log }_e}2}} satisfies the equation t2 - 9t + 8 = 0, then the value of
{{2\sin x} \over {\sin x + \sqrt 3 \cos x}}\left( {0 < x < {\pi \over 2}} \right) is :

Multiple Choice Options

  • A 3\sqrt 3
  • B 32{3 \over 2}
  • C 2 3\sqrt 3
  • D 12{1 \over 2}

Step-by-Step Verified Solution

e(cos⁡2x+cos⁡4x+...........∞)ln⁡2=2cos⁡2x+cos⁡4x+...........∞{e^{({{\cos }^2}x + {{\cos }^4}x + ...........\infty )\ln 2}} = {2^{{{\cos }^2}x + {{\cos }^4}x + ...........\infty }}

= 2cos⁡2x1−cos⁡2x{2^{{{{{\cos }^2}x} \over {1 - {{\cos }^2}x}}}}

=2cot⁡2x= {2^{{{\cot }^2}x}}

Given, t2−9t+8=0⇒t=1,8{t^2} - 9t + 8 = 0 \Rightarrow t = 1,8

⇒2cot⁡2x=1,8⇒cot2x=0,3\Rightarrow {2^{{{\cot }^2}x}} = 1,8 \Rightarrow co{t^2}x = 0,3

0 < x < {\pi \over 2} \Rightarrow \cot x = \sqrt 3

∴\therefore 2sin⁡xsin⁡x+3cos⁡x=21+3cot⁡x=24=12{{2\sin x} \over {\sin x + \sqrt 3 \cos x}} = {2 \over {1 + \sqrt 3 \cot x}} = {2 \over 4} = {1 \over 2}

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