Verified Solution • 2025–26 Session moderate

QJEEM18Ma5868: JEE Main 2018 Mathematics - Class 12 JEEM18Ma5868 Solution: equals :

lim⁡x→0xtan⁡2x−2xtan⁡x(1−cos⁡2x)2\mathop {\lim }\limits_{x \to 0} {{x\tan 2x - 2x\tan x} \over {{{\left( {1 - \cos 2x} \right)}^2}}} equals :

Multiple Choice Options

  • A 14{1 \over 4}
  • B 1
  • C 12{1 \over 2}
  • D −- 12{1 \over 2}

Step-by-Step Verified Solution

Let, L = lim⁡x→0(xtan⁡2x−2xtan⁡x)(1−cos⁡2x)2\mathop {\lim }\limits_{x \to 0} {{\left( {x\tan 2x - 2x\tan x} \right)} \over {{{\left( {1 - \cos 2x} \right)}^2}}} = lim⁡x→0K\mathop {\lim }\limits_{x \to 0} K (say)

⇒\Rightarrow K = x[2tan⁡x1−(tan⁡x)2]−2xtan⁡x(1−(1−2sin⁡2x))2{{x\left[ {{{2\tan x} \over {1 - {{\left( {\tan x} \right)}^2}}}} \right] - 2x\tan x} \over {{{\left( {1 - \left( {1 - 2{{\sin }^2}x} \right)} \right)}^2}}}

= 2xtan⁡x−[2xtan⁡x−2xtan⁡3x]4sin⁡4x×(1−tan⁡2x){{2x\tan x - \left[ {2x\tan x - 2x{{\tan }^3}x} \right]} \over {4{{\sin }^4}x \times (1 - {{\tan }^2}x)}}

= 2xtan⁡3x4sin⁡4x×(1−tan⁡2x){{2x{{\tan }^3}x} \over {4{{\sin }^4}x \times \left( {1 - {{\tan }^2}x} \right)}}

= 2xtan⁡3x4sin⁡4x×(cos⁡2x−sin⁡2xcos⁡2x){{2x{{\tan }^3}x} \over {4{{\sin }^4}x \times \left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {{{\cos }^2}x}}} \right)}}

= 2xsin⁡3xcos⁡3x4sin⁡4x×(cos⁡2x−sin⁡2xcos⁡2x){{2x{{{{\sin }^3}x} \over {{{\cos }^3}x}}} \over {4{{\sin }^4}x \times \left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {{{\cos }^2}x}}} \right)}}

⇒\Rightarrow K = x2sin⁡x×(cos⁡2x−sin⁡2x)cos⁡x{x \over {2\sin x \times ({{\cos }^2}x - {{\sin }^2}x)\cos x}}

∴   \therefore\,\,\, L = lim⁡x→0\mathop {\lim }\limits_{x \to 0} x2sin⁡x×lim⁡x→01cos⁡x(cos⁡2x−sin⁡2x){x \over {2\sin x}} \times \mathop {\lim }\limits_{x \to 0} {1 \over {\cos x({{\cos }^2}x - {{\sin }^2}x)}}

= lim⁡x→0\mathop {\lim }\limits_{x \to 0} x2sin⁡x×lim⁡x→01cos⁡0(cos⁡20−sin⁡20){x \over {2\sin x}} \times \mathop {\lim }\limits_{x \to 0} {1 \over {\cos 0\left( {{{\cos }^2}0 - {{\sin }^2}0} \right)}}

= 12{1 \over 2}

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