Verified Solution • 2025–26 Session moderate

QJEEM19Maac91: JEE Main 2019 Mathematics - Class 12 JEEM19Maac91 Solution: is :

lim⁡x→π/4cot⁡3x−tan⁡xcos⁡(x+π4)\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}} is :

Multiple Choice Options

  • A 828\sqrt 2
  • B 4
  • C 424\sqrt 2
  • D 8

Step-by-Step Verified Solution

lim⁡x→π/4cot⁡3x−tan⁡xcos⁡(x+π4)\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}

= lim⁡x→π41tan⁡3x−tan⁡xcos⁡(x+π4)\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{1 \over {{{\tan }^3}x}} - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}

= lim⁡x→π41−tan⁡4x(tan⁡3x)cos⁡(x+π4)\mathop {\lim }\limits_{x \to {\pi \over 4}} {{1 - {{\tan }^4}x} \over {\left( {{{\tan }^3}x} \right)\cos \left( {x + {\pi \over 4}} \right)}}

= lim⁡x→π4(1−tan⁡2x)(1+tan⁡2x)(tan⁡3x)cos⁡(x+π4)\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + {{\tan }^2}x} \right)} \over {\left( {{{\tan }^3}x} \right)\cos \left( {x + {\pi \over 4}} \right)}}

= lim⁡x→π4(1−tan⁡2x)(1+tan⁡2π4)(tan⁡3π4)cos⁡(x+π4)\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + {{\tan }^2}{\pi \over 4}} \right)} \over {\left( {{{\tan }^3}{\pi \over 4}} \right)\cos \left( {x + {\pi \over 4}} \right)}}

= lim⁡x→π4(1−tan⁡2x)(1+1)1.cos⁡(x+π4)\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + 1} \right)} \over {1.\cos \left( {x + {\pi \over 4}} \right)}}

= lim⁡x→π42(1−tan⁡2x)(cos⁡x−sin⁡x2)\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\left( {1 - {{\tan }^2}x} \right)} \over {\left( {{{\cos x - \sin x} \over {\sqrt 2 }}} \right)}}

= lim⁡x→π422(1−sin⁡2xcos⁡2x)(cos⁡x−sin⁡x)\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {1 - {{{{\sin }^2}x} \over {{{\cos }^2}x}}} \right)} \over {\left( {\cos x - \sin x} \right)}}

= lim⁡x→π422(cos⁡2x−sin⁡2x)cos⁡2x(cos⁡x−sin⁡x)\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {{{\cos }^2}x - {{\sin }^2}x} \right)} \over {{{\cos }^2}x\left( {\cos x - \sin x} \right)}}

= lim⁡x→π422(cos⁡x+sin⁡x)(cos⁡x−sin⁡x)cos⁡2x(cos⁡x−sin⁡x)\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos x + \sin x} \right)\left( {\cos x - \sin x} \right)} \over {{{\cos }^2}x\left( {\cos x - \sin x} \right)}}

= lim⁡x→π422(cos⁡x+sin⁡x)cos⁡2x\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos x + \sin x} \right)} \over {{{\cos }^2}x}}

= lim⁡x→π422(cos⁡π4+sin⁡π4)cos⁡2π4\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos {\pi \over 4} + \sin {\pi \over 4}} \right)} \over {{{\cos }^2}{\pi \over 4}}}

= 22(12+12)12{{2\sqrt 2 \left( {{1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }}} \right)} \over {{1 \over 2}}}

= 22(22)12{{2\sqrt 2 \left( {{2 \over {\sqrt 2 }}} \right)} \over {{1 \over 2}}}

= 8

Other Method :

lim⁡x→π/4cot⁡3x−tan⁡xcos⁡(x+π4)\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}

This is in 00{0 \over 0} form so L' Hospital rule is applicable.

= lim⁡x→π43cot⁡3x(−cos⁡ec2x)−sec⁡2x−sin⁡(x+π4)\mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\cot }^3}x\left( { - \cos e{c^2}x} \right) - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}

= 3×1(−(2)2)−(2)2−1{{3 \times 1\left( { - {{\left( {\sqrt 2 } \right)}^2}} \right) - {{\left( {\sqrt 2 } \right)}^2}} \over { - 1}}

= −6−2−1{{ - 6 - 2} \over { - 1}} = 8

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