Verified Solution • 2025–26 Session moderate

QJEEM20Ma1fe5: JEE Main 2020 Mathematics - Class 12 JEEM20Ma1fe5 Solution: The derivative of with respect to at x = is :

The derivative of
tan⁡−1(1+x2−1x){\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right) with
respect to tan⁡−1(2x1−x21−2x2){\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right) at x = 12{1 \over 2} is :

Multiple Choice Options

  • A 233{{2\sqrt 3 } \over 3}
  • B 235{{2\sqrt 3 } \over 5}
  • C 310{{\sqrt 3 } \over {10}}
  • D 312{{\sqrt 3 } \over {12}}

Step-by-Step Verified Solution

Let f = tan⁡−1(1+x2−1x){\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)

Put x = tan θ\theta ⇒\Rightarrow θ\theta = tan–1 x

f = tan⁡−1(sec⁡θ−1tan⁡θ){\tan ^{ - 1}}\left( {{{\sec \theta - 1} \over {\tan \theta }}} \right)

⇒\Rightarrow f = tan⁡−1(1−cos⁡θsin⁡θ){\tan ^{ - 1}}\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right) = θ2{\theta \over 2}

⇒\Rightarrow f = tan⁡−1x2{{{{\tan }^{ - 1}}x} \over 2}

∴\therefore dfdx{{df} \over {dx}} = 12(1+x2){1 \over {2\left( {1 + {x^2}} \right)}} ....(1)

Let g = tan⁡−1(2x1−x21−2x2){\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)

Put x = sin θ\theta ⇒\Rightarrow θ\theta = sin–1 x

⇒\Rightarrow g = tan⁡−1(2sin⁡θcos⁡θ1−2sin⁡2θ){\tan ^{ - 1}}\left( {{{2\sin \theta \cos \theta } \over {1 - 2{{\sin }^2}\theta }}} \right)

⇒\Rightarrow g = tan–1 (tan 2 θ\theta ) = 2 θ\theta

⇒\Rightarrow g = 2sin-1 x

⇒\Rightarrow dgdx{{dg} \over {dx}} = 21−x2{2 \over {\sqrt {1 - {x^2}} }} ...(2)

Using (i) and (ii),

∴\therefore dfdg{{df} \over {dg}} = 12(1+x2)1−x22{1 \over {2\left( {1 + {x^2}} \right)}}{{\sqrt {1 - {x^2}} } \over 2}

At x = 12{1 \over 2} , (dfdg)x=12{\left( {{{df} \over {dg}}} \right)_{x = {1 \over 2}}} = 310{{\sqrt 3 } \over {10}}

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