Verified Solution • 2025–26 Session moderate

QJEEM17Ma737d: JEE Main 2017 Mathematics - Class 12 JEEM17Ma737d Solution: If for , the derivatives of is , then equals

If for x∈(0,14)x \in \left( {0,{1 \over 4}} \right) , the derivatives of

tan⁡−1(6xx1−9x3){\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right) is x.g(x)\sqrt x .g\left( x \right) , then g(x)g\left( x \right) equals

Multiple Choice Options

  • A 3xx1−9x3{{{3x\sqrt x } \over {1 - 9{x^3}}}}
  • B 3x1−9x3{{{3x} \over {1 - 9{x^3}}}}
  • C 31+9x3{{3 \over {1 + 9{x^3}}}}
  • D 91+9x3{{9 \over {1 + 9{x^3}}}}

Step-by-Step Verified Solution

Let y = tan⁡−1(6xx1−9x3){\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)

= tan⁡−1[2.(3x32)1−(3x32)2]{\tan ^{ - 1}}\left[ {{{2.\left( {3{x^{{3 \over 2}}}} \right)} \over {1 - {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}} \right]

= 2 tan⁡−1(3x32){\tan ^{ - 1}}\left( {3{x^{{3 \over 2}}}} \right)

∴\therefore dydx=2.11+(3x32)2.3×32(x)12{{dy} \over {dx}} = 2.{1 \over {1 + {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}.3 \times {3 \over 2}{\left( x \right)^{{1 \over 2}}}

= 91+9x3.x{9 \over {1 + 9{x^3}}}.\sqrt x

∴\therefore g(x) = 91+9x3{9 \over {1 + 9{x^3}}}

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