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QJEEM10Mac8e2: JEE Main 2010 Mathematics - Class 12 JEEM10Mac8e2 Solution: Let be a differentiable function with and . Let . Then

Let f:(−1,1)→Rf:\left( { - 1,1} \right) \to R be a differentiable function with f(0)=−1f\left( 0 \right) = - 1 and f′(0)=1f'\left( 0 \right) = 1 . Let g(x)=[f(2f(x)+2)]2g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2} . Then g′(0)=g'\left( 0 \right) =

Multiple Choice Options

  • A −4-4
  • B 00
  • C −2-2
  • D 44

Step-by-Step Verified Solution

g′(x)=2(f(2f(x)+2))g'\left( x \right) = 2\left( {f\left( {2f\left( x \right) + 2} \right)} \right)

                           (ddx(f(2f(x)+2)))\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{d \over {dx}}\left( {f\left( {2f\left( x \right) + 2} \right)} \right)} \right)

=2f(2f(x)+2)f′(2f(x))= 2f\left( {2f\left( x \right) + 2} \right)f'\left( {2f\left( x \right)} \right)

                           +2).(2f′(x))\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left. 2 \right).\left( {2f'\left( x \right)} \right)

⇒g′(0)=2f(2f(0)+2).\Rightarrow g'\left( 0 \right) = 2f\left( {2f\left( 0 \right) + 2} \right).

                           f′(2f(0)+2).2f′(0)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( {2f\left( 0 \right) + 2} \right).2f'\left( 0 \right)

=4f(0)(f′(0))2= 4f\left( 0 \right){\left( {f'\left( 0 \right)} \right)^2}

=4(−1)(1)2=−4= 4\left( { - 1} \right){\left( 1 \right)^2} = - 4

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