JEE Main 2019DifferentiationDifferentiationMediumQuestionLet f:(−1,1)→Rf:\left( { - 1,1} \right) \to Rf:(−1,1)→R be a differentiable function with f(0)=−1f\left( 0 \right) = - 1f(0)=−1 and f′(0)=1f'\left( 0 \right) = 1f′(0)=1. Let g(x)=[f(2f(x)+2)]2g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}g(x)=[f(2f(x)+2)]2. Then g′(0)=g'\left( 0 \right) = g′(0)=OptionsA−4-4−4B000C−2-2−2D444Check AnswerHide SolutionSolutiong′(x)=2(f(2f(x)+2))g'\left( x \right) = 2\left( {f\left( {2f\left( x \right) + 2} \right)} \right)g′(x)=2(f(2f(x)+2)) (ddx(f(2f(x)+2)))\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{d \over {dx}}\left( {f\left( {2f\left( x \right) + 2} \right)} \right)} \right)(dxd(f(2f(x)+2))) =2f(2f(x)+2)f′(2f(x)) = 2f\left( {2f\left( x \right) + 2} \right)f'\left( {2f\left( x \right)} \right)=2f(2f(x)+2)f′(2f(x)) +2).(2f′(x))\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left. 2 \right).\left( {2f'\left( x \right)} \right)+2).(2f′(x)) ⇒g′(0)=2f(2f(0)+2). \Rightarrow g'\left( 0 \right) = 2f\left( {2f\left( 0 \right) + 2} \right).⇒g′(0)=2f(2f(0)+2). f′(2f(0)+2).2f′(0)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( {2f\left( 0 \right) + 2} \right).2f'\left( 0 \right)f′(2f(0)+2).2f′(0) =4f(0)(f′(0))2 = 4f\left( 0 \right){\left( {f'\left( 0 \right)} \right)^2}=4f(0)(f′(0))2 =4(−1)(1)2=−4 = 4\left( { - 1} \right){\left( 1 \right)^2} = - 4=4(−1)(1)2=−4