JEE Main 2023DifferentiationDifferentiationHardQuestionLet g:R→Rg: \mathbf{R} \rightarrow \mathbf{R}g:R→R be a non constant twice differentiable function such that g′(12)=g′(32)\mathrm{g}^{\prime}\left(\frac{1}{2}\right)=\mathrm{g}^{\prime}\left(\frac{3}{2}\right)g′(21)=g′(23). If a real valued function fff is defined as f(x)=12[g(x)+g(2−x)]f(x)=\frac{1}{2}[g(x)+g(2-x)]f(x)=21[g(x)+g(2−x)], thenOptionsAf′′(x)=0f^{\prime \prime}(x)=0f′′(x)=0 for atleast two xxx in (0,2)(0,2)(0,2)Bf′(32)+f′(12)=1f^{\prime}\left(\frac{3}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)=1f′(23)+f′(21)=1Cf′′(x)=0f^{\prime \prime}(x)=0f′′(x)=0 for no xxx in (0,1)(0,1)(0,1)Df′′(x)=0f^{\prime \prime}(x)=0f′′(x)=0 for exactly one xxx in (0,1)(0,1)(0,1)Check AnswerHide SolutionSolutionf′(x)=g′(x)−g′(2−x)2,f′(32)=g′(32)−g′(12)2=0f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(2-x)}{2}, f^{\prime}\left(\frac{3}{2}\right)=\frac{g^{\prime}\left(\frac{3}{2}\right)-g^{\prime}\left(\frac{1}{2}\right)}{2}=0f′(x)=2g′(x)−g′(2−x),f′(23)=2g′(23)−g′(21)=0 Also f′(12)=g′(12)−g′(32)2=0,f′(12)=0\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\frac{\mathrm{g}^{\prime}\left(\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{3}{2}\right)}{2}=0, \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0f′(21)=2g′(21)−g′(23)=0,f′(21)=0 ⇒f′(32)=f′(12)=0\Rightarrow \mathrm{f}^{\prime}\left(\frac{3}{2}\right)=\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0⇒f′(23)=f′(21)=0 ⇒rootsin(12,1)\Rightarrow \operatorname{rootsin}\left(\frac{1}{2}, 1\right)⇒rootsin(21,1) and (1,32)\left(1, \frac{3}{2}\right)(1,23) ⇒f′′(x)\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})⇒f′′(x) is zero at least twice in (12,32)\left(\frac{1}{2}, \frac{3}{2}\right)(21,23)