JEE Main 2024DifferentiationDifferentiationMediumQuestionLet f(x)=2x+tan−1xf(x) = 2x + {\tan ^{ - 1}}xf(x)=2x+tan−1x and g(x)=loge(1+x2+x),x∈[0,3]g(x) = {\log _e}(\sqrt {1 + {x^2}} + x),x \in [0,3]g(x)=loge(1+x2+x),x∈[0,3]. ThenOptionsAthere exists x^∈[0,3]\widehat x \in [0,3]x∈[0,3] such that f′(x^)<g′(x^)f'(\widehat x) < g'(\widehat x)f′(x)<g′(x)Bthere exist 0<x1<x2<30 < {x_1} < {x_2} < 30<x1<x2<3 such that f(x)<g(x),∀x∈(x1,x2)f(x) < g(x),\forall x \in ({x_1},{x_2})f(x)<g(x),∀x∈(x1,x2)Cminf′(x)=1+maxg′(x)\min f'(x) = 1 + \max g'(x)minf′(x)=1+maxg′(x)Dmaxf(x)>maxg(x)\max f(x) > \max g(x)maxf(x)>maxg(x)Check AnswerHide SolutionSolutionf′(x)=2+11+x2,g′(x)=1x2+1f′′(x)=−2x(1+x2)2<0g′′(x)=−12(x2+1)−3/2⋅2x<0f′(x)∣min=f′(3)=2+110=2110g′(x)∣max=g′(0)=1f′(x)∣max=f(3)=2+tan−13g(x)∣max=g(3)=ln(3+10)<ln<7<2\begin{aligned} & f^{\prime}(x)=2+\frac{1}{1+x^2}, g^{\prime}(x)=\frac{1}{\sqrt{x^2+1}} \\\\ & f^{\prime \prime}(x)=-\frac{2 x}{\left(1+x^2\right)^2}<0 \\\\ & g^{\prime \prime}(x)=-\frac{1}{2}\left(x^2+1\right)^{-3 / 2} \cdot 2 x<0 \\\\ & \left.f^{\prime}(x)\right|_{\min }=f^{\prime}(3)=2+\frac{1}{10}=\frac{21}{10} \\\\ & \left.g^{\prime}(x)\right|_{\max }=g^{\prime}(0)=1 \\\\ & \left.f^{\prime}(x)\right|_{\max }=f(3)=2+\tan ^{-1} 3 \\\\ & \left.g(x)\right|_{\max }=g(3)=\ln (3+\sqrt{10})<\ln <7<2 \end{aligned}f′(x)=2+1+x21,g′(x)=x2+11f′′(x)=−(1+x2)22x<0g′′(x)=−21(x2+1)−3/2⋅2x<0f′(x)∣min=f′(3)=2+101=1021g′(x)∣max=g′(0)=1f′(x)∣max=f(3)=2+tan−13g(x)∣max=g(3)=ln(3+10)<ln<7<2