JEE Main 20231 Feb, Morning shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002Let 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]. ThenAOption A. there 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)BOption B. there 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)COption C. minf′(x)=1+maxg′(x)\min f'(x) = 1 + \max g'(x)minf′(x)=1+maxg′(x)DOption D. maxf(x)>maxg(x)\max f(x) > \max g(x)maxf(x)>maxg(x)CheckSkip · show solution
