JEE Main 202427 Jan, Evening shiftMathematicsHardLimit, Continuity and Differentiability420× since 2002Consider the function f:(0,2)→Rf:(0,2) \rightarrow \mathbf{R}f:(0,2)→R defined by f(x)=x2+2xf(x)=\frac{x}{2}+\frac{2}{x}f(x)=2x+x2 and the function g(x)g(x)g(x) defined by g(x)={min⌊f(t)},0<t≤x and 0<x≤132+x,1<x<2. Then, g(x)=\left\{\begin{array}{ll} \min \lfloor f(t)\}, & 0<\mathrm{t} \leq x \text { and } 0 < x \leq 1 \\ \frac{3}{2}+x, & 1 < x < 2 \end{array} .\right. \text { Then, }g(x)={min⌊f(t)},23+x,0<t≤x and 0<x≤11<x<2. Then, AOption A. ggg is continuous but not differentiable at x=1x=1x=1BOption B. ggg is continuous and differentiable for all x∈(0,2)x \in(0,2)x∈(0,2)COption C. ggg is not continuous for all x∈(0,2)x \in(0,2)x∈(0,2)DOption D. ggg is neither continuous nor differentiable at x=1x=1x=1CheckSkip · show solution
