JEE Main 202118 Mar, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002If f(x)={1∣x∣; ∣x∣ ≥1ax2+b; ∣x∣ <1f(x) = \left\{ {\matrix{ {{1 \over {|x|}}} & {;\,|x|\, \ge 1} \cr {a{x^2} + b} & {;\,|x|\, < 1} \cr } } \right.f(x)={∣x∣1ax2+b;∣x∣≥1;∣x∣<1 is differentiable at every point of the domain, then the values of a and b are respectively :AOption A. 12,12{1 \over 2},{1 \over 2}21,21BOption B. 12,−32{1 \over 2}, - {3 \over 2}21,−23COption C. 52,−32{5 \over 2}, - {3 \over 2}25,−23DOption D. −12,32- {1 \over 2},{3 \over 2}−21,23CheckSkip · show solution
