JEE Main 201911 Jan, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002Let f(x)={−1−2≤x<0x2−1,0≤x≤2f\left( x \right) = \left\{ {\matrix{ { - 1} & { - 2 \le x < 0} \cr {{x^2} - 1,} & {0 \le x \le 2} \cr } } \right.f(x)={−1x2−1,−2≤x<00≤x≤2 and g(x)=∣f(x)∣+f(∣x∣).g(x) = \left| {f\left( x \right)} \right| + f\left( {\left| x \right|} \right).g(x)=∣f(x)∣+f(∣x∣). Then, in the interval (–2, 2), g is :AOption A. non continuousBOption B. differentiable at all pointsCOption C. not differentiable at two pointsDOption D. not differentiable at one pointCheckSkip · show solution
