Calculus > Continuity & Differentiability
Medium
core
Match List-I with List-II $\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) } f(x) = |x| & \text{(I) Not differentiable at } x=-2 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) } f(x) = |x+2| & \text{(II) Not differentiable at } x=0 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) } f(x) = |x^2-4| & \text{(III) Not differentiable at } x=2 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) } f(x) = |x-2| & \text{(IV) Not differentiable at } x=2,-2 \text{ only} \\[1.2ex] \hline \end{array}$ Choose the correct answer from the options given below:
✅ Correct Option: 2
Related questions:
2025: 21 May Shift 2
2023: 15 June Shift 2
2026: 29th May Shift 1
2023: 23 May Shift 3
2025: 16 May Shift 1
2025: 21 May Shift 1