Differential Calculus
Non-differentiable points of abs-value function
MJMT_Full_Test_04
Grade 12
Question:
The set of all real values of $x$ where $f(x)=\sin|x|-|x|+2(x-\pi)\cos|x|$ is not differentiable is
Null set
$\{\pi\}$
$\{0\}$
$\{0,\pi\}$
Step-by-Step Solution
Key Concept: For $x>0$ and $x<0$, simplify $f$ separately. At $x=\pi$: $\cos|\pi|=\cos\pi=-1$; both branches agree and LHD=RHD. At $x=0$: check LHD vs RHD.
Non-differentiable set $=\{0\}$.
Correct Answer: 3