JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Continuity and Differentiability

Rewrite $f(x)=|x-1|\cos |x-2|\sin |x-1|+(x-3)|x-1||x-4|$.

Factor out $|x-1|$:


At $x=1$: The factor $|x-1|$ makes the function potentially non-differentiable. Check if the bracketed expression is zero at $x=1$: $\cos 1\cdot \sin 0+(-2)\cdot 3=-6\ne 0$. So $f$ is non-differentiable at $x=1$.

At $x=2$: $f(2)=1\cdot \cos 0\cdot \sin 1+(-1)\cdot 2=\sin 1-2$. The function is smooth at $x=2$ (the absolute value $|x-2|$ inside cosine is differentiable when the argument is non-zero).

At $x=4$: $f(4)=3[\cos 2\sin 3+1\cdot 0]=3\cos 2\sin 3$. Check differentiability: the factor $|x-4|$ multiplied by $(x-3)$ and $|x-1|$ creates a non-differentiability at $x=4$.

Therefore, the function is non-differentiable at exactly 2 points: $x=1$ and $x=4$.