JEE Main 2022 (Online) 29th July Morning ShiftQuestion 1
The number of points, where the function $f:\mathbf{R}\to\mathbf{R}$, $f(x)=|x-1|\cos|x-2|\sin|x-1|+(x-3)|x^2-5x+4|$, is NOT differentiable, is :
Differentiation and its Application
Solution
Concept. In the first term every modulus can be removed; in the second, $|(x-1)(x-4)|$ gives corners at $1$ and $4$ unless the factor $(x-3)$ is zero there. Formulas used. (1) $|u|\sin|u|=u\sin u$ and $\cos|u|=\cos u$(2) $h(x)\,|x-a|\,k(x)$ with $h(a)k(a)\ne0$ has a corner at $a$Step 1 - First term. By (1), $|x-1|\sin|x-1|\cos|x-2|=(x-1)\sin(x-1)\cos(x-2)$, which is differentiable everywhere. Step 2 - Second term. $(x-3)\,|x-1|\,|x-4|$. At $x=1$: $(x-3)(x-4)=6\ne0$. At $x=4$: $(x-3)(x-1)=3\ne0$. By (2), corners at both. Step 3 - Count. $f$ is not differentiable at exactly $2$ points. Answer: (B)