JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Continuity and Differentiability

Write $f(x)=|2x+1|-3|x+2|+|(x+2)(x-1)|$. The only candidates for a corner are the zeros of the moduli, $x=-2$, $x=-\frac{1}{2}$ and $x=1$. At $x=-\frac{1}{2}$ only $|2x+1|$ turns, so the one-sided derivatives differ by $4$ and $f$ is not differentiable there. At $x=1$ only $|(x+2)(x-1)|$ turns, and its derivative jumps from $-(2x+1)=-3$ to $2x+1=3$, so $f$ is not differentiable there either. At $x=-2$ two moduli turn together: $-3|x+2|$ has derivative $+3$ on the left and $-3$ on the right, while $|(x+2)(x-1)|$ has derivative $2x+1=-3$ on the left and $-(2x+1)=3$ on the right. The two jumps cancel and $f^{\prime }(-2^{-})=f^{\prime }(-2^{+})=-2$, so $f$ is differentiable at $x=-2$. The number of points of non-differentiability is $2$.