JEE Advanced 2020 — Paper 2 — Question 38

Compute $f(x) = |2x-1|+|2x+1|$ piecewise: $f(x) = \begin{cases} -4x, & x \le -\frac12 \\ 2, & -\frac12 \le x \le \frac12 \\ 4x, & x \ge \frac12 \end{cases}$. $g(x) = x-[x] = \{x\}$ on $(-1,1)$: $g(x) = \begin{cases} x+1, & -1<x<0 \\ x, & 0 \le x <1 \end{cases}$. Composite $f(g(x)) = \begin{cases} 2, & -1<x<-\frac12 \\ 4x+4, & -\frac12<x<0 \\ 2, & 0\le x<\frac12 \\ 4x, & \frac12<x<1 \end{cases}$. Check continuity at $x=-\frac12,0,\frac12$. At $x=0$ left limit $4$ eq right limit $2$ so discontinuous. At $x=-\frac12$ and $\frac12$ limits equal function values, so continuous. Thus $c=1$ (point $x=0$). Check differentiability: $f(g)$ is differentiable where it is continuous and derivative is same from both sides. At $x=-\frac12$ left derivative $0$ eq right derivative $4$, so not differentiable. At $x=0$ discontinuous so not differentiable. At $x=\frac12$ left derivative $0$ eq right derivative $4$, so not differentiable. Thus $d=3$ (points $x=-\frac12,0,\frac12$). Hence $c+d = 1+3 = 4$.