GATE CS 2026 Set 2 Q54 — Calculus: Limits, Continuity, Differentiability

The value of $f(r)$ depends only on the second decimal digit $d_2$ of $r = 0.d_1 d_2 d_3 \dots$ (using standard decimal representation without infinite trailing $9$s).

On any open interval of the form $(0.d_1 d_2, \, 0.d_1 (d_2+1))$, the second digit $d_2$ is constant. Thus, $f(r)$ is constant and continuous on these intervals. The only possible points of discontinuity in $(0,1)$ are the boundary points of the form $x = 0.d_1 d_2$ where $d_1 \in \{0, 1, \dots, 9\}$ and $d_2 \in \{0, 1, \dots, 9\}$.

For any boundary point $x = 0.d_1 d_2$:

• For $r$ slightly larger than $x$, the second digit is $d_2$, so $\lim_{r \to x^+} f(r) = f(x)$.
• For $r$ slightly smaller than $x$, the second digit is $d_2 - 1$ (or $9$ if $d_2 = 0$, with the first digit being $d_1 - 1$).

A discontinuity occurs at $x$ if and only if the value of $f$ changes across this boundary. Let us analyze the transitions of the second digit:

• For $d_2 = 0$: The transition is from $9$ to $0$. Since both $9, 0 \notin \{2,3,6,7\}$, $f(r)$ remains $0$ on both sides. Thus, $f$ is continuous at $0.d_1 0$.
• For $d_2 \in \{1, \dots, 9\}$ within a fixed $d_1$:
  • $0 \to 1$: $f$ goes $0 \to 0$ (continuous)
  • $1 \to 2$: $f$ goes $0 \to 1$ (discontinuous)
  • $2 \to 3$: $f$ goes $1 \to 1$ (continuous)
  • $3 \to 4$: $f$ goes $1 \to 0$ (discontinuous)
  • $4 \to 5$: $f$ goes $0 \to 0$ (continuous)
  • $5 \to 6$: $f$ goes $0 \to 1$ (discontinuous)
  • $6 \to 7$: $f$ goes $1 \to 1$ (continuous)
  • $7 \to 8$: $f$ goes $1 \to 0$ (discontinuous)
  • $8 \to 9$: $f$ goes $0 \to 0$ (continuous)

Thus, for each of the $10$ possible values of $d_1 \in \{0, 1, \dots, 9\}$, there are exactly $4$ points of discontinuity corresponding to $d_2 \in \{2, 4, 6, 8\}$.

The total number of points of discontinuity in $(0,1)$ is: $$10 \times 4 = 40$$