GATE CS 2026 Set 2 Q.64: Calculus

Consider a function $f: (0,1) \rightarrow \{0, 1\}$ defined as follows.

For a real number $r \in (0,1)$ , $f(r) = 1$ if the second digit after the decimal point in $r$ is one of the four digits 2, 3, 6 and 7. Otherwise, $f(r)$ is equal to 0.

The number of points in $(0,1)$ at which $f$ is discontinuous is ___________. (answer in integer)

Your answer

Official answer and solution

Official answer: 40
From: Official GATE 2026 answer key

Solution

Identify the interval boundaries in $(0, 1)$ where the second decimal digit transitions into or out of $\{2, 3\}$ or $\{6, 7\}$ .

For each of the $10$ choices of the first decimal digit $d_1 \in \{0, 1, \dots, 9\}$ , the second decimal digit $d_2$ belongs to $\{2, 3, 6, 7\}$ on two disjoint sub-intervals of $[0.d_1, 0.d_1 + 0.1)$ :

$[0.d_1 2, \; 0.d_1 4)$ , where $d_2 \in \{2, 3\}$ , which has 2 boundary points ( $0.d_1 2$ and $0.d_1 4$ ).
$[0.d_1 6, \; 0.d_1 8)$ , where $d_2 \in \{6, 7\}$ , which has 2 boundary points ( $0.d_1 6$ and $0.d_1 8$ ).

Inside each open interval, $f$ is constant ( $1$ or $0$ ), and at each of the $4$ boundary points per first digit $d_1$ , $f$ jumps between $0$ and $1$ . All $10 \times 4 = 40$ boundary points lie strictly inside $(0, 1)$ (the smallest is $0.02$ and the largest is $0.98$ ). Thus $f$ is discontinuous at $40$ points.

Final answer

Common trap

Counting every single digit transition ( $2, 3, 4, 6, 7, 8$ ) without noticing that $2$ and $3$ are adjacent ( $f=1$ across $0.d_1 3$ ) and $6$ and $7$ are adjacent ( $f=1$ across $0.d_1 7$ ).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2026 Computer Science and Information Technology - Session 2 (CS2)
Question: 64 of 65
Section: Computer Science and Information Technology

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