How many points of discontinuity does f have in (0,1) in GATE CS 2026 Set-2 Q64?

The answer is 1. The function f(r) depends on the second digit after the decimal point. For most real numbers, this digit is well-defined and f is locally constant (hence continuous). The only problematic points are those where the second decimal digit is ambiguous due to the dual decimal representation — numbers of the form r = k/100 for integer k, where the trailing 9-repeating representation causes ambiguity. Among all such boundary points in (0,1), exactly 1 is a genuine point of discontinuity.

Why is the second decimal digit ambiguous for some real numbers?

Some real numbers have two decimal representations. For example, 0.13 = 0.12999... Both are valid representations of the same number, but they give different second digits (3 vs 2). This dual representation occurs at rational numbers of the form p/100 where p is an integer. At such boundary points, f(r) may be defined differently depending on which representation is used, potentially causing discontinuity.

What is a point of discontinuity in real analysis?

A function f is discontinuous at point r if the limit of f as x approaches r does not equal f(r), or if the limit does not exist. For piecewise or digit-based functions, discontinuities occur at boundary points where arbitrarily nearby values of r give different function values.

Discontinuity of f(r) Second Decimal Digit | GATE CS 2026 Set-2 Q64

Computer Sciences > GATE 2026 SET-2 > Engineering Mathematics

Consider a function 𝑓:(0,1)→{0,1} defined as follows.
For a real number 𝑟∈(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)

Correct : 40

Propagation delay one-way = 3000 × 1000 m × 5 ns/m = 15 ms, so RTT = 30 ms.
Transmission time per byte = 8 bits ÷ 10⁸ bps = 80 ns.
For max utilization, window size W ≥ 1 + 2a, where a = 15 ms ÷ 80 ns = 187,500 → W = 375,001 → needs ⌈log₂(375,001)⌉ = 19 bits.
For no wrap-around in 60 seconds, bytes sent = 10⁸ ÷ 8 × 60 = 750,000,000 → needs ⌈log₂(750,000,000)⌉ = 30 bits.
N must satisfy both conditions, so N = max(19, 30) = 30.
The correct answer is N = 30.

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