Consider a function 𝑓: (0,1) β†’ {0, 1} defined as follows. For a real number…

2026

Consider a function 𝑓: (0,1) β†’ {0, 1} defined as follows. For a real number π‘Ÿ ∈ (0,1) , 𝑓(π‘Ÿ) = 1 if the second digit after the decimal point in π‘Ÿ is one of the four digits 2, 3, 6 and 7. Otherwise, 𝑓(π‘Ÿ) is equal to 0. The number of points in (0,1) at which 𝑓 is discontinuous is ___________. (answer in integer)

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Correct answer: 40

Write a number r in (0, 1) as 0.d1d2d3... . The value of f depends only on the second digit d2.

On every open interval (k/100, (k+1)/100), the first two decimal digits are fixed, so f is constant there. Therefore discontinuities can occur only at the boundary points k/100, where k = 1, 2, ..., 99.

Across such a boundary, the second decimal digit changes from (k - 1) mod 10 on the left to k mod 10 on the right. The set for which f = 1 is {2, 3, 6, 7}. The function value changes exactly at the transitions
1 -> 2, 3 -> 4, 5 -> 6, and 7 -> 8.

Thus, in each block of ten hundredths there are 4 discontinuity points. There are 10 such blocks from 0.00 to 1.00, so the total number of discontinuities is 10 x 4 = 40.

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