Continuity

For each number, the digits can either be 0 or 9, and there are 7 digits at most.
We calculate the total number of such numbers for 1, 2, ..., 7 digits.
- For 1 digit: Only 1 possibility (9).
- For 2 digits: 2 choices for the first digit (9 or 0, but first digit cannot be 0), and 2 choices for the second (9 or 0), so .
- For 3 digits: Similarly, , and so on.
The total number of such numbers is .
Thus, the correct answer is (B).

We are given the composite function , where and .
To find the points of discontinuity of , we need to examine where the individual functions and are discontinuous.
1. The function is discontinuous where the denominator is zero, i.e., at .
2. The function is discontinuous where the denominator is zero, i.e., at .
Thus, is discontinuous where or .
- gives , leading to , which is a valid solution.
- gives , leading to , which also affects the discontinuity.
Hence, the discontinuities occur at and .