Let [t] denote the greatest integer less than or equal to t. Let f:[0, ∞) → R be a function defined by f(x)=[x/2+3]-[√x]. Let S be the set of all points in the interval [0,8] at which f is not continuous. Then Σaes a is equal to __________.

Solution

Maths sir · 4 steps

List every point where either floor jumps, then check whether the two jumps cancel each other.

1. Pull out the 3. Since 3 is an integer, [x/2+3] = [x/2] + 3. So f(x) = [x/2] - [√x] + 3, and the 3 cannot cause any break.
2. Find the candidates. [x/2] jumps up by 1 when x/2 is an integer, at x = 2, 4, 6, 8. [√x] jumps up by 1 when √x is an integer, at x = 1, 4 in [0, 8]. At x = 0 the domain starts, and both terms are right-continuous there, so 0 is not a break.

   Point	Jump in [x/2]	Jump in [√x]	Net jump in f
   x = 1	0	+1	-1, discontinuous
   x = 2	+1	0	+1, discontinuous
   x = 4	+1	+1	0, continuous
   x = 6	+1	0	+1, discontinuous
   x = 8	+1	0	+1 from the left, discontinuous

3. Check x = 4. Just left of 4, f = 1 - 1 + 3 = 3. At 4 and just right of it, f = 2 - 2 + 3 = 3. The two jumps cancel, so f is continuous at 4.
4. So. S = {1, 2, 6, 8} and the sum is 1 + 2 + 6 + 8 = 17.

The common slip: 21. Counting x = 4 as a discontinuity gives 1 + 2 + 4 + 6 + 8 = 21. Both floors jump by 1 at 4, so in the difference the jumps cancel.

The idea to keep. For a sum or difference of floor functions, a jump in one term is not automatically a break. If two terms jump at the same point, compute the left limit and the value there to see whether they cancel.

Answer from the NTA final answer key. Solution by PrepWiser.