Let f(x) = x.[ x/2 ], for -10< x < 10, where [t] denotes the greatest integer function. Then the number of points of discontinuity of f is equal to _____.

Solution

Maths sir · 4 steps

The floor part jumps at every even integer, but at x = 0 the factor x kills the jump.

1. Where the jumps are. [x/2] changes value only when x/2 crosses an integer, that is at even x. In -10 < x < 10 these are x = -8, -6, -4, -2, 0, 2, 4, 6, 8, which makes 9 candidate points.
2. Size of the jump. At x = 2k, the value of [x/2] is k-1 just to the left and k at and to the right. So f goes from 2k(k-1) to 2k · k.

   f(2k⁺) - f(2k⁻) = 2k · k - 2k(k-1) = 2k

3. Check each point. For k ≠ 0 the jump 2k is not zero, so f is discontinuous at the 8 points ± 2, ± 4, ± 6, ± 8. At x = 0 the jump is 0: on the left f(x) = -x → 0, on the right f(x) = 0, and f(0) = 0, so f is continuous there.
4. So. The number of points of discontinuity is 8.

The common slip: 9. Counting every even integer, including x = 0, gives 9. At x = 0 the factor x is zero, so the jump vanishes and f stays continuous.

The idea to keep. When a greatest integer function is multiplied by another factor, list the jump points first, then check each one. If the factor is zero at a jump point, the jump disappears.

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