Let f(x) = [ 2x² + 1 ] and g(x) = { 2x - 3,, x < 0; 2x + 3,, x ≥ 0 , where [t] is the greatest integer ≤ t. Then, in the open interval (-1, 1), the number of points where fog is discontinuous is equal to ______________.

Solution

Maths sir · 5 steps

Pull the 1 out of the bracket, then count how many integers 2g² crosses on each side of zero.

1. f(g(x)) = [2g(x)² + 1] = 1 + [2g(x)²]

2. Where it breaks. The greatest integer function jumps exactly where its inside crosses an integer. So count the points where 2g² passes through a whole number. Also check x = 0, where g itself jumps.
3. Right piece. For 0 ≤ x < 1, g = 2x + 3 runs over [3, 5). So 2g² rises steadily from 18 towards 50, never reaching 50. It passes each of 19, 20, …, 49 exactly once, which gives 31 points.
4. Left piece. For -1 < x < 0, g = 2x - 3 runs over (-5, -3). So g² falls steadily from 25 to 9, and 2g² covers (18, 50). It passes each of 19, …, 49 exactly once, which gives another 31 points.
5. Check x = 0. g jumps from -3 to 3, but squaring removes the sign, so g² → 9 from both sides. Just left of 0, just right of 0 and at 0, [2g²] = 18 and fog = 19. So fog is continuous at x = 0.
6. So. The number of points of discontinuity is 31 + 31 = 62.

The common slip: 63. Counting x = 0 as a discontinuity because g jumps there gives 63. Since g² = 9 on both sides, fog does not break at 0.

The idea to keep. For [h(x)], count the integers that h crosses, piece by piece. At a jump of the inner function, always recheck: an even power like g² can glue the jump back together.

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