Let ,f:[-1,2] → R be given by f(x)=2 x²+x+[x²]-[x], where [t] denotes the greatest integer less than or equal to t. The number of points, where f is not continuous, is :

Solution

Maths sir · 3 steps

List every point where a box can jump, then check which jumps cancel each other out.

1. Find candidates. 2x² + x is continuous, so only the two box functions can break f. [x] jumps at x = 0, 1, 2. [x²] jumps where x² is an integer, that is at x = -1, 0, 1, √2, √3, 2 inside [-1, 2].
2. Watch the signs. [x²] enters with a plus sign and [x] with a minus sign. Where both rise by 1 at the same point, the jumps cancel and f stays continuous.

   Point	Jump in [x²]	Jump in -[x]	Is f continuous?
   x = -1	1 → 0 on the right	none ([x] = -1)	No: f(-1) = 3, right limit 2
   x = 0	none (x² stays below 1)	+1 → 0	No: left limit 1, f(0) = 0
   x = 1	+1	-1	Yes: both sides and f(1) equal 3
   x = √2	+1	none ([x] = 1)	No
   x = √3	+1	none ([x] = 1)	No
   x = 2	+1	-1	Yes: left limit and f(2) both equal 12

3. So. f is discontinuous only at x = -1, 0, √2, √3, which is 4 points. The answer is (c).

The idea to keep. With a sum of box functions, the jump points are only candidates. Check each one, because opposite jumps can cancel, and at an endpoint check only the one side that exists.

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