Let [t] denote the greatest integer ≤ t. The number of points where the function f(x) = [x]| x² - 1 | + sin ( π/([x] + 3) ) - [x + 1],x ∈ ( - 2,2) is not continuous is _____________.

Solution

Maths sir · 3 steps

Only integers can break this function, so check x = -1, 0, 1 one at a time.

1. Where to look. Inside each interval [n, n+1) the value [x] is constant, so f is a continuous expression there. Trouble can only happen where [x] jumps, that is at x = -1, 0, 1. Also note that [x+1] = [x] + 1.
2. Piece by piece. On (-2,-1), [x] = -2 and sin π = 0, so f = -2(x²-1) + 1 = 3 - 2x². On [-1,0), [x] = -1 and sin π/2 = 1, so f = -(1-x²) + 1 = x². On [0,1), f = √3/2 - 1. On [1,2), f = (x²-1) + 1/√2 - 2.

   Point	Left-hand limit	f at the point (= right-hand limit)	Continuous?
   x = -1	3 - 2 = 1	(-1)² = 1	yes
   x = 0	0² = 0	√3/2 - 1 ≈ -0.13	no
   x = 1	√3/2 - 1 ≈ -0.13	1/√2 - 2 ≈ -1.29	no

3. So. At x = -1 the jumps cancel, because |x² - 1| = 0 there. The function breaks only at x = 0 and x = 1, giving 2 points.

The common slip: 3. Counting every integer in (-2, 2) as a break gives 3. At x = -1 the factor |x² - 1| is zero and the other jumps cancel, so f stays continuous there.

The idea to keep. With [x] in a function, write out a separate formula for each unit interval, then compare the left limit with the value at each integer. Do not assume every integer is a break.

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