Let f(x) = |(x - 1)(x² - 2x - 3)| + x - 3, x ∈ R. If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ____________.

Solution

Maths sir · 4 steps

Open the modulus piece by piece, then watch where f' changes sign, including at the corners.

1. Factor first. x² - 2x - 3 = (x-3)(x+1), so the expression inside the modulus is g(x) = (x+1)(x-1)(x-3) = x³ - 3x² - x + 3. On (0, 4), g is positive on (0,1), negative on (1,3) and positive on (3,4).

   Interval	f(x)	f'(x)	Behaviour
   (0, 1)	g + x - 3 = x³ - 3x²	3x(x-2) < 0	decreasing
   (1, 3)	-g + x - 3 = -x³ + 3x² + 2x - 6	-3x² + 6x + 2	increasing, then decreasing
   (3, 4)	g + x - 3 = x³ - 3x²	3x(x-2) > 0	increasing

2. Inside (1, 3). f' = 0 gives 3x² - 6x - 2 = 0, so x = 1 + √15/3 ≈ 2.29, which lies in (1, 3). f'(1⁺) = 5 > 0 and f'(3⁻) = -7 < 0, so f rises and then falls. This point is a local maximum.
3. The two corners. At x = 1, f goes from decreasing to increasing, so it is a local minimum. At x = 3, f goes from decreasing to increasing again, so it is another local minimum. Here f is not differentiable, but the sign change of f' still decides the extremum.
4. So. m = 2 (at x = 1 and x = 3) and M = 1 (at x = 1 + √15/3), so m + M = 3.

The common slip: 1. Counting only the points where f' = 0 finds just the maximum near x = 2.29 and gives 1. That misses the two corner minima at x = 1 and x = 3.

The idea to keep. With a modulus, extrema can sit at the zeros inside the modulus, where f' does not exist. Check the sign of f' on both sides of every corner, not only where f' = 0.

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