Question

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 ______.

Options

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Answer 1

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 3.

Explanation:

f(x) = `{{:((x^2 - 1)(x - 3) + (x - 3)",", x ∈ (0, 1] ∪ [3, 4)),(-(x^2 - 1)(x - 3) + (x - 3)",", x ∈ [1, 3]):}`

differentiate w.r.t ‘x’.

`\implies` f(x) = `{{:(3x^2 - 6x",", x ∈ (0, 1) ∪ (3, 4)),(-3x^2 + 6x + 2",", x ∈ (1, 3)):}`

f(x) is not differentiable at x = 1 and x = 3.

Take, f'(x) = 0 at x = `1 ± sqrt(5/3)`

Again differentiable w.r.t. x.

f"(x) = `{{:(6x - 6",", x ∈ (0, 1) ∪ (3, 4)),(-6x + 6",", x ∈ (1, 3)):}`

At x = 2 function has local maximum and x = `1 + sqrt(5/3)` is the point of local minimum.

Then, m + M = 2 + 1 = 3.