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The line y = mx + 2 intersects the parabola 2y = x² at points (x1, y1) and (x2, y2) with x1 < x2. Find the value of m for which the integral from x1 to x2 of (mx + 2 - x²/2) dx is minimum.

1. 0
2. -1
3. 1
4. -2

Correct answer: 0

Solution

The integral of (mx+2 - x²/2) from x1 to x2 gives the area between the line y = mx+2 and the parabola 2y=x². From intersection: x² = 2mx+4, so x²-2mx-4=0. By Vieta: x1+x2=2m, x1*x2=-4. (x2-x1)² = (x1+x2)² - 4x1x2 = 4m²+16. The integral value = (1/6)(x2-x1)³ = (1/6)((4m²+16)^(3/2))... wait, more precisely integral = (x2-x1)³/6. (x2-x1)² = 4m²+16. This is minimized when m=0, giving (x2-x1)² = 16, minimum area. So m=0.

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