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When solving the differential equation d²y/dx² = 6x - 4, the integration gives dy/dx = 3x² - 4x + A. If dy/dx becomes zero at x = 1, what is the value of A?
- 1
- 2
- 3
- 4
Correct answer: 1
Solution
Given dy/dx = 3x^2 - 4x + A with dy/dx = 0 at x = 1: 3(1)^2 - 4(1) + A = 0, so 3 - 4 + A = 0, giving A = 1. The correct value A=1 is option index 0, so the stored index 1 (value 2) is wrong.
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