A-LevelMathematicsIntegration9709/12 · Oct/Nov 2012 · Q108 Marks
A curve is defined for x > 0 and is such that dy/dx = x + 4/x². The point P (4, 8) lies on the curve.
- (i)Find the equation of the curve.[4]
- (ii)Show that the gradient of the curve has a minimum value when x = 2 and state this minimum value.[4]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2012). The mark or mark type is shown in brackets.
- [B1] ∫ x dx = x²/2
- [B1] ∫ 4/x² dx = -4/x
- [M1] Uses (4, 8) in y = x²/2 - 4/x + c (for constant c)
- [A1] → c = 1
- [B1] d²y/dx² = 1 - 8/x³
- [B1] d²y/dx² = 0 when x = 2
- [B1] Verification that it's a minimum (e.g., using 3rd differential +ve, or 2nd diff goes -0+)
- [A1] → gradient of 3
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