A-LevelMathematicsDifferentiation9709/12 · May/June 2016 · Q1012 Marks

The diagram shows the part of the curve y = 8/x + 2x for x > 0, and the minimum point M. [Figure 10.1]

(i)Find expressions for dy/dx, d²y/dx² and ∫y² dx.[5]
(ii)Find the coordinates of M and determine the coordinates and nature of the stationary point on the part of the curve for which x < 0.[5]
(iii)Find the volume obtained when the region bounded by the curve, the x-axis and the lines x = 1 and x = 2 is rotated through 360° about the x-axis.[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2016). The mark or mark type is shown in brackets.

[B1] dy/dx = -8x⁻² + 2 unsimplified ok
[B1] d²y/dx² = 16x⁻³ unsimplified ok
[3 × B1] ∫y²dx = -64x⁻¹ + 32x + 4x³ / 3 oe (+c) B1 for each term – unsimplified ok
[M1] sets dy/dx to 0 → x = ±2 Sets to 0 and attempts to solve
[A1] →M(2, 8) Any pair of correct values A1
[A1] Other turning point is (-2, -8) Second pair of values A1
[M1] If x = -2, d²y/dx² < 0 Using their d²y/dx² if kx⁻³ and x<0
[A1] Maximum
[M1] Vol = π × [ part (i)] from 1 to 2 Evidence of using limits 1&2 in their integral of y² (ignore π)
[A1] 220π/3, 73.3π, 230

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