The diagram shows part of the curve y = 4√x – x. The curve has a maximum point at M and meets the x-axis at O and A. [Figure X.X]

Mark scheme

Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.

1. [5] (i) At A, 4√x − x = 0 → x(4/√x − 1) = 0 → √x(4 − √x) = 0 → x=0 or x=16; B1 co - independent of working.; A(16, 0); dy/dx = 2x⁻¹ᐟ² - 1; B1 B1 B1 for each part.; dy/dx = 0 when x = 4 → (4, 4); M1 A1 Sets to 0 and solves his eqn. co; M(4, 4)
2. [6] (ii) Vol = π ∫y² dx; M1 Use of correct formula + attempt at integration; y² = (4√x - x)² = 16x + x² - 8x^(3/2); A3,2,1 One mark for each term – unsimplified; ∫y² dx = ∫(16x + x² - 8x^(3/2)) dx = 8x² + x³/3 - (16/5)x^(5/2); DM1 Correct use of his limits.; Limits 0 to 16 → π [8(16)² + (16)³/3 - (16/5)(16)^(5/2)]; A1 co - (429 ok); = 136.5π. (or 137π)