The diagram shows part of the curve y = x/2 + 6/x. The line y = 4 intersects the curve at the points P and Q. [Figure 18]

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

1. [B1 B1] (i) First, find the x-coordinates of P and Q by equating y = x/2 + 6/x and y = 4: x/2 + 6/x = 4. Multiply by 2x: x² + 12 = 8x. x² - 8x + 12 = 0. Factorising: (x - 2)(x - 6) = 0. So, x = 2 or x = 6. Let P be (2, 4) and Q be (6, 4). Inspection or guesswork OK.
2. [B1] (i) Find the derivative dy/dx: y = ½x + 6x⁻¹. dy/dx = ½ - 6x⁻² = ½ - 6/x². Unsimplified OK.
3. [*M1] (i) Find the equations of the tangents at P and Q. At P (x = 2): dy/dx = ½ - 6/(2)² = ½ - 6/4 = ½ - 3/2 = -1. Tangent equation: y - 4 = -1(x - 2) → y - 4 = -x + 2 → y = -x + 6. At Q (x = 6): dy/dx = ½ - 6/(6)² = ½ - 6/36 = ½ - 1/6 = 3/6 - 1/6 = 2/6 = 1/3. Tangent equation: y - 4 = (1/3)(x - 6) → 3y - 12 = x - 6 → 3y = x + 6. Correct method for either tangent.
4. [DM1] (i) Find the intersection of the two tangents: y = -x + 6 and 3y = x + 6. Substitute y = -x + 6 into the second equation: 3(-x + 6) = x + 6. -3x + 18 = x + 6. 12 = 4x, so x = 3. Substitute x = 3 into y = -x + 6: y = -3 + 6 = 3. The tangents meet at (3, 3). Could solve BOTH equations separately with y = x and get x = 3 both times.
5. [A1] (i) Since x = 3 and y = 3, the point (3, 3) lies on the line y = x. Statement about y = x not required.
6. [*M1] (ii) The volume obtained when the shaded region is rotated through 360° about the x-axis is given by π∫(y_curve)² dx - π∫(y_line)² dx, evaluated between x=2 and x=6. (y_curve)² = (x/2 + 6/x)² = x²/4 + 2(x/2)(6/x) + 36/x² = x²/4 + 6 + 36/x². Integrate using π∫ y²dx (doesn't need π or dx). Allow incorrect squaring. Not awarded for π∫{4 - (x/2 + 6/x)²} dx. Integration indicated by increase in any power by 1.
7. [A2,1] (ii) ∫(x²/4 + 6 + 36/x²) dx = x³/12 + 6x - 36/x. 3 things wanted -1 each error, allow + C. (Doesn't need π).
8. [DM1] (ii) Evaluate the integral with limits from 2 to 6: [x³/12 + 6x - 36/x] from 2 to 6. At x = 6: (6³/12 + 6(6) - 36/6) = (216/12 + 36 - 6) = (18 + 36 - 6) = 48. At x = 2: (2³/12 + 6(2) - 36/2) = (8/12 + 12 - 18) = (2/3 - 6) = -16/3. Difference = 48 - (-16/3) = 48 + 16/3 = 144/3 + 16/3 = 160/3. So, Volume_curve = π(160/3). Evidence of their values 6 and 2 from (i) substituted into their integrand and then subtracted. 48 - 16/3 is enough.
9. [M1] (ii) The volume from rotating the line y = 4 is a cylinder with radius 4 and height (6-2)=4. Volume_line = πr²h = π(4)²(6 - 2) = π(16)(4) = 64π. Use of πr²h or integration of 4² (could be from ∫4²dx).
10. [A1] (ii) Volume of shaded region = Volume_curve - Volume_line = π(160/3) - 64π = π(160/3 - 192/3) = π(-32/3). The volume must be positive, so the shaded region is below the line. This means we should subtract the curve volume from the cylinder volume. Volume = 64π - π(160/3) = π(192/3 - 160/3) = 32π/3. Subtracts → 10π oe (e.g. 32/3 π, 33.5 awrt).