Figure 2 shows a sketch of the curve C with equation y = 1/2x²-3x+8. The points P and Q lie on C such that the tangent to C at P is parallel to the tangent to C at Q. The x-coordinate of P is $1$.

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

1. Part (a) M1: Substitutes x=1 into y=1/2x²-3x+8.
2. Part (a) A1: y=5.5 or 11/2.
3. Part (b) B1: For (d y)/(~d x) = x-3.
4. Part (b) M1: Substitutes x=1 into their (d y)/(~d x) to find the gradient. E.g. m = 1-3 = -2.
5. Part (b) M1: Uses their P(1, 5.5) and their gradient m in a correct line equation formula. E.g. y-5.5 = -2(x-1).
6. Part (b) A1: y=-2x+7.5 (oe).
7. Part (c) M1: For finding the axis of symmetry x=3 (e.g. from -b/2a) or implicitly by using derivative x-3.
8. Part (c) A1: Correct x-coordinate of Q using symmetry x_(Q)=5.
9. Part (c) A1: Final answer x=5.
10. Part (d) B1: Finds y_(Q): Substitutes x_(Q)=5 into y=1/2x²-3x+8 ⇒ y_(Q) = 11/2 or $5.5$.
11. Part (d) M1: Attempts to find the midpoint of PQ. Correct use of formula for both x and y. E.g. x_(mid) = (1+5)/2=3, y_(mid) = (5.5+5.5)/2=5.5.
12. Part (d) A1: For correctly substituting the midpoint $(3, 5.5)$ into y=2x-1, leading to $5.5=5$.
13. Part (d) A1: States coordinates of Q as $(5, 5.5)$. (Implies candidate ignores the contradiction from the line equation and provides the expected point based on the value for x_(Q) and y_(Q) from calculations.)