The curve C has equation y = x²(3-x). The point P lies on C and has x-coordinate $1$. (a) Find the equation of the tangent to C at P. The normal to C at P has equation x - 5y + 9 = 0. (b) Find the exact area of the triangle formed by the normal to C at P and the coordinate axes.

Mark scheme

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

1. (a) M1: Expands y = 3x² - x³ (or uses product rule implicitly).
2. (a) A1: Differentiates to get dy/dx = 6x - 3x².
3. (a) M1: Substitutes x=1 into their dy/dx to find the gradient of the tangent.
4. (a) A1: Gradient $= 3$.
5. (a) B1: Finds the y-coordinate of P by substituting x=1 into y = x²(3-x). (y=2).
6. (a) M1: Uses y-y₁ = m(x-x₁) with their gradient and point P(1,2).
7. (a) A1: y = 3x - 1 (any equivalent form).
8. (b) M1: Finds the intercepts of the normal x - 5y + 9 = 0 with the axes. (Sets x=0 to find y-intercept and y=0 to find x-intercept).
9. (b) A1: Intercepts are (0, 9/5) and $(-9, 0)$ (both correct).
10. (b) M1: Uses the formula for the area of a triangle: 1/2 × base × height. (Can be implied by values).
11. (b) A1: Area = 81/10 or $8.1$ (must be exact, positive value).