| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Curve properties and tangent/normal |
| Difficulty | Standard +0.3 This is a straightforward integration question with standard follow-up parts. Part (i) requires basic integration and using a point to find the constant. Part (ii) involves factorizing a cubic and using discriminant/derivative to distinguish touching from cutting. Part (iii) requires finding gradients and checking perpendicularity. All techniques are routine C2 material with clear signposting, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y = x^3 - 6x^2 + 9x + c\) Substitute \(x = 2, y = -2 \Rightarrow c = -4\) \(\Rightarrow y = x^3 - 6x^2 + 9x - 4\) | M1 A1 A1 A1 M1 A1 | 5 marks |
| (ii) Factorising Gives \(y = (x-1)^2(x-4)\) Double root at \(x = 1\) indicates touching \(x\)-axis, so A is (1, 0) B is (4, 0) | M1 A1 B1 B1 | 4 marks |
| (iii) At C(0, 4), \(\frac{dy}{dx} = 9\) At B(4, 0), \(\frac{dy}{dx} = 3 \times 16 - 12 \times 4 + 9 = 9\) So tangents are parallel so the normal of one is perpendicular to the tangent of the other. | B1 M1 E1 | 3 marks |
**(i)** $y = x^3 - 6x^2 + 9x + c$ Substitute $x = 2, y = -2 \Rightarrow c = -4$ $\Rightarrow y = x^3 - 6x^2 + 9x - 4$ | M1 A1 A1 A1 M1 A1 | 5 marks | Integration For c
**(ii)** Factorising Gives $y = (x-1)^2(x-4)$ Double root at $x = 1$ indicates touching $x$-axis, so A is (1, 0) B is (4, 0) | M1 A1 B1 B1 | 4 marks
**(iii)** At C(0, 4), $\frac{dy}{dx} = 9$ At B(4, 0), $\frac{dy}{dx} = 3 \times 16 - 12 \times 4 + 9 = 9$ So tangents are parallel so the normal of one is perpendicular to the tangent of the other. | B1 M1 E1 | 3 marks9 The gradient of a curve is given by $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } - 12 x + 9$. The curve passes through the point $( 2 , - 2 )$.
\begin{enumerate}[label=(\roman*)]
\item Find the equation of the curve.
\item Show that the curve touches the $x$-axis at one point (A) and cuts it at another (B). State the coordinates of A and B.
\item The curve cuts the $y$-axis at C . Show that the tangent at C is perpendicular to the normal at B.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 Q9 [12]}}