| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Moderate -0.3 This is a standard C2 differentiation application requiring finding a normal (gradient = -1/derivative), solving a quadratic for intersection, and computing area between curve and line using integration. All steps are routine textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = 1 - 2x\), gradient \(= 1 - 2 = -1\). Gradient of normal \(= \frac{-1}{-1} = 1\). Equation: \(y - 5 = 1(x-1)\), so \(y = x + 4\) | M1, A1, M1, M1, A1 | |
| (ii) \(5 + x - x^2 = x + 4 \therefore x^2 - 1 = 0\), so \(x = 1\) (at \(P\)) or \(x = -1 \therefore Q(-1, 3)\) | M1, A1 | |
| (iii) Area \(= \int_{-1}^{1} [(5 + x - x^2) - (x + 4)] \, dx = \int_{-1}^{1} (1 - x^2) \, dx = [x - \frac{1}{3}x^3]_{-1}^{1} = (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) = \frac{4}{3}\) | M1, M1, A1, A1 | (12) |
| Answer | Marks |
|---|---|
| Total | (72) |
**(i)** $\frac{dy}{dx} = 1 - 2x$, gradient $= 1 - 2 = -1$. Gradient of normal $= \frac{-1}{-1} = 1$. Equation: $y - 5 = 1(x-1)$, so $y = x + 4$ | M1, A1, M1, M1, A1 |
**(ii)** $5 + x - x^2 = x + 4 \therefore x^2 - 1 = 0$, so $x = 1$ (at $P$) or $x = -1 \therefore Q(-1, 3)$ | M1, A1 |
**(iii)** Area $= \int_{-1}^{1} [(5 + x - x^2) - (x + 4)] \, dx = \int_{-1}^{1} (1 - x^2) \, dx = [x - \frac{1}{3}x^3]_{-1}^{1} = (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) = \frac{4}{3}$ | M1, M1, A1, A1 | **(12)**
---
**Total** | **(72)** |9.\\
\includegraphics[max width=\textwidth, alt={}, center]{33f9663f-26bb-445e-af6e-ca5ca927f7dd-3_638_757_1064_493}
The diagram shows the curve with equation $y = 5 + x - x ^ { 2 }$ and the normal to the curve at the point $P ( 1,5 )$.\\
(i) Find an equation for the normal to the curve at $P$ in the form $y = m x + c$.\\
(ii) Find the coordinates of the point $Q$, where the normal to the curve at $P$ intersects the curve again.\\
(iii) Show that the area of the shaded region bounded by the curve and the straight line $P Q$ is $\frac { 4 } { 3 }$.
\hfill \mbox{\textit{OCR C2 Q9 [12]}}