| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Standard +0.3 This is a standard C2 calculus question combining differentiation, normals, and integration. Part (a) requires finding dy/dx and the perpendicular gradient (routine). Part (b) involves solving a quadratic equation. Part (c) requires setting up and evaluating a definite integral for area between curve and line. All techniques are textbook exercises with no novel insight required, though the multi-step nature and 13 total marks make it slightly above average difficulty. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dy}{dx} = 1 - 2x\) | M1 | |
| \(\text{grad} = 1 - 2 = -1\) | A1 | |
| \(\text{grad of normal} = \frac{-1}{-1} = 1\) | M1 | |
| \(y - 5 = 1(x-1)\) | M1 | |
| \(y = x + 4\) | A1 | |
| (b) \(5 + x - x^2 = x + 4\) | M1 | |
| \(x^2 - 1 = 0\) | M1 | |
| \(x = 1\) (at \(P\)) or \(-1\) \(\therefore Q(-1, 3)\) | A1 | |
| (c) \(\text{area under curve} = \int_{-1}^{1}(5 + x - x^2)\,dx\) | ||
| \(= [5x + \frac{1}{2}x^2 - \frac{1}{3}x^3]_{-1}^{1}\) | M1 A1 | |
| \(= (5 + \frac{1}{2} - \frac{1}{3}) - (-5 + \frac{1}{2} + \frac{1}{3}) = 9\frac{1}{3}\) | M1 | |
| \(\text{area of trapezium} = \frac{1}{2} \times (3+5) \times 2 = 8\) | B1 | |
| \(\text{shaded area} = 9\frac{1}{3} - 8 = \frac{4}{3}\) | M1 A1 | (13) |
**(a)** $\frac{dy}{dx} = 1 - 2x$ | M1 |
$\text{grad} = 1 - 2 = -1$ | A1 |
$\text{grad of normal} = \frac{-1}{-1} = 1$ | M1 |
$y - 5 = 1(x-1)$ | M1 |
$y = x + 4$ | A1 |
**(b)** $5 + x - x^2 = x + 4$ | M1 |
$x^2 - 1 = 0$ | M1 |
$x = 1$ (at $P$) or $-1$ $\therefore Q(-1, 3)$ | A1 |
**(c)** $\text{area under curve} = \int_{-1}^{1}(5 + x - x^2)\,dx$ | |
$= [5x + \frac{1}{2}x^2 - \frac{1}{3}x^3]_{-1}^{1}$ | M1 A1 |
$= (5 + \frac{1}{2} - \frac{1}{3}) - (-5 + \frac{1}{2} + \frac{1}{3}) = 9\frac{1}{3}$ | M1 |
$\text{area of trapezium} = \frac{1}{2} \times (3+5) \times 2 = 8$ | B1 |
$\text{shaded area} = 9\frac{1}{3} - 8 = \frac{4}{3}$ | M1 A1 | (13)\includegraphics{figure_2}
Figure 2 shows the curve with equation $y = 5 + x - x^2$ and the normal to the curve at the point $P(1, 5)$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for the normal to the curve at $P$ in the form $y = mx + c$. [5]
\item Find the coordinates of the point $Q$, where the normal to the curve at $P$ intersects the curve again. [2]
\item Show that the area of the shaded region bounded by the curve and the straight line $PQ$ is $\frac{4}{3}$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q8 [13]}}