| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 14 |
| 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 C3 differentiation question involving product rule with exponential, finding stationary points, normals, and numerical methods. All techniques are routine for C3 level—product rule, solving dy/dx=0, finding normal equations, sign-change for interval verification, and iteration. The iterative formula is given (not derived), making part (d) mechanical. Slightly above average due to multiple parts requiring careful execution, but no novel insight needed. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks |
|---|---|
| (a) \(\frac{dy}{dx} = 2 \times e^x + (2x+3) \times (-e^{-x}) = -(2x+1)e^{-x}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(x = -\frac{1}{2} \therefore(-\frac{1}{2}, 2e^{\frac{1}{2}})\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(\therefore y = x + 3\) | M1 A1 |
| Answer | Marks |
|---|---|
| sign change, \(f(x)\) continuous \(\therefore\) root | M1 M1 A1 |
| (d) \(x_1 = -1.1619, x_2 = -1.2218, x_3 = -1.2408, x_4 = -1.2465 = -1.25\) (2dp) | M1 A2 |
| Answer | Marks | Guidance |
|---|---|---|
| sign change, \(f(x)\) continuous \(\therefore\) root | M1 A1 | (14 marks) |
**(a)** $\frac{dy}{dx} = 2 \times e^x + (2x+3) \times (-e^{-x}) = -(2x+1)e^{-x}$ | M1 A1 |
SP: $-(2x+1)e^{-x} = 0$
$x = -\frac{1}{2} \therefore(-\frac{1}{2}, 2e^{\frac{1}{2}})$ | M1 A1 |
**(b)** $x = 0, y = 3$, grad $= -1$, grad of normal $= 1$
$\therefore y = x + 3$ | M1 A1 |
**(c)** $x + 3 = (2x+3)e^{-x}$
$x + 3 - (2x+3)e^{-x} = 0$
let $f(x) = x + 3 - (2x+3)e^{-x}$
$f(-2) = 8.4, f(-1) = -0.72 \therefore$ root
sign change, $f(x)$ continuous $\therefore$ root | M1 M1 A1 |
**(d)** $x_1 = -1.1619, x_2 = -1.2218, x_3 = -1.2408, x_4 = -1.2465 = -1.25$ (2dp) | M1 A2 |
**(e)** $f(-1.255) = 0.026, f(-1.245) = -0.016$
sign change, $f(x)$ continuous $\therefore$ root | M1 A1 | (14 marks)
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**Total: 75 marks**A curve has the equation $y = (2x + 3)e^{-x}$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact coordinates of the stationary point of the curve. [4]
\end{enumerate}
The curve crosses the $y$-axis at the point $P$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find an equation for the normal to the curve at $P$. [2]
\end{enumerate}
The normal to the curve at $P$ meets the curve again at $Q$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show that the $x$-coordinate of $Q$ lies in the interval $[-2, -1]$. [3]
\item Use the iterative formula
$$x_{n+1} = \frac{3 - 3e^{x_n}}{e^{x_n} - 2}$$
with $x_0 = -1$, to find $x_1$, $x_2$, $x_3$ and $x_4$. Give the value of $x_4$ to 2 decimal places. [3]
\item Show that your value for $x_4$ is the $x$-coordinate of $Q$ correct to 2 decimal places. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q8 [14]}}