| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| 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 C3 differentiation question involving the product rule with exponential functions. Part (i) requires routine differentiation and solving for stationary points. Parts (ii)-(v) involve finding a normal, showing a root lies in an interval using sign changes, applying a given iterative formula, and verifying accuracy—all standard techniques. While multi-part with 13 marks total, each component is textbook-level with no novel insight required, making it slightly easier than the average A-level question. |
| 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 |
|---|---|
| \(x = -\frac{1}{2}\) \(\therefore \left(-\frac{1}{2}, 2e^{\frac{1}{2}}\right)\) | M1 A1, M1 A1 |
| Answer | Marks |
|---|---|
| \(\therefore y = x + 3\) | M1, A1 |
| Answer | Marks |
|---|---|
| sign change, \(f(x)\) continuous \(\therefore\) root | M1, M1, A1 |
| Answer | Marks |
|---|---|
| \(x_1 = -1.1619\), \(x_2 = -1.2218\), \(x_3 = -1.2408\), \(x_4 = -1.2465 = -1.25\) (2dp) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| sign change, \(f(x)\) continuous \(\therefore\) root | M1, A1 | (13) |
## (i)
$\frac{dy}{dx} = 2 \times e^{-x} + (2x+3) \times (-e^{-x}) = -(2x+1)e^{-x}$
SP: $-(2x+1)e^{-x} = 0$
$x = -\frac{1}{2}$ $\therefore \left(-\frac{1}{2}, 2e^{\frac{1}{2}}\right)$ | M1 A1, M1 A1 |
## (ii)
$x = 0$, $y = 3$, grad $= -1$, grad of normal $= 1$
$\therefore y = x + 3$ | M1, A1 |
## (iii)
$x + 3 = (2x+3)e^{-x}$
$x + 3 - (2x+3)e^{-x} = 0$
let $f(x) = x + 3 - (2x+3)e^{-x}$
$f(-2) = 8.4$, $f(-1) = -0.72$
sign change, $f(x)$ continuous $\therefore$ root | M1, M1, A1 |
## (iv)
$x_1 = -1.1619$, $x_2 = -1.2218$, $x_3 = -1.2408$, $x_4 = -1.2465 = -1.25$ (2dp) | M1 A1 |
## (v)
$f(-1.255) = 0.026$, $f(-1.245) = -0.016$
sign change, $f(x)$ continuous $\therefore$ root | M1, A1 | (13)
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**Total: (72)**A curve has the equation $y = (2x + 3)\mathrm{e}^{-x}$.
\begin{enumerate}[label=(\roman*)]
\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=(\roman*)]
\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=(\roman*)]
\setcounter{enumi}{2}
\item Show that the $x$-coordinate of $Q$ lies between $-2$ and $-1$. [3]
\item Use the iterative formula
$$x_{n+1} = \frac{3 - 3\mathrm{e}^{x_n}}{\mathrm{e}^{x_n} - 2},$$
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. [2]
\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{OCR C3 Q9 [13]}}