| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of normal line |
| Difficulty | Standard +0.3 This is a standard C3 tangent/normal question with routine differentiation using chain rule, followed by straightforward algebraic manipulation and iteration. While multi-part with several marks, each step follows predictable patterns: find derivative, find normal equation, substitute to show inequalities, rearrange equation, and apply given iterative formula. Slightly easier than average due to clear scaffolding and standard techniques throughout. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dy}{dx} = 2 - \frac{3}{2x+5} \times 2 = 2 - \frac{6}{2x+5}\) | M1 | |
| grad \(= -4\), grad of normal \(= \frac{1}{4}\) | A1 | |
| \(\therefore y + 4 = \frac{1}{4}(x + 2)\) [\(y = \frac{1}{4}x - \frac{7}{2}\)] | M1 A1 | |
| (b) \(\frac{1}{4}x - \frac{7}{2} = 2x - 3 \ln(2x + 5)\) | M1 | |
| \(\frac{7}{4}x + \frac{7}{2} - 3 \ln(2x + 5) = 0\), let \(f(x) = \frac{7}{4}x + \frac{7}{2} - 3 \ln(2x + 5)\) | M1 | |
| \(f(1) = -0.59\), \(f(2) = 0.41\) | M1 | |
| sign change, \(f(x)\) continuous \(\therefore\) root | A1 | |
| (c) \(\frac{7}{4}x + \frac{7}{2} - 3 \ln(2x + 5) = 0\) | M1 | |
| \(7x + 14 - 12 \ln(2x + 5) = 0\) | A1 | |
| \(7x = 12 \ln(2x + 5) - 14\) | M1 | |
| \(x = \frac{12}{7} \ln(2x + 5) - 2\) | A1 | |
| (d) \(x_1 = 1.5648, x_2 = 1.5923, x_3 = 1.6039, x_4 = 1.6087, x_5 = 1.6107\) | M1 A1 | |
| \(q = 1.61\) (3sf) | A1 | |
| \(f(1.605) = -0.0073\), \(f(1.615) = 0.0029\) | M1 | |
| sign change, \(f(x)\) continuous \(\therefore\) root \(\therefore q = 1.61\) (3sf) | A1 | (14 marks) |
**(a)** $\frac{dy}{dx} = 2 - \frac{3}{2x+5} \times 2 = 2 - \frac{6}{2x+5}$ | M1
grad $= -4$, grad of normal $= \frac{1}{4}$ | A1
$\therefore y + 4 = \frac{1}{4}(x + 2)$ [$y = \frac{1}{4}x - \frac{7}{2}$] | M1 A1
**(b)** $\frac{1}{4}x - \frac{7}{2} = 2x - 3 \ln(2x + 5)$ | M1
$\frac{7}{4}x + \frac{7}{2} - 3 \ln(2x + 5) = 0$, let $f(x) = \frac{7}{4}x + \frac{7}{2} - 3 \ln(2x + 5)$ | M1
$f(1) = -0.59$, $f(2) = 0.41$ | M1
sign change, $f(x)$ continuous $\therefore$ root | A1
**(c)** $\frac{7}{4}x + \frac{7}{2} - 3 \ln(2x + 5) = 0$ | M1
$7x + 14 - 12 \ln(2x + 5) = 0$ | A1
$7x = 12 \ln(2x + 5) - 14$ | M1
$x = \frac{12}{7} \ln(2x + 5) - 2$ | A1
**(d)** $x_1 = 1.5648, x_2 = 1.5923, x_3 = 1.6039, x_4 = 1.6087, x_5 = 1.6107$ | M1 A1
$q = 1.61$ (3sf) | A1
$f(1.605) = -0.0073$, $f(1.615) = 0.0029$ | M1
sign change, $f(x)$ continuous $\therefore$ root $\therefore q = 1.61$ (3sf) | A1 | (14 marks)8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d17a1b86-d758-4470-834a-b32a41f90c89-4_478_937_251_450}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the curve with equation $y = 2 x - 3 \ln ( 2 x + 5 )$ and the normal to the curve at the point $P ( - 2 , - 4 )$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for the normal to the curve at $P$.
The normal to the curve at $P$ intersects the curve again at the point $Q$ with $x$-coordinate $q$.
\item Show that $1 < q < 2$.
\item Show that $q$ is a solution of the equation
$$x = \frac { 12 } { 7 } \ln ( 2 x + 5 ) - 2 .$$
\item Use the iterative formula
$$x _ { n + 1 } = \frac { 12 } { 7 } \ln \left( 2 x _ { n } + 5 \right) - 2 ,$$
with $x _ { 0 } = 1.5$, to find the value of $q$ to 3 significant figures and justify the accuracy of your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q8 [14]}}