| 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 +0.8 This is a substantial multi-part question requiring chain rule differentiation, finding normals, algebraic manipulation to derive an iteration formula, numerical iteration, convergence verification, and analysis of a divergent iteration. While individual techniques are C3 standard, the extended reasoning across five parts (especially analyzing why one iteration converges and another fails) elevates this above typical textbook exercises. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - 4e^{1-4x}\) | M1 | |
| grad \(= -3\), grad of normal \(= \frac{1}{3}\) | A1 | |
| \(\therefore y - \frac{2}{3} = \frac{1}{3}(x - \frac{1}{4})\) | M1 A1 | [4x - 12y + 17 = 0] |
| (b) SP: \(\frac{1}{2}x^{-\frac{1}{2}} - 4e^{1-4x} = 0\), \(\frac{1}{2\sqrt{x}} = 4e^{1-4x}\) | M1 | |
| \(\frac{1}{8\sqrt{x}} = e^{1-4x}\) | M1 | |
| \(8\sqrt{x} = e^{4x-1}\) | M1 | |
| \(4x - 1 = \ln 8\sqrt{x}\) | M1 | |
| \(x = \frac{1}{4}(1 + \ln 8\sqrt{x})\) | A1 | |
| (c) \(x_1 = 0.7699, x_2 = 0.7372, x_3 = 0.7317, x_4 = 0.7308 = 0.731\) (3dp) | M1 A2 | |
| (d) let \(f(x) = \frac{1}{2}x^{-\frac{1}{2}} - 4e^{1-4x}\) | M1 | |
| \(f(0.7305) = -0.00025\), \(f(0.7315) = 0.0017\) | M1 | |
| sign change, \(f(x)\) continuous \(\therefore\) root | A1 | |
| (e) \(x_1 = 6.304, x_2 = 1.683 \times 10^{19}\) | B2 | |
| diverges rapidly away from root | (14 marks) |
**(a)** $\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - 4e^{1-4x}$ | M1 |
grad $= -3$, grad of normal $= \frac{1}{3}$ | A1 |
$\therefore y - \frac{2}{3} = \frac{1}{3}(x - \frac{1}{4})$ | M1 A1 | [4x - 12y + 17 = 0] |
**(b)** SP: $\frac{1}{2}x^{-\frac{1}{2}} - 4e^{1-4x} = 0$, $\frac{1}{2\sqrt{x}} = 4e^{1-4x}$ | M1 |
$\frac{1}{8\sqrt{x}} = e^{1-4x}$ | M1 |
$8\sqrt{x} = e^{4x-1}$ | M1 |
$4x - 1 = \ln 8\sqrt{x}$ | M1 |
$x = \frac{1}{4}(1 + \ln 8\sqrt{x})$ | A1 |
**(c)** $x_1 = 0.7699, x_2 = 0.7372, x_3 = 0.7317, x_4 = 0.7308 = 0.731$ (3dp) | M1 A2 |
**(d)** let $f(x) = \frac{1}{2}x^{-\frac{1}{2}} - 4e^{1-4x}$ | M1 |
$f(0.7305) = -0.00025$, $f(0.7315) = 0.0017$ | M1 |
sign change, $f(x)$ continuous $\therefore$ root | A1 |
**(e)** $x_1 = 6.304, x_2 = 1.683 \times 10^{19}$ | B2 |
diverges rapidly away from root | (14 marks)
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**Total: 75 marks**8. The curve $C$ has the equation $y = \sqrt { x } + \mathrm { e } ^ { 1 - 4 x } , x \geq 0$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for the normal to the curve at the point $\left( \frac { 1 } { 4 } , \frac { 3 } { 2 } \right)$.
The curve $C$ has a stationary point with $x$-coordinate $\alpha$ where $0.5 < \alpha < 1$.
\item Show that $\alpha$ is a solution of the equation
$$x = \frac { 1 } { 4 } [ 1 + \ln ( 8 \sqrt { x } ) ]$$
\item Use the iteration formula
$$x _ { n + 1 } = \frac { 1 } { 4 } \left[ 1 + \ln \left( 8 \sqrt { x _ { n } } \right) \right]$$
with $x _ { 0 } = 1$ to find $x _ { 1 } , x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving the value of $x _ { 4 }$ to 3 decimal places.
\item Show that your value for $x _ { 4 }$ is the value of $\alpha$ correct to 3 decimal places.
\item Another attempt to find $\alpha$ is made using the iteration formula
$$x _ { n + 1 } = \frac { 1 } { 64 } \mathrm { e } ^ { 8 x _ { n } - 2 }$$
with $x _ { 0 } = 1$. Describe the outcome of this attempt.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q8 [14]}}