| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Show root in interval |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard A-level techniques: (a) solving an exponential equation using logarithms (routine manipulation), (b) verifying a root using sign change method (standard procedure with clear guidance), and (c) applying a given iterative formula (calculator work). All parts are well-scaffolded with no novel problem-solving required, making this easier than average. |
| Spec | 1.06b Gradient of e^(kx): derivative and exponential model1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
(a) $5e^{x-1} + 3 = 18 \Rightarrow e^{x-1} = 3$ M1
$x = \ln 3 + 1$ or $e^x = 3e$ A1
$x = \ln 3e$ A1
(3 marks)
(b) $5e^{x-1} + 3 = 10 - x^2$ A1
Sets and proceeds to find and use a suitable function. Eg $f(x) = 7 - x^2 - 5e^{x-1}$ B1
Attempts $f(1.1335) = 0.001$ and $f(1.1345) = -0.007$ M1
Correct values with reason (change of sign and continuous) and conclusion, hence $\alpha$ is $1.134$ to 3 d.p. A1
(3 marks)
(c) $x = -\frac{7 - 5e^{-3-1}}{2} = -2.628388$ M1 A1
$\beta = -2.620330$ A1
(3 marks)
(9 marks)6.
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\caption{Figure 3}
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Figure 3 shows a sketch of curve $C _ { 1 }$ with equation $y = 5 \mathrm { e } ^ { x - 1 } + 3$\\
and curve $C _ { 2 }$ with equation $y = 10 - x ^ { 2 }$\\
The point $P$ lies on $C _ { 1 }$ and has $y$ coordinate 18
\begin{enumerate}[label=(\alph*)]
\item Find the $x$ coordinate of $P$, writing your answer in the form $\ln k$, where $k$ is a constant to be found.
The curve $C _ { 1 }$ meets the curve $C _ { 2 }$ at $x = \alpha$ and at $x = \beta$, as shown in Figure 3.
\item Using a suitable interval and a suitable function that should be stated, show that to 3 decimal places $\alpha = 1.134$
The iterative equation
$$x _ { n + 1 } = - \sqrt { 7 - 5 \mathrm { e } ^ { x _ { n } - 1 } }$$
is used to find an approximation to $\beta$.
Using this iterative formula with $x _ { 1 } = - 3$
\item find the value of $x _ { 2 }$ and the value of $\beta$, giving each answer to 6 decimal places.
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\hfill \mbox{\textit{Edexcel P3 2020 Q6 [9]}}