| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Sign Change with Function Evaluation |
| Difficulty | Standard +0.3 This is a multi-part question covering standard C3 techniques: finding roots using a calculator (part a), product rule differentiation (part b), algebraic manipulation to show a given result (part c), straightforward iteration (part d), and sign change verification (part e). While it involves several steps and the exponential function, each individual part uses routine methods with no novel insight required. The 'show that' and 'prove' parts involve guided algebraic work rather than independent problem-solving. This is slightly easier than average due to the structured, guided nature of the question. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.06a Exponential function: a^x and e^x graphs and properties1.06b Gradient of e^(kx): derivative and exponential model1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x) = 0 \Rightarrow x^2 + 3x + 1 = 0 \Rightarrow x = \frac{-3 \pm \sqrt{5}}{2}\) awrt \(-0.382, -2.618\) | M1, A1 | M1: solves by completing square or formula, producing two non-integer answers (do not accept factorisation); A1: answers correct, accept awrt \(-0.382, -2.618\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(vu' + uv'\): \(f'(x) = e^{x^2}(2x+3) + (x^2+3x+1)e^{x^2} \times 2x\) | M1, A1, A1 | M1: applies product rule to \((x^2+3x+1)e^{x^2}\); A1: one correct term; A1: fully correct unsimplified answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(e^{x^2}(2x+3) + (x^2+3x+1)e^{x^2} \times 2x = 0\) | M1 | Sets \(f'(x) = 0\) and factorises out or cancels \(e^{x^2}\) to produce polynomial in \(x\) |
| \(\Rightarrow e^{x^2}\{2x^3 + 6x^2 + 4x + 3\} = 0\) | ||
| \(\Rightarrow x(2x^2+4) = -3(2x^2+1)\) | M1 | Rearranges cubic to \(Ax^3 + Bx = Cx^2 + D\) and factorises to reach \(x(Ax^2+B) = Cx^2+D\) |
| \(\Rightarrow x = -\frac{3(2x^2+1)}{2(x^2+2)}\) | A1* | Correctly proceeds to given answer (given answer mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sub \(x_0 = -2.4\) into \(x_{n+1} = -\frac{3(2x_n^2+1)}{2(x_n^2+2)}\): \(x_1\) awrt \(-2.420\), \(x_2\) awrt \(-2.427\), \(x_3\) awrt \(-2.430\) | M1, A1, A1 | M1 and first A1 for \(x_1\); second A1 for \(x_2\) and \(x_3\) both correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sub \(x = -2.425\) and \(x = -2.435\) into \(f'(x)\) and compare signs: \(f'(-2.425) = +22.4\), \(f'(-2.435) = -15.02\) | M1 | Substitutes both values into \(f'(x)\) (or adapted cubic part) and compares signs |
| Change in sign, hence \(f'(x) = 0\) in between, therefore \(\alpha = -2.43\) (2dp) | A1 | Correct conclusion with correct values shown |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = 0 \Rightarrow x^2 + 3x + 1 = 0 \Rightarrow x = \frac{-3 \pm \sqrt{5}}{2}$ awrt $-0.382, -2.618$ | M1, A1 | M1: solves by completing square or formula, producing two non-integer answers (do not accept factorisation); A1: answers correct, accept awrt $-0.382, -2.618$ |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $vu' + uv'$: $f'(x) = e^{x^2}(2x+3) + (x^2+3x+1)e^{x^2} \times 2x$ | M1, A1, A1 | M1: applies product rule to $(x^2+3x+1)e^{x^2}$; A1: one correct term; A1: fully correct unsimplified answer |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $e^{x^2}(2x+3) + (x^2+3x+1)e^{x^2} \times 2x = 0$ | M1 | Sets $f'(x) = 0$ and factorises out or cancels $e^{x^2}$ to produce polynomial in $x$ |
| $\Rightarrow e^{x^2}\{2x^3 + 6x^2 + 4x + 3\} = 0$ | | |
| $\Rightarrow x(2x^2+4) = -3(2x^2+1)$ | M1 | Rearranges cubic to $Ax^3 + Bx = Cx^2 + D$ and factorises to reach $x(Ax^2+B) = Cx^2+D$ |
| $\Rightarrow x = -\frac{3(2x^2+1)}{2(x^2+2)}$ | A1* | Correctly proceeds to given answer (given answer mark) |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sub $x_0 = -2.4$ into $x_{n+1} = -\frac{3(2x_n^2+1)}{2(x_n^2+2)}$: $x_1$ awrt $-2.420$, $x_2$ awrt $-2.427$, $x_3$ awrt $-2.430$ | M1, A1, A1 | M1 and first A1 for $x_1$; second A1 for $x_2$ and $x_3$ both correct |
## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sub $x = -2.425$ and $x = -2.435$ into $f'(x)$ and compare signs: $f'(-2.425) = +22.4$, $f'(-2.435) = -15.02$ | M1 | Substitutes both values into $f'(x)$ (or adapted cubic part) and compares signs |
| Change in sign, hence $f'(x) = 0$ in between, therefore $\alpha = -2.43$ (2dp) | A1 | Correct conclusion with correct values shown |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a80a71cb-42e0-4587-8f8e-bacd69b8d07a-11_481_858_228_552}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$ where
$$\mathrm { f } ( x ) = \left( x ^ { 2 } + 3 x + 1 \right) \mathrm { e } ^ { x ^ { 2 } }$$
The curve cuts the $x$-axis at points $A$ and $B$ as shown in Figure 2 .
\begin{enumerate}[label=(\alph*)]
\item Calculate the $x$ coordinate of $A$ and the $x$ coordinate of $B$, giving your answers to 3 decimal places.
\item Find $\mathrm { f } ^ { \prime } ( x )$.
The curve has a minimum turning point at the point $P$ as shown in Figure 2.
\item Show that the $x$ coordinate of $P$ is the solution of
$$x = - \frac { 3 \left( 2 x ^ { 2 } + 1 \right) } { 2 \left( x ^ { 2 } + 2 \right) }$$
\item Use the iteration formula
$$x _ { n + 1 } = - \frac { 3 \left( 2 x _ { n } ^ { 2 } + 1 \right) } { 2 \left( x _ { n } ^ { 2 } + 2 \right) } , \quad \text { with } x _ { 0 } = - 2.4$$
to calculate the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$, giving your answers to 3 decimal places.
The $x$ coordinate of $P$ is $\alpha$.
\item By choosing a suitable interval, prove that $\alpha = - 2.43$ to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2013 Q7 [13]}}