| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Challenging +1.2 This is a multi-part question requiring quotient rule differentiation with exponentials, algebraic manipulation to match a given form, finding turning points, and numerical methods (iteration and change of sign). While it involves several techniques and careful algebra, each part follows standard A-level procedures without requiring novel insight. The 'show that' structure provides scaffolding, and the iteration/numerical work is routine Further Maths content. Slightly above average due to the algebraic complexity and multiple steps, but well within expected Further Maths capability. |
| Spec | 1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(...xe^x + ...e^x\) | M1 | Attempts product rule on \(xe^x\), achieving form \(...xe^x \pm ...e^x\). If quotient rule applied instead, score M0 |
| \(k(xe^x + e^x)\) | A1 | e.g. \(7(xe^x + e^x)\), may be unsimplified, may be implied by further work |
| \(\frac{d}{dx}(\sqrt{e^{3x}-2}) = \frac{1}{2} \times 3e^{3x}(e^{3x}-2)^{-\frac{1}{2}}\) | B1 | Simplified or unsimplified |
| \((f'(x)=)\frac{(e^{3x}-2)^{\frac{1}{2}}("7"xe^x+"7"e^x)-"\frac{3}{2}"e^{3x}(e^{3x}-2)^{-\frac{1}{2}} \times "7"xe^x}{e^{3x}-2}\) | dM1 | Attempts quotient rule (dependent on previous M1). Do not be concerned by constants for "7" or "\(\frac{3}{2}\)" which may both be 1 |
| \(f'(x) = \frac{7e^x(e^{3x}(2-x)-4x-4)}{2(e^{3x}-2)^{\frac{3}{2}}}\) | A1 | Following fully correct differentiated expression. Check if (a) continued after other parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^{3x}(2-x)-4x-4=0 \Rightarrow x(...e^{3x}\pm...)=...e^{3x}\pm...\) | M1 | Sets numerator to zero, collects \(x\) terms on one side, attempts to factorise with \(x\) as common factor. Condone sign slips. Allow \(A\) and \(B\) instead of \(-4\) |
| \(\Rightarrow x = \frac{2e^{3x}-4}{e^{3x}+4}\) * | A1* | Achieves printed answer with no errors including invisible brackets |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Draws vertical line \(x=1\) up to curve then across to line \(y=x\), then up to curve finishing at root (minimum of 2 vertical and horizontal lines tending to root) | B1 | Starting at \(x_1=1\), look for at least 2 sets of vertical and horizontal lines. Condone lack of arrows but sequence should finish at intersection. Condone initial vertical line not starting from \(x\)-axis |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_2 = \frac{2e^3-4}{e^3+4} = 1.5017756...\) | M1 | Substitutes 1 into iterative formula. Values embedded in formula sufficient. May be implied by awrt 1.50 |
| \(x_2 =\) awrt \(1.502\) | A1 | isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\beta = 1.968\) | dB1 | cao, can only be scored if M1 scored in (d)(i). If (d)(i) rounded to 1.50 then allow 1.97 to score M1A0dB1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h(x) = \frac{2e^{3x}-4}{e^{3x}+4} - x\); \(h(0.4315)=-0.000297...\), \(h(0.4325)=0.000947...\) | M1 | Substitutes \(x=0.4315\) and \(x=0.4325\) into suitable function, gets one value correct (rounded or truncated to 1sf). Tighter interval containing root 0.4317388728... also allowable |
| Both calculations correct AND states: change of sign, \(f'(x)\) is continuous, \(\alpha=0.432\) (to 3dp) | A1cao | Requires both calculations correct (to 1sf), statement of change of sign AND continuity of their function (not \(f(x)\)), and minimal conclusion e.g. "hence \(\alpha=0.432\)". Do not allow "hence root" |
## Question 15:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $...xe^x + ...e^x$ | M1 | Attempts product rule on $xe^x$, achieving form $...xe^x \pm ...e^x$. If quotient rule applied instead, score M0 |
| $k(xe^x + e^x)$ | A1 | e.g. $7(xe^x + e^x)$, may be unsimplified, may be implied by further work |
| $\frac{d}{dx}(\sqrt{e^{3x}-2}) = \frac{1}{2} \times 3e^{3x}(e^{3x}-2)^{-\frac{1}{2}}$ | B1 | Simplified or unsimplified |
| $(f'(x)=)\frac{(e^{3x}-2)^{\frac{1}{2}}("7"xe^x+"7"e^x)-"\frac{3}{2}"e^{3x}(e^{3x}-2)^{-\frac{1}{2}} \times "7"xe^x}{e^{3x}-2}$ | dM1 | Attempts quotient rule (dependent on previous M1). Do not be concerned by constants for "7" or "$\frac{3}{2}$" which may both be 1 |
| $f'(x) = \frac{7e^x(e^{3x}(2-x)-4x-4)}{2(e^{3x}-2)^{\frac{3}{2}}}$ | A1 | Following fully correct differentiated expression. Check if (a) continued after other parts |
**Total: 5 marks**
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{3x}(2-x)-4x-4=0 \Rightarrow x(...e^{3x}\pm...)=...e^{3x}\pm...$ | M1 | Sets numerator to zero, collects $x$ terms on one side, attempts to factorise with $x$ as common factor. Condone sign slips. Allow $A$ and $B$ instead of $-4$ |
| $\Rightarrow x = \frac{2e^{3x}-4}{e^{3x}+4}$ * | A1* | Achieves printed answer with no errors including invisible brackets |
**Total: 2 marks**
---
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Draws vertical line $x=1$ up to curve then across to line $y=x$, then up to curve finishing at root (minimum of 2 vertical and horizontal lines tending to root) | B1 | Starting at $x_1=1$, look for at least 2 sets of vertical and horizontal lines. Condone lack of arrows but sequence should finish at intersection. Condone initial vertical line not starting from $x$-axis |
**Total: 1 mark**
---
### Part (d)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_2 = \frac{2e^3-4}{e^3+4} = 1.5017756...$ | M1 | Substitutes 1 into iterative formula. Values embedded in formula sufficient. May be implied by awrt 1.50 |
| $x_2 =$ awrt $1.502$ | A1 | isw |
### Part (d)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\beta = 1.968$ | dB1 | cao, can only be scored if M1 scored in (d)(i). If (d)(i) rounded to 1.50 then allow 1.97 to score M1A0dB1 |
**Total: 3 marks**
---
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h(x) = \frac{2e^{3x}-4}{e^{3x}+4} - x$; $h(0.4315)=-0.000297...$, $h(0.4325)=0.000947...$ | M1 | Substitutes $x=0.4315$ and $x=0.4325$ into suitable function, gets one value correct (rounded or truncated to 1sf). Tighter interval containing root 0.4317388728... also allowable |
| Both calculations correct AND states: change of sign, $f'(x)$ is continuous, $\alpha=0.432$ (to 3dp) | A1cao | Requires both calculations correct (to 1sf), statement of change of sign AND continuity of their function (not $f(x)$), and minimal conclusion e.g. "hence $\alpha=0.432$". Do not allow "hence root" |
**Total: 2 marks**
**Question 15 Total: 13 marks**\begin{enumerate}
\item A curve has equation $y = \mathrm { f } ( x )$, where
\end{enumerate}
$$\mathrm { f } ( x ) = \frac { 7 x \mathrm { e } ^ { x } } { \sqrt { \mathrm { e } ^ { 3 x } - 2 } } \quad x > \ln \sqrt [ 3 ] { 2 }$$
(a) Show that
$$\mathrm { f } ^ { \prime } ( x ) = \frac { 7 \mathrm { e } ^ { x } \left( \mathrm { e } ^ { 3 x } ( 2 - x ) + A x + B \right) } { 2 \left( \mathrm { e } ^ { 3 x } - 2 \right) ^ { \frac { 3 } { 2 } } }$$
where $A$ and $B$ are constants to be found.\\
(b) Hence show that the $x$ coordinates of the turning points of the curve are solutions of the equation
$$x = \frac { 2 \mathrm { e } ^ { 3 x } - 4 } { \mathrm { e } ^ { 3 x } + 4 }$$
The equation $x = \frac { 2 \mathrm { e } ^ { 3 x } - 4 } { \mathrm { e } ^ { 3 x } + 4 }$ has two positive roots $\alpha$ and $\beta$ where $\beta > \alpha$\\
A student uses the iteration formula
$$x _ { n + 1 } = \frac { 2 \mathrm { e } ^ { 3 x _ { n } } - 4 } { \mathrm { e } ^ { 3 x _ { n } } + 4 }$$
in an attempt to find approximations for $\alpha$ and $\beta$\\
Diagram 1 shows a plot of part of the curve with equation $y = \frac { 2 \mathrm { e } ^ { 3 x } - 4 } { \mathrm { e } ^ { 3 x } + 4 }$ and part of the line with equation $y = x$
Using Diagram 1 on page 42\\
(c) draw a staircase diagram to show that the iteration formula starting with $x _ { 1 } = 1$ can be used to find an approximation for $\beta$
Use the iteration formula with $x _ { 1 } = 1$, to find, to 3 decimal places,\\
(d) (i) the value of $x _ { 2 }$\\
(ii) the value of $\beta$
Using a suitable interval and a suitable function that should be stated\\
(e) show that $\alpha = 0.432$ to 3 decimal places.
Only use the copy of Diagram 1 if you need to redraw your answer to part (c).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0839eb5f-2850-4d77-baf7-a6557d71076e-42_736_812_372_143}
\captionsetup{labelformat=empty}
\caption{Diagram 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0839eb5f-2850-4d77-baf7-a6557d71076e-42_738_815_370_1114}
\captionsetup{labelformat=empty}
\caption{copy of Diagram 1}
\end{center}
\end{figure}
\hfill \mbox{\textit{Edexcel Paper 1 2023 Q15 [13]}}