| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Show that derivative equals expression |
| Difficulty | Standard +0.3 This is a multi-part question involving standard chain rule differentiation, equation of tangent through origin, sign-change verification, algebraic rearrangement, and iterative calculation. All parts use routine A-level techniques with no novel insight required. Part (a) is straightforward chain rule, parts (b)-(d) involve guided algebraic manipulation, and part (e) is calculator work. Slightly easier than average due to the heavily scaffolded structure. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \sqrt{3+4e^{x^2}} = \left(3+4e^{x^2}\right)^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \frac{1}{2}\left(3+4e^{x^2}\right)^{-\frac{1}{2}} \times 8xe^{x^2}\) | M1 | Differentiates using chain rule to obtain \(kxe^{x^2}\left(3+4e^{x^2}\right)^{-\frac{1}{2}}\) OR \(ke^{x^2}\left(3+4e^{x^2}\right)^{-\frac{1}{2}}\); if using implicit differentiation accept \(\frac{kxe^{x^2}}{y}\) or \(\frac{ke^{x^2}}{y}\) |
| \(= 4xe^{x^2}\left(3+4e^{x^2}\right)^{-\frac{1}{2}}\) | A1 | Correct derivative in simplest form. Also award for \(\frac{dy}{dx} = \frac{4xe^{x^2}}{y}\) o.e. simplified answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\left(3+4e^{x^2}\right)^{\frac{1}{2}}}{x} = 4xe^{x^2}\left(3+4e^{x^2}\right)^{-\frac{1}{2}}\) | M1 | Sets their \(\frac{dy}{dx} = \frac{\left(3+4e^{x^2}\right)^{\frac{1}{2}}}{x}\) or equivalent such as use of \(y=mx\) with \(m=\) their \(\frac{dy}{dx}\) and \(y=\left(3+4e^{x^2}\right)^{\frac{1}{2}}\); allow with another variable \(x \to \alpha\) |
| \(\frac{\left(3+4e^{x^2}\right)}{x} = 4xe^{x^2}\) | dM1 | Multiplies up to eliminate the square root; allow consistent use of another variable \(x \to \alpha\); dependent on previous M1 and a suitable \(\frac{dy}{dx}\) of the form that scored M1 in part (a) |
| \(4x^2e^{x^2} - 4e^{x^2} - 3 = 0\) * | A1* | Correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = 4x^2e^{x^2} - 4e^{x^2} - 3 \Rightarrow f(1) = -3\) AND \(f(2) = 652\cdots\) | M1 | Attempts both \(f(1)\) and \(f(2)\) (or tighter) and obtains at least one value correct to 1sf rounded or truncated; condone \(f(2)=12e^4-3\); if \(f\) or \(y\) not stated assume \(f(x)=4x^2e^{x^2}-4e^{x^2}-3\) |
| Change of sign and \(f(x)\) is continuous hence root in \((1,2)\) | A1 | Requires: both values correct or rounded/truncated to at least 1sf; reference to sign change e.g. \(f(1)=-3<0\), \(f(2)=652>0\) or \(f(1)\times f(2)<0\); mention of continuity; minimal conclusion e.g. hence root |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4x^2e^{x^2}-4e^{x^2}-3=0 \Rightarrow x^2 = \frac{4e^{x^2}+3}{4e^{x^2}} = \frac{4+3e^{-x^2}}{4} \Rightarrow x = \frac{1}{2}\sqrt{4+3e^{-x^2}}\) * | B1* | Correct proof; look for: the line \(4x^2e^{x^2}-4e^{x^2}-3=0\); evidence of rearrangement e.g. \(x^2=\frac{4e^{x^2}+3}{4e^{x^2}}\) or \(x^2=1+\frac{3}{4e^{x^2}}\) or \(4x^2=\frac{4e^{x^2}+3}{e^{x^2}}\); correct square root work leading to given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_1=1 \Rightarrow x_2 = \frac{1}{2}\sqrt{4+3e^{-1}}\) | M1 | Attempts to use the iteration formula; may be implied by awrt 1.13 for \(x_2\) or awrt 1.10 for \(x_3\) or sight of a correctly embedded value; cannot be awarded for just the value of \(\alpha\) |
| \(x_3 = 1.0997\) | A1 | awrt 1.0997 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\alpha = 1.1051\) | A1 | 1.1051 cao following the award of M1 |
# Question 9:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \sqrt{3+4e^{x^2}} = \left(3+4e^{x^2}\right)^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \frac{1}{2}\left(3+4e^{x^2}\right)^{-\frac{1}{2}} \times 8xe^{x^2}$ | M1 | Differentiates using chain rule to obtain $kxe^{x^2}\left(3+4e^{x^2}\right)^{-\frac{1}{2}}$ OR $ke^{x^2}\left(3+4e^{x^2}\right)^{-\frac{1}{2}}$; if using implicit differentiation accept $\frac{kxe^{x^2}}{y}$ or $\frac{ke^{x^2}}{y}$ |
| $= 4xe^{x^2}\left(3+4e^{x^2}\right)^{-\frac{1}{2}}$ | A1 | Correct derivative in simplest form. Also award for $\frac{dy}{dx} = \frac{4xe^{x^2}}{y}$ o.e. simplified answer |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\left(3+4e^{x^2}\right)^{\frac{1}{2}}}{x} = 4xe^{x^2}\left(3+4e^{x^2}\right)^{-\frac{1}{2}}$ | M1 | Sets their $\frac{dy}{dx} = \frac{\left(3+4e^{x^2}\right)^{\frac{1}{2}}}{x}$ or equivalent such as use of $y=mx$ with $m=$ their $\frac{dy}{dx}$ and $y=\left(3+4e^{x^2}\right)^{\frac{1}{2}}$; allow with another variable $x \to \alpha$ |
| $\frac{\left(3+4e^{x^2}\right)}{x} = 4xe^{x^2}$ | dM1 | Multiplies up to eliminate the square root; allow consistent use of another variable $x \to \alpha$; dependent on previous M1 and a suitable $\frac{dy}{dx}$ of the form that scored M1 in part (a) |
| $4x^2e^{x^2} - 4e^{x^2} - 3 = 0$ * | A1* | Correct proof |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = 4x^2e^{x^2} - 4e^{x^2} - 3 \Rightarrow f(1) = -3$ AND $f(2) = 652\cdots$ | M1 | Attempts both $f(1)$ and $f(2)$ (or tighter) and obtains at least one value correct to 1sf rounded or truncated; condone $f(2)=12e^4-3$; if $f$ or $y$ not stated assume $f(x)=4x^2e^{x^2}-4e^{x^2}-3$ |
| Change of sign and $f(x)$ is continuous hence root in $(1,2)$ | A1 | Requires: both values correct or rounded/truncated to at least 1sf; reference to sign change e.g. $f(1)=-3<0$, $f(2)=652>0$ or $f(1)\times f(2)<0$; mention of continuity; minimal conclusion e.g. hence root |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4x^2e^{x^2}-4e^{x^2}-3=0 \Rightarrow x^2 = \frac{4e^{x^2}+3}{4e^{x^2}} = \frac{4+3e^{-x^2}}{4} \Rightarrow x = \frac{1}{2}\sqrt{4+3e^{-x^2}}$ * | B1* | Correct proof; look for: the line $4x^2e^{x^2}-4e^{x^2}-3=0$; evidence of rearrangement e.g. $x^2=\frac{4e^{x^2}+3}{4e^{x^2}}$ or $x^2=1+\frac{3}{4e^{x^2}}$ or $4x^2=\frac{4e^{x^2}+3}{e^{x^2}}$; correct square root work leading to given answer |
## Part (e)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_1=1 \Rightarrow x_2 = \frac{1}{2}\sqrt{4+3e^{-1}}$ | M1 | Attempts to use the iteration formula; may be implied by awrt 1.13 for $x_2$ or awrt 1.10 for $x_3$ or sight of a correctly embedded value; cannot be awarded for just the value of $\alpha$ |
| $x_3 = 1.0997$ | A1 | awrt 1.0997 |
## Part (e)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha = 1.1051$ | A1 | 1.1051 cao following the award of M1 |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5abaa077-1da4-4023-b442-194f6972095b-26_659_783_287_641}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve $C$ with equation
$$y = \sqrt { 3 + 4 \mathrm { e } ^ { x ^ { 2 } } } \quad x \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, giving your answer in simplest form.
The point $P$ with $x$ coordinate $\alpha$ lies on $C$.\\
Given that the tangent to $C$ at $P$ passes through the origin, as shown in Figure 3,
\item show that $x = \alpha$ is a solution of the equation
$$4 x ^ { 2 } e ^ { x ^ { 2 } } - 4 e ^ { x ^ { 2 } } - 3 = 0$$
\item Hence show that $\alpha$ lies between 1 and 2
\item Show that the equation in part (b) can be written in the form
$$x = \frac { 1 } { 2 } \sqrt { 4 + 3 \mathrm { e } ^ { - x ^ { 2 } } }$$
The iteration formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \sqrt { 4 + 3 \mathrm { e } ^ { - x _ { n } ^ { 2 } } }$$
with $x _ { 1 } = 1$ is used to find an approximation for $\alpha$.
\item Use the iteration formula to find, to 4 decimal places, the value of
\begin{enumerate}[label=(\roman*)]
\item $X _ { 3 }$
\item $\alpha$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2023 Q9 [11]}}