Question 5 - A-Level Maths

Edexcel P3 2022 January — Question 5 9 marks

Exam Board	Edexcel
Module	P3 (Pure Mathematics 3)
Year	2022
Session	January
Marks	9
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 straightforward multi-part question testing standard P3 techniques: sign change verification requires simple substitution, the iteration is mechanical calculator work, and finding the maximum requires routine differentiation of ln and polynomial terms followed by solving a linear equation. All parts are textbook exercises with no novel problem-solving required.
Spec	1.07n Stationary points: find maxima, minima using derivatives 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates 1.09a Sign change methods: locate roots 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

5. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{f3272b4c-d8dc-4f32-add9-153de90f4d0a-10_620_622_210_662} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 shows a sketch of part of the curve with equation $$y = 6 \ln ( 2 x + 3 ) - \frac { 1 } { 2 } x ^ { 2 } + 4 \quad x > - \frac { 3 } { 2 }$$ The curve cuts the negative $x$-axis at the point $P$, as shown in Figure 1.

Show that the $x$ coordinate of $P$ lies in the interval $[ - 1.25 , - 1.2 ]$ The curve cuts the positive $x$-axis at the point $Q$, also shown in Figure 1.
Using the iterative formula $$x _ { n + 1 } = \sqrt { 12 \ln \left( 2 x _ { n } + 3 \right) + 8 } \text { with } x _ { 1 } = 6$$
1. find, to 4 decimal places, the value of $x _ { 2 }$
2. find, by continued iteration, the $x$ coordinate of $Q$. Give your answer to 4 decimal places. The curve has a maximum turning point at $M$, as shown in Figure 1.
Using calculus and showing each stage of your working, find the $x$ coordinate of $M$.

Show mark scheme Show mark scheme source

Question 5:

Part (a):

Answer	Marks	Guidance
Answer	Marks	Guidance
Attempts $y$ at $-1.25$ and $-1.2$, one correct to 1sf	M1
$y(-1.25)=-0.9$ and $y(-1.2)=0.2$; sign change and continuous, hence root	A1	Acceptable reasons: sign change and continuous, $y(-1.25)\times y(-1.2)<0$ and continuous; minimal conclusions: "hence proven", "hence root"

Part (b)(i):

Answer	Marks	Guidance
Answer	Marks	Guidance
Attempts $\sqrt{12\ln(15)+8}$	M1	Accept $\sqrt{12\ln(15)+8}=\ldots$ or AWRT 6.4
$x_2=$ AWRT $6.3637$	A1

Part (b)(ii):

Answer	Marks	Guidance
Answer	Marks	Guidance
$x=6.4142$	B1	CAO; must show evidence of $x_2$ or continued iteration

Part (c):

Answer	Marks	Guidance
Answer	Marks	Guidance
$\frac{dy}{dx}=\frac{12}{2x+3}-x$	M1 A1	M1: $\ln(2x+3)\to\frac{\ldots}{2x+3}$; A1: correct derivative
Sets $\frac{12}{2x+3}=x \Rightarrow 2x^2+3x-12=0$	dM1	Sets $\frac{\ldots}{2x+3}\pm\ldots x=0$, proceeds to value of $x$ via correct method for 3TQ
$x=\frac{-3+\sqrt{105}}{4}$ or AWRT $1.81$ ONLY	A1	If $x=\frac{-3-\sqrt{105}}{4}$ also written, it must be rejected

## Question 5:

### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempts $y$ at $-1.25$ and $-1.2$, one correct to 1sf | M1 | |
| $y(-1.25)=-0.9$ and $y(-1.2)=0.2$; sign change and continuous, hence root | A1 | Acceptable reasons: sign change and continuous, $y(-1.25)\times y(-1.2)<0$ and continuous; minimal conclusions: "hence proven", "hence root" |

### Part (b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempts $\sqrt{12\ln(15)+8}$ | M1 | Accept $\sqrt{12\ln(15)+8}=\ldots$ or AWRT 6.4 |
| $x_2=$ AWRT $6.3637$ | A1 | |

### Part (b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x=6.4142$ | B1 | CAO; must show evidence of $x_2$ or continued iteration |

### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx}=\frac{12}{2x+3}-x$ | M1 A1 | M1: $\ln(2x+3)\to\frac{\ldots}{2x+3}$; A1: correct derivative |
| Sets $\frac{12}{2x+3}=x \Rightarrow 2x^2+3x-12=0$ | dM1 | Sets $\frac{\ldots}{2x+3}\pm\ldots x=0$, proceeds to value of $x$ via correct method for 3TQ |
| $x=\frac{-3+\sqrt{105}}{4}$ or AWRT $1.81$ ONLY | A1 | If $x=\frac{-3-\sqrt{105}}{4}$ also written, it must be rejected |

---

Show LaTeX source

5.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f3272b4c-d8dc-4f32-add9-153de90f4d0a-10_620_622_210_662}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a sketch of part of the curve with equation

$$y = 6 \ln ( 2 x + 3 ) - \frac { 1 } { 2 } x ^ { 2 } + 4 \quad x > - \frac { 3 } { 2 }$$

The curve cuts the negative $x$-axis at the point $P$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$ coordinate of $P$ lies in the interval $[ - 1.25 , - 1.2 ]$

The curve cuts the positive $x$-axis at the point $Q$, also shown in Figure 1.\\
Using the iterative formula

$$x _ { n + 1 } = \sqrt { 12 \ln \left( 2 x _ { n } + 3 \right) + 8 } \text { with } x _ { 1 } = 6$$
\item \begin{enumerate}[label=(\roman*)]
\item find, to 4 decimal places, the value of $x _ { 2 }$
\item find, by continued iteration, the $x$ coordinate of $Q$. Give your answer to 4 decimal places.

The curve has a maximum turning point at $M$, as shown in Figure 1.
\end{enumerate}\item Using calculus and showing each stage of your working, find the $x$ coordinate of $M$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel P3 2022 Q5 [9]}}

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