| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.8 This question requires applying the product rule to a composite function involving trigonometric terms, then algebraically manipulating the stationary point condition (dy/dx = 0) into the given form x = 2arctan(4/x), which involves non-trivial rearrangement. While the iteration part (b) is routine calculator work, the derivation in part (a) demands careful differentiation and algebraic insight beyond standard textbook exercises. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 2x\cos\left(\frac{1}{2}x\right) - \frac{1}{2}x^2\sin\left(\frac{1}{2}x\right)\) | M1 A1 | M1: Differentiates using product rule to obtain form \(Ax\cos\left(\frac{1}{2}x\right) + Bx^2\sin\left(\frac{1}{2}x\right)\). Product rule must be correct if quoted. A1: Correct simplified or unsimplified derivative. |
| Sets \(\frac{dy}{dx} = 0 \Rightarrow \tan\left(\frac{1}{2}x\right) = \frac{4}{x} \Rightarrow x = 2\arctan\left(\frac{4}{x}\right)\)* | dM1 A1* | dM1: Sets \(\frac{dy}{dx}=0\) and proceeds to equation involving \(\tan\left(\frac{1}{2}x\right)\). May be implied. M1 must have been scored. A1*: Must see intermediate line \(\tan\left(\frac{1}{2}x\right)=\frac{4}{x}\) or \(\frac{\sin\left(\frac{1}{2}x\right)}{\cos\left(\frac{1}{2}x\right)}=\frac{2x}{\frac{1}{2}x^2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_2 = 2\arctan\left(\frac{4}{2}\right)\) | M1 | Attempts to use iteration formula. Award for \(x_2 = 2\arctan\left(\frac{4}{2}\right)\) or awrt 2.21. Note: may be implied by awrt 127 if using degrees. |
| \(x_2 =\) awrt \(2.214\) | A1 | |
| \(x_6 = 2.155\) cao | A1 | Note: this is not awrt |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 2x\cos\left(\frac{1}{2}x\right) - \frac{1}{2}x^2\sin\left(\frac{1}{2}x\right)$ | M1 A1 | M1: Differentiates using product rule to obtain form $Ax\cos\left(\frac{1}{2}x\right) + Bx^2\sin\left(\frac{1}{2}x\right)$. Product rule must be correct if quoted. A1: Correct simplified or unsimplified derivative. |
| Sets $\frac{dy}{dx} = 0 \Rightarrow \tan\left(\frac{1}{2}x\right) = \frac{4}{x} \Rightarrow x = 2\arctan\left(\frac{4}{x}\right)$* | dM1 A1* | dM1: Sets $\frac{dy}{dx}=0$ and proceeds to equation involving $\tan\left(\frac{1}{2}x\right)$. May be implied. **M1 must have been scored.** A1*: Must see intermediate line $\tan\left(\frac{1}{2}x\right)=\frac{4}{x}$ or $\frac{\sin\left(\frac{1}{2}x\right)}{\cos\left(\frac{1}{2}x\right)}=\frac{2x}{\frac{1}{2}x^2}$ |
**(4 marks)**
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## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_2 = 2\arctan\left(\frac{4}{2}\right)$ | M1 | Attempts to use iteration formula. Award for $x_2 = 2\arctan\left(\frac{4}{2}\right)$ or awrt 2.21. Note: may be implied by awrt 127 if using degrees. |
| $x_2 =$ awrt $2.214$ | A1 | |
| $x_6 = 2.155$ cao | A1 | Note: this is **not** awrt |
**(3 marks) — Total: 7 marks**
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = x ^ { 2 } \cos \left( \frac { 1 } { 2 } x \right) \quad 0 < x \leqslant \pi$$
The curve has a stationary point at the point $P$.\\
(a) Show, using calculus, that the $x$ coordinate of $P$ is a solution of the equation
$$x = 2 \arctan \left( \frac { 4 } { x } \right)$$
Using the iteration formula
$$x _ { n + 1 } = 2 \arctan \left( \frac { 4 } { x _ { n } } \right) \quad x _ { 1 } = 2$$
(b) find the value of $x _ { 2 }$ and the value of $x _ { 6 }$, giving your answers to 3 decimal places.\\
\hfill \mbox{\textit{Edexcel P3 2021 Q1 [7]}}