| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Solve equation involving derivatives |
| Difficulty | Standard +0.8 This question requires quotient rule differentiation of trigonometric functions, setting the derivative to zero, and algebraic manipulation to reach the given form. Part (b) adds numerical solving. The algebraic manipulation from dy/dx = 0 to the target equation requires careful handling of trigonometric identities and is non-trivial, placing it above average difficulty but not exceptionally hard for P3 level. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09a Sign change methods: locate roots |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{(1+\sin x)(-2\sin x) - (1+2\cos x)\cos x}{(1+\sin x)^2}\) | M1A1 | Attempts quotient or product rule; M1 for numerator structure, A1 for correct expression |
| \((1+\sin x)(-2\sin x) - (1+2\cos x)\cos x = 0 \Rightarrow -2\sin x - \cos x - 2 = 0\) | dM1 | Sets numerator to zero, multiplies brackets, uses \(\sin^2 x + \cos^2 x = 1\); no squared terms remain |
| \(2\sin x + \cos x = -2\) * | A1* | No errors or missing brackets; dependent on first M mark |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(R = \sqrt{2^2 + 1^2} = \sqrt{5}\); writes \(2\sin x + \cos x\) in form \(R\sin(x+\alpha)\) or \(R\cos(x-\alpha)\) | M1M1A1 | First M1: \(R=\sqrt{5}\); Second M1: \(\alpha = \tan^{-1}\!\left(\tfrac{1}{2}\right)\) or \(\tan^{-1}(2)\); A1: \(\sqrt{5}\sin(x+0.464)=-2\) or \(\sqrt{5}\cos(x-1.107)=-2\) |
| \(\sqrt{5}\sin(x+0.464) = -2\) or \(\sqrt{5}\cos(x-1.107) = -2 \Rightarrow x = \ldots\) | dM1 | Attempts to solve, finds value for \(x\); dependent on both previous M marks |
| \(x =\) awrt \(3.79\) | A1 | Only one solution required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t^2 + 4t + 3 = 0\) or equivalent | A1 | The \(= 0\) may be implied by later work and the three terms do not need to be on the same side of the equation |
| Correctly rearranges their equation to find a value for \(x\): \(t^2 + 4t + 3 = 0 \Rightarrow t = -3 \Rightarrow x = 2\big(\arctan("-3") + \pi\big)\) | dM1 | Ignore any reference to \(t = -1\). It is dependent on both of the previous method marks. |
| awrt \(3.79\) | A1 |
# Question 9:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{(1+\sin x)(-2\sin x) - (1+2\cos x)\cos x}{(1+\sin x)^2}$ | M1A1 | Attempts quotient or product rule; M1 for numerator structure, A1 for correct expression |
| $(1+\sin x)(-2\sin x) - (1+2\cos x)\cos x = 0 \Rightarrow -2\sin x - \cos x - 2 = 0$ | dM1 | Sets numerator to zero, multiplies brackets, uses $\sin^2 x + \cos^2 x = 1$; no squared terms remain |
| $2\sin x + \cos x = -2$ * | A1* | No errors or missing brackets; dependent on first M mark |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $R = \sqrt{2^2 + 1^2} = \sqrt{5}$; writes $2\sin x + \cos x$ in form $R\sin(x+\alpha)$ or $R\cos(x-\alpha)$ | M1M1A1 | First M1: $R=\sqrt{5}$; Second M1: $\alpha = \tan^{-1}\!\left(\tfrac{1}{2}\right)$ or $\tan^{-1}(2)$; A1: $\sqrt{5}\sin(x+0.464)=-2$ or $\sqrt{5}\cos(x-1.107)=-2$ |
| $\sqrt{5}\sin(x+0.464) = -2$ or $\sqrt{5}\cos(x-1.107) = -2 \Rightarrow x = \ldots$ | dM1 | Attempts to solve, finds value for $x$; dependent on both previous M marks |
| $x =$ awrt $3.79$ | A1 | Only one solution required |
## Mark Scheme Content
**Question (continued):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t^2 + 4t + 3 = 0$ or equivalent | A1 | The $= 0$ may be implied by later work and the three terms do not need to be on the same side of the equation |
| Correctly rearranges their equation to find a value for $x$: $t^2 + 4t + 3 = 0 \Rightarrow t = -3 \Rightarrow x = 2\big(\arctan("-3") + \pi\big)$ | dM1 | Ignore any reference to $t = -1$. It is dependent on both of the previous method marks. |
| awrt $3.79$ | A1 | |
**There may be other variants of these methods, but the mark scheme should still be able to be applied. If in doubt send to review.**
**Answer only scores 0 marks**9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{44035bf8-f54c-472a-b969-b4fa4fa3d203-30_773_775_255_587}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Figure 5 shows the curve with equation
$$y = \frac { 1 + 2 \cos x } { 1 + \sin x } \quad - \frac { \pi } { 2 } < x < \frac { 3 \pi } { 2 }$$
The point $M$, shown in Figure 5, is the minimum point on the curve.
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$ coordinate of $M$ is a solution of the equation
$$2 \sin x + \cos x = - 2$$
\item Hence find, to 3 significant figures, the $x$ coordinate of $M$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2022 Q9 [9]}}