| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Small angle approximation |
| Type | Estimate root of equation |
| Difficulty | Standard +0.3 This question requires knowing that stationary points occur when f'(x)=0, applying the small angle approximation cos(x)≈1-x²/2, and solving a simple quadratic. Part (b) is routine tangent-finding. The small angle approximation is a standard A-level technique, and the algebraic manipulation is straightforward, making this slightly easier than average. |
| Spec | 1.05e Small angle approximations: sin x ~ x, cos x ~ 1-x^2/2, tan x ~ x1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\alpha + \frac{1}{2}\left(1 - \frac{\alpha^2}{2}\right)\) | M1 | Fully substitutes \(\cos x = 1 - \frac{x^2}{2}\) into the derivative |
| \(2\alpha + \frac{1}{2}\left(1 - \frac{\alpha^2}{2}\right) = 0 \Rightarrow 2\alpha + \frac{1}{2} - \frac{\alpha^2}{4} = 0\) | dM1 | Attempts to multiply out to achieve a 3TQ \((=0)\) and attempts to find a value for \(\alpha\). Condone slips. Allow solving quadratic via any method. If they use a calculator you may need to check this. |
| \(\alpha = -0.243\) (3dp) only | A1 | \((\alpha =) -0.243\) only cao. Can only be scored provided a correct 3TQ is seen. If both roots found then the other must be rejected (or a choice made of \(-0.243\)). Condone \(x = -0.243\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(0) = \frac{1}{2}\cos 0 \Rightarrow \ldots \Rightarrow y = \ldots x + 3\) | M1 | Attempts to find the gradient when \(x = 0\) and achieves equation of form \(y = \text{"f}'(0)\text{"}x + 3\). \(x=0\) must be fully substituted. Do not allow if they use gradient of normal. Also allow small angle approximation: \(f'(x) \approx 2x + \frac{1}{2}\left(1-\frac{x^2}{2}\right)\) when \(x=0\), \(f'(0) = \text{"}\frac{1}{2}\text{"} \Rightarrow y = \text{"f}'(0)\text{"}x+3\) |
| \(y = \frac{1}{2}x + 3\) | A1 | \(y = \frac{1}{2}x + 3\) or equivalent in form \(y = mx + c\). Stating just values \(m = 0.5\), \(c = 3\) without correct equation is A0 |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\alpha + \frac{1}{2}\left(1 - \frac{\alpha^2}{2}\right)$ | M1 | Fully substitutes $\cos x = 1 - \frac{x^2}{2}$ into the derivative |
| $2\alpha + \frac{1}{2}\left(1 - \frac{\alpha^2}{2}\right) = 0 \Rightarrow 2\alpha + \frac{1}{2} - \frac{\alpha^2}{4} = 0$ | dM1 | Attempts to multiply out to achieve a 3TQ $(=0)$ and attempts to find a value for $\alpha$. Condone slips. Allow solving quadratic via any method. If they use a calculator you may need to check this. |
| $\alpha = -0.243$ (3dp) only | A1 | $(\alpha =) -0.243$ only cao. Can only be scored provided a correct 3TQ is seen. If both roots found then the other must be rejected (or a choice made of $-0.243$). Condone $x = -0.243$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(0) = \frac{1}{2}\cos 0 \Rightarrow \ldots \Rightarrow y = \ldots x + 3$ | M1 | Attempts to find the gradient when $x = 0$ and achieves equation of form $y = \text{"f}'(0)\text{"}x + 3$. $x=0$ must be fully substituted. Do not allow if they use gradient of normal. Also allow small angle approximation: $f'(x) \approx 2x + \frac{1}{2}\left(1-\frac{x^2}{2}\right)$ when $x=0$, $f'(0) = \text{"}\frac{1}{2}\text{"} \Rightarrow y = \text{"f}'(0)\text{"}x+3$ |
| $y = \frac{1}{2}x + 3$ | A1 | $y = \frac{1}{2}x + 3$ or equivalent in form $y = mx + c$. Stating just values $m = 0.5$, $c = 3$ without correct equation is A0 |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}
The curve $C$ has equation $y = \mathrm { f } ( x )$ where $x \in \mathbb { R }$\\
Given that
\begin{itemize}
\item $\mathrm { f } ^ { \prime } ( x ) = 2 x + \frac { 1 } { 2 } \cos x$
\item the curve has a stationary point with $x$ coordinate $\alpha$
\item $\alpha$ is small\\
(a) use the small angle approximation for $\cos x$ to estimate the value of $\alpha$ to 3 decimal places.
\end{itemize}
The point $P ( 0,3 )$ lies on $C$\\
(b) Find the equation of the tangent to the curve at $P$, giving your answer in the form $y = m x + c$, where $m$ and $c$ are constants to be found.
\hfill \mbox{\textit{Edexcel Paper 1 2023 Q4 [5]}}