Edexcel Paper 1 2019 June — Question 2 5 marks

Exam BoardEdexcel
ModulePaper 1 (Paper 1)
Year2019
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSmall angle approximation
TypeEstimate root of equation
DifficultyModerate -0.3 This is a straightforward application of small angle approximation requiring students to sketch a line on a given graph, identify one intersection point, then substitute cos x ≈ 1 - x²/2 into the equation and solve a simple quadratic. While it involves multiple steps, each step is routine and the question explicitly guides students through the process with clear instructions.
Spec1.05e Small angle approximations: sin x ~ x, cos x ~ 1-x^2/2, tan x ~ x1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases

2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{91a2f26a-add2-4b58-997d-2ae229548217-04_670_1447_212_333} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a plot of part of the curve with equation \(y = \cos x\) where \(x\) is measured in radians. Diagram 1, on the opposite page, is a copy of Figure 1.
  1. Use Diagram 1 to show why the equation $$\cos x - 2 x - \frac { 1 } { 2 } = 0$$ has only one real root, giving a reason for your answer. Given that the root of the equation is \(\alpha\), and that \(\alpha\) is small,
  2. use the small angle approximation for \(\cos x\) to estimate the value of \(\alpha\) to 3 decimal places.
    \includegraphics[max width=\textwidth, alt={}]{91a2f26a-add2-4b58-997d-2ae229548217-05_664_1452_246_333}
    \section*{Diagram 1}

Question 2:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Draws \(y = 2x + \frac{1}{2}\) on Figure 1/Diagram 1 with correct gradient and interceptB1 Look for straight line with intercept \(\approx \frac{1}{2}\) and further point at \(\approx\left(\frac{1}{2}, 1\frac{1}{2}\right)\). Tolerance of 0.25 of a square. Must appear in quadrants 1, 2 and 3.
States there is only one intersection, therefore only one rootB1 Requires a reason and minimal conclusion. Allowable linear graph must have intercept \(\pm\frac{1}{2}\) with one intersection with \(\cos x\) OR gradient \(\pm 2\) with one intersection with \(\cos x\)
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\cos x - 2x - \frac{1}{2} = 0 \Rightarrow 1 - \frac{x^2}{2} - 2x - \frac{1}{2} = 0\)M1 Attempts small angle approximation \(\cos x = 1 - \frac{x^2}{2}\) in given equation
Solves their \(x^2 + 4x - 1 = 0\)dM1 Proceeds to 3TQ in single variable and attempts to solve. Allow completion of square, formula or calculator. Do not allow factorisation unless equation does factorise.
\(x \approx 0.236\), accept \(-2 + \sqrt{5}\)A1 Do not allow if another root given and it is not obvious that 0.236 has been chosen
## Question 2:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Draws $y = 2x + \frac{1}{2}$ on Figure 1/Diagram 1 with correct gradient and intercept | B1 | Look for straight line with intercept $\approx \frac{1}{2}$ and further point at $\approx\left(\frac{1}{2}, 1\frac{1}{2}\right)$. Tolerance of 0.25 of a square. Must appear in quadrants 1, 2 and 3. |
| States there is only one intersection, therefore only one root | B1 | Requires a reason and minimal conclusion. Allowable linear graph must have intercept $\pm\frac{1}{2}$ with one intersection with $\cos x$ OR gradient $\pm 2$ with one intersection with $\cos x$ |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos x - 2x - \frac{1}{2} = 0 \Rightarrow 1 - \frac{x^2}{2} - 2x - \frac{1}{2} = 0$ | M1 | Attempts small angle approximation $\cos x = 1 - \frac{x^2}{2}$ in given equation |
| Solves their $x^2 + 4x - 1 = 0$ | dM1 | Proceeds to 3TQ in single variable and attempts to solve. Allow completion of square, formula or calculator. Do not allow factorisation unless equation does factorise. |
| $x \approx 0.236$, accept $-2 + \sqrt{5}$ | A1 | Do not allow if another root given and it is not obvious that 0.236 has been chosen |

---
2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{91a2f26a-add2-4b58-997d-2ae229548217-04_670_1447_212_333}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a plot of part of the curve with equation $y = \cos x$ where $x$ is measured in radians. Diagram 1, on the opposite page, is a copy of Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Use Diagram 1 to show why the equation

$$\cos x - 2 x - \frac { 1 } { 2 } = 0$$

has only one real root, giving a reason for your answer.

Given that the root of the equation is $\alpha$, and that $\alpha$ is small,
\item use the small angle approximation for $\cos x$ to estimate the value of $\alpha$ to 3 decimal places.

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{91a2f26a-add2-4b58-997d-2ae229548217-05_664_1452_246_333}
\end{center}

\section*{Diagram 1}
\end{enumerate}

\hfill \mbox{\textit{Edexcel Paper 1 2019 Q2 [5]}}