| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Small angle approximation |
| Type | Approximate area under curve |
| Difficulty | Challenging +1.2 This is a structured multi-part question on small angle approximations that guides students through standard techniques. Part (a) requires routine application of double angle formula and small angle approximations. Part (b) involves straightforward integration of a power function after substitution. Parts (c)(i) and (c)(ii) test conceptual understanding of when approximations are valid, but the reasoning required is relatively straightforward (recognizing that 6.3-6.4 radians are not 'small' angles). While it spans multiple techniques, each step is scaffolded and uses standard A-level methods without requiring novel insight. |
| Spec | 1.05e Small angle approximations: sin x ~ x, cos x ~ 1-x^2/2, tan x ~ x1.05l Double angle formulae: and compound angle formulae1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses small angle approximation for \(\sin x\) at least once | B1 | AO 1.1b |
| Replaces \(\cos 2x\) with \(1 - \frac{(2x)^2}{2}\); or uses double angle identity with small angle approximations | M1 | AO 3.1a — condone sign error or missing brackets |
| \(\sin x - \sin x \cos 2x \approx x - x\!\left(1 - \frac{4x^2}{2}\right) \approx 2x^3\) | R1 | AO 2.1 — completes rigorous argument; condone "=" instead of "\(\approx\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Forms \(\int_0^{0.25} \sqrt{8 \times 2x^3}\,dx\) or better | M1 | AO 3.1a — condone missing limits and \(dx\) |
| Simplifies integrand to \(Bx^{3/2}\) | M1 | AO 1.1a |
| Integrates \(Bx^{3/2}\) correctly: \(= 4\!\left[\frac{2x^{5/2}}{5}\right]_0^{0.25}\) | A1F | AO 1.1b |
| \(= \frac{8}{5} \times 0.25^{5/2} = \frac{8}{5}\times\!\left(\frac{1}{2}\right)^5 = 2^{-2} \times 5^{-1}\) | R1 | AO 2.1 — substitutes correct limits and completes argument in correct form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Explains that the limits 6.4 and 6.3 are not small | E1 | AO 2.4 — the approximation is only valid for small values of \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Explains how limits can be changed: \(\sin x - \sin x\cos 2x\) is periodic (period \(2\pi\)), so evaluating over a different interval gives the same value; reduce limits by \(2\pi\) | E1 | AO 2.4 |
| Deduces \(a = 6.3 - 2\pi \approx 0.017\) and \(b = 6.4 - 2\pi \approx 0.117\) | R1 | AO 2.2a |
## Question 15(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses small angle approximation for $\sin x$ at least once | B1 | AO 1.1b |
| Replaces $\cos 2x$ with $1 - \frac{(2x)^2}{2}$; or uses double angle identity with small angle approximations | M1 | AO 3.1a — condone sign error or missing brackets |
| $\sin x - \sin x \cos 2x \approx x - x\!\left(1 - \frac{4x^2}{2}\right) \approx 2x^3$ | R1 | AO 2.1 — completes rigorous argument; condone "=" instead of "$\approx$" |
## Question 15(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Forms $\int_0^{0.25} \sqrt{8 \times 2x^3}\,dx$ or better | M1 | AO 3.1a — condone missing limits and $dx$ |
| Simplifies integrand to $Bx^{3/2}$ | M1 | AO 1.1a |
| Integrates $Bx^{3/2}$ correctly: $= 4\!\left[\frac{2x^{5/2}}{5}\right]_0^{0.25}$ | A1F | AO 1.1b |
| $= \frac{8}{5} \times 0.25^{5/2} = \frac{8}{5}\times\!\left(\frac{1}{2}\right)^5 = 2^{-2} \times 5^{-1}$ | R1 | AO 2.1 — substitutes correct limits and completes argument in correct form |
## Question 15(c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Explains that the limits 6.4 and 6.3 are not small | E1 | AO 2.4 — the approximation is only valid for small values of $x$ |
## Question 15(c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Explains how limits can be changed: $\sin x - \sin x\cos 2x$ is periodic (period $2\pi$), so evaluating over a different interval gives the same value; reduce limits by $2\pi$ | E1 | AO 2.4 |
| Deduces $a = 6.3 - 2\pi \approx 0.017$ and $b = 6.4 - 2\pi \approx 0.117$ | R1 | AO 2.2a |15
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\sin x - \sin x \cos 2 x \approx 2 x ^ { 3 }$$
for small values of $x$.\\
15
\item Hence, show that the area between the graph with equation
$$y = \sqrt { 8 ( \sin x - \sin x \cos 2 x ) }$$
the positive $x$-axis and the line $x = 0.25$ can be approximated by
$$\text { Area } \approx 2 ^ { m } \times 5 ^ { n }$$
where $m$ and $n$ are integers to be found.\\
15
\item (i) Explain why
$$\int _ { 6.3 } ^ { 6.4 } 2 x ^ { 3 } \mathrm {~d} x$$
is not a suitable approximation for
$$\int _ { 6.3 } ^ { 6.4 } ( \sin x - \sin x \cos 2 x ) d x$$
Question 15 continues on the next page
15 (c) (ii) Explain how
$$\int _ { 6.3 } ^ { 6.4 } ( \sin x - \sin x \cos 2 x ) d x$$
may be approximated by
$$\int _ { a } ^ { b } 2 x ^ { 3 } \mathrm {~d} x$$
for suitable values of $a$ and $b$.\\
\includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-31_2492_1721_217_150}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{042e248a-9efa-4844-957d-f05715900ffc-36_2486_1719_221_150}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2021 Q15 [10]}}