| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Convert sin/cos ratio to tan |
| Difficulty | Moderate -0.3 This is a multi-part C2 question combining numerical integration (trapezium rule with clear instructions), a basic transformation, and a routine trig equation solved by dividing to get tan x. All parts are standard textbook exercises requiring straightforward application of techniques with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = 0.5\) | B1 | \(h = 0.5\) stated or used. PI by \(x\)-values 0, 0.5, 1, 1.5, 2 provided no contradiction |
| \(f(x) = \sin x\); \(I \approx \frac{h}{2}\{\ldots\}\); \(\{\ldots\} = f(0) + f(2) + 2[f(0.5) + f(1) + f(1.5)]\) | M1 | OE summing of areas of 'trapezia' |
| \(\{\ldots\} = 0 + 0.90929\ldots + 2[0.4794\ldots + 0.84147\ldots + 0.99749\ldots]\) | A1 | Min. of 2dp values rounded or truncated. Can be implied by later correct work provided \(>1\) term or a single term which rounds to 1.39 |
| \((I \approx)\ 0.25[5.546\ldots] = 1.3865\ldots = 1.39\) (to 3sf) | A1 (Total: 4) | CAO Must be 1.39 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Stretch(I) in \(y\)-direction(II) scale factor 2(III) | M1, A1 (Total: 2) | Need (I) and either (II) or (III); All correct needs (I) and (II) and (III); [\(>1\) transformation scores 0/2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\sin x}{\cos x} = \frac{1}{2}\); \(\tan x = 0.5\) | M1 | \(\frac{\sin x}{\cos x} = \tan x\) used to get \(\tan x = k\), or identity \(\cos^2 x + \sin^2 x = 1\) used to get either \(\sin^2 x = p\) or \(\cos^2 x = q\) (\(p\) and \(q\) must be between 0 and 1) |
| \(\tan x = 0.5\) | A1 | Either \(\tan x = \frac{1}{2}\) or \(\cos x = \pm\sqrt{\frac{4}{5}}\) \((= \pm 0.894\ldots)\) or \(\sin x = \pm\sqrt{\frac{1}{5}}\) \((= \pm 0.447\ldots)\) |
| \(x = \alpha\) or \(\pi + \alpha\) where \(\alpha = \tan^{-1}(k)\) | m1 | Correct method to find 2nd angle. Any wrong ft quadrants then m0. Squaring method candidates must have rejected extra 'quadrants' for m1. Condone degrees or mixture |
| \(x = 0.464, \ 3.61\) | A1 (Total: 4) | Both. Condone \(>3\)sf [0.463(6..), 3.60(5..or 6.)]; Accept pair of truncated values [0.463, 3.60]; Ignore answers outside interval 0 to 6.28 |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = 0.5$ | B1 | $h = 0.5$ stated or used. PI by $x$-values 0, 0.5, 1, 1.5, 2 provided no contradiction |
| $f(x) = \sin x$; $I \approx \frac{h}{2}\{\ldots\}$; $\{\ldots\} = f(0) + f(2) + 2[f(0.5) + f(1) + f(1.5)]$ | M1 | OE summing of areas of 'trapezia' |
| $\{\ldots\} = 0 + 0.90929\ldots + 2[0.4794\ldots + 0.84147\ldots + 0.99749\ldots]$ | A1 | Min. of 2dp values rounded or truncated. Can be implied by later correct work provided $>1$ term or a single term which rounds to 1.39 |
| $(I \approx)\ 0.25[5.546\ldots] = 1.3865\ldots = 1.39$ (to 3sf) | A1 (Total: 4) | CAO Must be 1.39 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Stretch(I) in $y$-direction(II) scale factor 2(III) | M1, A1 (Total: 2) | Need (I) and either (II) or (III); All correct needs (I) and (II) and (III); [$>1$ transformation scores 0/2] |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\sin x}{\cos x} = \frac{1}{2}$; $\tan x = 0.5$ | M1 | $\frac{\sin x}{\cos x} = \tan x$ used to get $\tan x = k$, or identity $\cos^2 x + \sin^2 x = 1$ used to get either $\sin^2 x = p$ or $\cos^2 x = q$ ($p$ and $q$ must be between 0 and 1) |
| $\tan x = 0.5$ | A1 | Either $\tan x = \frac{1}{2}$ or $\cos x = \pm\sqrt{\frac{4}{5}}$ $(= \pm 0.894\ldots)$ or $\sin x = \pm\sqrt{\frac{1}{5}}$ $(= \pm 0.447\ldots)$ |
| $x = \alpha$ or $\pi + \alpha$ where $\alpha = \tan^{-1}(k)$ | m1 | Correct method to find 2nd angle. Any wrong ft quadrants then m0. Squaring method candidates must have rejected extra 'quadrants' for m1. Condone degrees or mixture |
| $x = 0.464, \ 3.61$ | A1 (Total: 4) | Both. Condone $>3$sf [0.463(6..), 3.60(5..or 6.)]; Accept pair of truncated values [0.463, 3.60]; Ignore answers outside interval 0 to 6.28 |
---\begin{enumerate}[label=(\alph*)]
\item The area of the shaded region is given by $\int _ { 0 } ^ { 2 } \sin x \mathrm {~d} x$, where $x$ is in radians.
Use the trapezium rule with five ordinates (four strips) to find an approximate value for the area of the shaded region, giving your answer to three significant figures.
\item Describe the geometrical transformation that maps the graph of $y = \sin x$ onto the graph of $y = 2 \sin x$.
\item Use a trigonometrical identity to solve the equation
$$2 \sin x = \cos x$$
in the interval $0 \leqslant x \leqslant 2 \pi$, giving your solutions in radians to three significant figures.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q6 [10]}}