| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Graphs & Exact Values |
| Type | Find general solution of trig equation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring standard technique: solve for the compound angle, find principal values using exact trig values, write general solution, then substitute to find a specific solution. While it's FP1 (inherently slightly harder), the method is routine and well-practiced, making it slightly easier than average overall. |
| Spec | 1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\) | B1 | Correct angle in 1st or 2nd quadrant for \(\sin^{-1}(\sqrt{3}/2)\) |
| \(\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}\) | B1F | Correct ft angle in remaining quadrant; B0F if angle in incorrect quadrant |
| \(2x+\frac{\pi}{4} = 2n\pi + \frac{\pi}{3}\); \(2x+\frac{\pi}{4} = 2n\pi + \frac{2\pi}{3}\) | M1 | OE. Either. Ft on c's \(\sin^{-1}(\sqrt{3}/2)\) |
| \(x = \frac{1}{2}\left(2n\pi+\frac{\pi}{3}-\frac{\pi}{4}\right)\); \(x = \frac{1}{2}\left(2n\pi+\frac{2\pi}{3}-\frac{\pi}{4}\right)\) | m1 | Correct rearrangement of \(2x+\frac{\pi}{4} = 2n\pi+\alpha\) to \(x=\ldots\) |
| GS: \(x = n\pi + \frac{\pi}{24}\); \(x = n\pi + \frac{5\pi}{24}\) | A2,1,0 | Both in ACF; exact and in terms of \(\pi\) for A2. A1 if decimal approx used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(n=5\) gives greatest solution \(<6\pi\): \(= 5\pi + \frac{5\pi}{24}\) | M1 | Applying correct value of \(n\) giving greatest solution \(<6\pi\) |
| \(= \frac{125\pi}{24}\) | A1 | Dep on correct full GS |
## Question 3(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ | B1 | Correct angle in 1st or 2nd quadrant for $\sin^{-1}(\sqrt{3}/2)$ |
| $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$ | B1F | Correct ft angle in remaining quadrant; B0F if angle in incorrect quadrant |
| $2x+\frac{\pi}{4} = 2n\pi + \frac{\pi}{3}$; $2x+\frac{\pi}{4} = 2n\pi + \frac{2\pi}{3}$ | M1 | OE. Either. Ft on c's $\sin^{-1}(\sqrt{3}/2)$ |
| $x = \frac{1}{2}\left(2n\pi+\frac{\pi}{3}-\frac{\pi}{4}\right)$; $x = \frac{1}{2}\left(2n\pi+\frac{2\pi}{3}-\frac{\pi}{4}\right)$ | m1 | Correct rearrangement of $2x+\frac{\pi}{4} = 2n\pi+\alpha$ to $x=\ldots$ |
| GS: $x = n\pi + \frac{\pi}{24}$; $x = n\pi + \frac{5\pi}{24}$ | A2,1,0 | Both in ACF; exact and in terms of $\pi$ for A2. A1 if decimal approx used |
## Question 3(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $n=5$ gives greatest solution $<6\pi$: $= 5\pi + \frac{5\pi}{24}$ | M1 | Applying correct value of $n$ giving greatest solution $<6\pi$ |
| $= \frac{125\pi}{24}$ | A1 | Dep on correct full GS |
---3
\begin{enumerate}[label=(\alph*)]
\item Find the general solution of the equation
$$\sin \left( 2 x + \frac { \pi } { 4 } \right) = \frac { \sqrt { 3 } } { 2 }$$
giving your answer in terms of $\pi$.
\item Use your general solution to find the exact value of the greatest solution of this equation which is less than $6 \pi$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2013 Q3 [8]}}