| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | General solution — find all solutions |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard techniques for general solutions of trigonometric equations. Part (a) requires recognizing tan(π/6) = 1/√3 and applying the periodicity of tan with a simple substitution. Part (b) extends this by requiring students to consider both positive and negative square roots. While the half-angle and phase shift add minor complexity, these are routine manipulations for FP1 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Mark |
| (a) | \(\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}\) | B1 |
| \(\left(\frac{x}{2} - \frac{\pi}{4}\right) = n\pi + \frac{\pi}{6}\) | M1 | Correct use of either \(n\pi\) or \(2n\pi\). Eg either \(n\pi + \alpha\) or both \(2n\pi + \alpha\) and \(2n\pi + \pi + \alpha\) OE where \(\alpha\) is c's \(\tan^{-1}(1/\sqrt{3})\). Condone degrees/decimals/mixture |
| \(x = 2\left(n\pi + \frac{\pi}{6} + \frac{\pi}{4}\right)\) \(\left(= 2n\pi + \frac{5\pi}{6}\right)\) | m1 | Either correct rearrangement of \(\frac{x}{2} - \frac{\pi}{4} = n\pi + \alpha\) to \(x = ...\), or correct rearrangements of both the equivalents above in the M1 line involving \(2n\pi\), where \(\alpha\) is c's \(\tan^{-1}(1/\sqrt{3})\). Condone degrees/decimals/mixture |
| A1 | 4 marks; ACF, but must now be exact and in terms of \(\pi\) | |
| (b) | \(\tan\left(\frac{x}{2} - \frac{\pi}{4}\right) = \pm\frac{1}{\sqrt{3}}\) | M1 |
| \(\tan\left(\frac{x}{2} - \frac{\pi}{4}\right) = -\frac{1}{\sqrt{3}}\) | m1 | OE If not correct, ft on c's working in (a) with c's \(\alpha\) replaced by \(-\alpha\). Condones as in m1 above |
| \(\Rightarrow \frac{x}{2} - \frac{\pi}{4} = n\pi - \frac{\pi}{6}\) | ||
| \(x = 2\left(n\pi + \frac{\pi}{6} + \frac{\pi}{4}\right)\), \(x = 2\left(n\pi - \frac{\pi}{6} + \frac{\pi}{4}\right)\) \(\left\{x = 2n\pi + \frac{5\pi}{6}, x = 2n\pi + \frac{\pi}{6}\right\}\) | A1F | 3 marks; Any valid form, but only ft on c's exact value for \(\tan^{-1}(1/\sqrt{3})\) in terms of \(\pi\) |
| Part | Answer/Working | Mark | Guidance |
|------|---|---|---|
| (a) | $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$ | B1 | OE (PI) Stated or used. A correct angle in 1st or 3rd quadrant for $\tan^{-1}(1/\sqrt{3})$. Condone degrees/decimal equivs |
| | $\left(\frac{x}{2} - \frac{\pi}{4}\right) = n\pi + \frac{\pi}{6}$ | M1 | Correct use of either $n\pi$ or $2n\pi$. Eg either $n\pi + \alpha$ or both $2n\pi + \alpha$ and $2n\pi + \pi + \alpha$ OE where $\alpha$ is c's $\tan^{-1}(1/\sqrt{3})$. Condone degrees/decimals/mixture |
| | $x = 2\left(n\pi + \frac{\pi}{6} + \frac{\pi}{4}\right)$ $\left(= 2n\pi + \frac{5\pi}{6}\right)$ | m1 | Either correct rearrangement of $\frac{x}{2} - \frac{\pi}{4} = n\pi + \alpha$ to $x = ...$, or correct rearrangements of both the equivalents above in the M1 line involving $2n\pi$, where $\alpha$ is c's $\tan^{-1}(1/\sqrt{3})$. Condone degrees/decimals/mixture |
| | | A1 | 4 marks; ACF, but must now be exact and in terms of $\pi$ |
| (b) | $\tan\left(\frac{x}{2} - \frac{\pi}{4}\right) = \pm\frac{1}{\sqrt{3}}$ | M1 | PI. Taking square roots, must include the $\pm$ or evidence of its use |
| | $\tan\left(\frac{x}{2} - \frac{\pi}{4}\right) = -\frac{1}{\sqrt{3}}$ | m1 | OE If not correct, ft on c's working in (a) with c's $\alpha$ replaced by $-\alpha$. Condones as in m1 above |
| | $\Rightarrow \frac{x}{2} - \frac{\pi}{4} = n\pi - \frac{\pi}{6}$ | | |
| | $x = 2\left(n\pi + \frac{\pi}{6} + \frac{\pi}{4}\right)$, $x = 2\left(n\pi - \frac{\pi}{6} + \frac{\pi}{4}\right)$ $\left\{x = 2n\pi + \frac{5\pi}{6}, x = 2n\pi + \frac{\pi}{6}\right\}$ | A1F | 3 marks; Any valid form, but only ft on c's exact value for $\tan^{-1}(1/\sqrt{3})$ in terms of $\pi$ |6 Find the general solution of each of the following equations:
\begin{enumerate}[label=(\alph*)]
\item $\quad \tan \left( \frac { x } { 2 } - \frac { \pi } { 4 } \right) = \frac { 1 } { \sqrt { 3 } }$;
\item $\quad \tan ^ { 2 } \left( \frac { x } { 2 } - \frac { \pi } { 4 } \right) = \frac { 1 } { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2012 Q6 [7]}}