| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Expand compound angle then solve |
| Difficulty | Moderate -0.3 This is a multi-part question that combines routine trigonometric equation solving (part a), numerical verification (b)(i), basic chord gradient calculation (b)(ii), and applying the chord-gradient limit to find a derivative (b)(iii). While it spans several techniques, each individual step is straightforward and follows standard A-level procedures without requiring novel insight or complex problem-solving. The connection between chord gradient and derivative is a standard calculus concept. Slightly easier than average due to the guided, procedural nature. |
| Spec | 1.05o Trigonometric equations: solve in given intervals1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Particular solution, eg \(-\frac{\pi}{6}\) or \(\frac{5\pi}{6}\) | B1 | |
| Introduction of \(n\pi\) or \(2n\pi\) | M1 | |
| GS \(x = -\frac{\pi}{6} + n\pi\) | A1F | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(g(0.05) \approx 0.542 68\) | B1, B1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{g(h) - g(0)}{h} = \frac{\sqrt{3}}{2} - \frac{1}{4}h\) | M1A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| As \(h \to 0\) this gives \(g'(0) = \frac{\sqrt{3}}{2}\) | A1F | 1 mark |
### Part (a)
Particular solution, eg $-\frac{\pi}{6}$ or $\frac{5\pi}{6}$ | B1 | | Degrees or decimals penalised in 3rd mark only
Introduction of $n\pi$ or $2n\pi$ | M1 | |
GS $x = -\frac{\pi}{6} + n\pi$ | A1F | 3 marks | OE(accept unsimplified); ft incorrect first solution
### Part (b)(i)
$f(0.05) \approx 0.542 66$
$g(0.05) \approx 0.542 68$ | B1, B1 | 2 marks | either value AWRT 0.5427; both values correct to 4DP
### Part (b)(ii)
$\frac{g(h) - g(0)}{h} = \frac{\sqrt{3}}{2} - \frac{1}{4}h$ | M1A1 | 2 marks | M1A0 if num error made
### Part (b)(iii)
As $h \to 0$ this gives $g'(0) = \frac{\sqrt{3}}{2}$ | A1F | 1 mark | AWRT 0.866; ft num error
### **Total for Question 7: 8 marks**
---7 The function f is defined for all real numbers by
$$f ( x ) = \sin \left( x + \frac { \pi } { 6 } \right)$$
\begin{enumerate}[label=(\alph*)]
\item Find the general solution of the equation $\mathrm { f } ( x ) = 0$.
\item The quadratic function g is defined for all real numbers by
$$\mathrm { g } ( x ) = \frac { 1 } { 2 } + \frac { \sqrt { 3 } } { 2 } x - \frac { 1 } { 4 } x ^ { 2 }$$
It can be shown that $\mathrm { g } ( x )$ gives a good approximation to $\mathrm { f } ( x )$ for small values of $x$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\mathrm { g } ( 0.05 )$ and $\mathrm { f } ( 0.05 )$ are identical when rounded to four decimal places.
\item A chord joins the points on the curve $y = \mathrm { g } ( x )$ for which $x = 0$ and $x = h$. Find an expression in terms of $h$ for the gradient of this chord.
\item Using your answer to part (b)(ii), find the value of $\mathrm { g } ^ { \prime } ( 0 )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2007 Q7 [8]}}