| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Taylor series about π/3 or π/6 |
| Difficulty | Challenging +1.2 This is a structured Further Maths question requiring systematic differentiation of sec x, pattern recognition to find p and q, then routine application of Taylor series formula about π/3. While it involves multiple steps and Further Maths content (making it harder than typical A-level), the question is highly scaffolded with clear parts guiding students through standard techniques. The differentiation and series expansion are mechanical once the method is known, and part (c) is straightforward substitution. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y=\sec x \Rightarrow \frac{dy}{dx}=\sec x \tan x\); \(\frac{d^2y}{dx^2}=(\sec x\tan x)\tan x+\sec x(\sec^2 x)\) | M1 | Differentiates twice, obtains second derivative expression (sign errors allowed) |
| \(\frac{d^2y}{dx^2}=\sec x(\sec^2 x-1)+\sec^3 x = 2\sec^3 x-\sec x\) | ||
| \(\frac{d^3y}{dx^3}=6\sec^2 x(\sec x\tan x)-\sec x\tan x\) | dM1 | Dependent on M1. Differentiates again for third derivative (sign errors allowed) |
| \(\frac{d^3y}{dx^3}=\sec x\tan x(6\sec^2 x-1)\) | ddM1 A1 | ddM1: Correct form with values of \(p,q\); A1: Fully correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sec\!\left(\frac{\pi}{3}\right)=2,\;\tan\!\left(\frac{\pi}{3}\right)=\sqrt{3} \Rightarrow f\!\left(\frac{\pi}{3}\right)=2,\;f'\!\left(\frac{\pi}{3}\right)=2\sqrt{3},\;f''\!\left(\frac{\pi}{3}\right)=14,\;f'''\!\left(\frac{\pi}{3}\right)=46\sqrt{3}\) | M1 | Uses sec \(x\) and three derivatives to find all four values (at least 2 correct if no method shown) |
| \(f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''(a)+...\) \(\Rightarrow 2+2\sqrt{3}\!\left(x-\frac{\pi}{3}\right)+7\!\left(x-\frac{\pi}{3}\right)^2+\frac{23\sqrt{3}}{3}\!\left(x-\frac{\pi}{3}\right)^3\) | dM1 A1 | dM1: Correct Taylor series with all four values; A1: Correct series (allow \(\sqrt{12}\) for \(2\sqrt{3}\), condone absence of \(\sec x=\) or \(y=\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sec\frac{7\pi}{24} \approx 2+2\sqrt{3}\!\left(-\frac{\pi}{24}\right)+7\!\left(-\frac{\pi}{24}\right)^2+\frac{23\sqrt{3}}{3}\!\left(-\frac{\pi}{24}\right)^3=...\) | M1 | Evidence of substituting \(\frac{7\pi}{24}\) into series (powers of \(x-\frac{\pi}{3}\)). Score M0 if only a value given unless it matches correct 4 s.f. answer |
| \(\sec\frac{7\pi}{24}\approx 1.637\) | A1 | 1.637 only (not awrt) from a correct series |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y=\sec x \Rightarrow \frac{dy}{dx}=\sec x \tan x$; $\frac{d^2y}{dx^2}=(\sec x\tan x)\tan x+\sec x(\sec^2 x)$ | M1 | Differentiates twice, obtains second derivative expression (sign errors allowed) |
| $\frac{d^2y}{dx^2}=\sec x(\sec^2 x-1)+\sec^3 x = 2\sec^3 x-\sec x$ | | |
| $\frac{d^3y}{dx^3}=6\sec^2 x(\sec x\tan x)-\sec x\tan x$ | dM1 | Dependent on M1. Differentiates again for third derivative (sign errors allowed) |
| $\frac{d^3y}{dx^3}=\sec x\tan x(6\sec^2 x-1)$ | ddM1 A1 | ddM1: Correct form with values of $p,q$; A1: Fully correct expression |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sec\!\left(\frac{\pi}{3}\right)=2,\;\tan\!\left(\frac{\pi}{3}\right)=\sqrt{3} \Rightarrow f\!\left(\frac{\pi}{3}\right)=2,\;f'\!\left(\frac{\pi}{3}\right)=2\sqrt{3},\;f''\!\left(\frac{\pi}{3}\right)=14,\;f'''\!\left(\frac{\pi}{3}\right)=46\sqrt{3}$ | M1 | Uses sec $x$ and three derivatives to find all four values (at least 2 correct if no method shown) |
| $f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''(a)+...$ $\Rightarrow 2+2\sqrt{3}\!\left(x-\frac{\pi}{3}\right)+7\!\left(x-\frac{\pi}{3}\right)^2+\frac{23\sqrt{3}}{3}\!\left(x-\frac{\pi}{3}\right)^3$ | dM1 A1 | dM1: Correct Taylor series with all four values; A1: Correct series (allow $\sqrt{12}$ for $2\sqrt{3}$, condone absence of $\sec x=$ or $y=$) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sec\frac{7\pi}{24} \approx 2+2\sqrt{3}\!\left(-\frac{\pi}{24}\right)+7\!\left(-\frac{\pi}{24}\right)^2+\frac{23\sqrt{3}}{3}\!\left(-\frac{\pi}{24}\right)^3=...$ | M1 | Evidence of substituting $\frac{7\pi}{24}$ into series (powers of $x-\frac{\pi}{3}$). Score M0 if only a value given unless it matches correct 4 s.f. answer |
| $\sec\frac{7\pi}{24}\approx 1.637$ | A1 | 1.637 only (not awrt) from a correct series |
---\begin{enumerate}
\item Given that $y = \sec x$\\
(a) show that
\end{enumerate}
$$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \sec x \tan x \left( p \sec ^ { 2 } x + q \right)$$
where $p$ and $q$ are integers to be determined.\\
(b) Hence determine the Taylor series expansion about $\frac { \pi } { 3 }$ of sec $x$ in ascending powers of $\left( x - \frac { \pi } { 3 } \right)$, up to and including the term in $\left( x - \frac { \pi } { 3 } \right) ^ { 3 }$, giving each coefficient in simplest form.\\
(c) Use the answer to part (b) to determine, to four significant figures, an approximate value of $\sec \left( \frac { 7 \pi } { 24 } \right)$
\hfill \mbox{\textit{Edexcel F2 2023 Q6 [9]}}