Question 5 - A-Level Maths

Edexcel FP2 2006 June — Question 5 8 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2006
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Taylor series
Type	Taylor series about π/4
Difficulty	Challenging +1.2 This is a straightforward Taylor series question requiring systematic differentiation and substitution at x=π/4, followed by numerical evaluation. While it involves Further Maths content (making it inherently harder than Core), the process is mechanical with no conceptual surprises—students follow the standard Taylor formula, compute derivatives of cos(2x), evaluate at the expansion point, then substitute a numerical value. The 8 marks reflect routine execution rather than problem-solving insight.
Spec	4.08a Maclaurin series: find series for function

5. (a) Find the Taylor expansion of \(\cos 2 x\) in ascending powers of \(\left( x - \frac { \pi } { 4 } \right)\) up to and including the term in \(\left( x - \frac { \pi } { 4 } \right) ^ { 5 }\).
(b) Use your answer to (a) to obtain an estimate of \(\cos 2\), giving your answer to 6 decimal places.
(3)(Total 8 marks)

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(a) \(f(x) = \cos 2x\), \(f\left(\frac{\pi}{4}\right) = 0\)
\(f'(x) = -2\sin 2x\), \(f'\left(\frac{\pi}{4}\right) = -2\)	M1
\(f''(x) = -4\cos 2x\), \(f''\left(\frac{\pi}{4}\right) = 0\)
\(f'''(x) = 8\sin 2x\), \(f'''\left(\frac{\pi}{4}\right) = 8\)	A1
\(f^{(iv)}(x) = 16\cos 2x\), \(f^{(iv)}\left(\frac{\pi}{4}\right) = 0\)
\(f^{(v)}(x) = 32\sin 2x\), \(f^{(v)}\left(\frac{\pi}{4}\right) = -32\)	A1
\(\cos 2x = f\left(\frac{\pi}{4}\right) + f'\left(\frac{\pi}{4}\right)\left(x-\frac{\pi}{4}\right) + \frac{f''\left(\frac{\pi}{4}\right)}{2}\left(x-\frac{\pi}{4}\right)^2 + \frac{f'''\left(\frac{\pi}{4}\right)}{3!}\left(x-\frac{\pi}{4}\right)^3 + \ldots\)	M1
Three terms are sufficient to establish method
\(\cos 2x = -2\left(x - \frac{\pi}{4}\right) + \frac{4}{3}\left(x - \frac{\pi}{4}\right)^3 - \frac{4}{15}\left(x - \frac{\pi}{4}\right)^5 + \ldots\)	A1	5 marks
(b) Substitute \(x = 1\) (\(1 - \frac{\pi}{4} \approx 0.21460\))	B1
\(\cos 2 = -2\left(x - \frac{\pi}{4}\right) + \frac{4}{3}\left(x - \frac{\pi}{4}\right)^3 - \frac{4}{15}\left(x - \frac{\pi}{4}\right)^5 + \ldots\)
\(\approx -0.416147\) cao	M1 A1	3 marks [8]

**(a)** $f(x) = \cos 2x$, $f\left(\frac{\pi}{4}\right) = 0$ | |

$f'(x) = -2\sin 2x$, $f'\left(\frac{\pi}{4}\right) = -2$ | M1 |

$f''(x) = -4\cos 2x$, $f''\left(\frac{\pi}{4}\right) = 0$ | |

$f'''(x) = 8\sin 2x$, $f'''\left(\frac{\pi}{4}\right) = 8$ | A1 |

$f^{(iv)}(x) = 16\cos 2x$, $f^{(iv)}\left(\frac{\pi}{4}\right) = 0$ | |

$f^{(v)}(x) = 32\sin 2x$, $f^{(v)}\left(\frac{\pi}{4}\right) = -32$ | A1 |

$\cos 2x = f\left(\frac{\pi}{4}\right) + f'\left(\frac{\pi}{4}\right)\left(x-\frac{\pi}{4}\right) + \frac{f''\left(\frac{\pi}{4}\right)}{2}\left(x-\frac{\pi}{4}\right)^2 + \frac{f'''\left(\frac{\pi}{4}\right)}{3!}\left(x-\frac{\pi}{4}\right)^3 + \ldots$ | M1 |

Three terms are sufficient to establish method | |

$\cos 2x = -2\left(x - \frac{\pi}{4}\right) + \frac{4}{3}\left(x - \frac{\pi}{4}\right)^3 - \frac{4}{15}\left(x - \frac{\pi}{4}\right)^5 + \ldots$ | A1 | 5 marks

**(b)** Substitute $x = 1$ ($1 - \frac{\pi}{4} \approx 0.21460$) | B1 |

$\cos 2 = -2\left(x - \frac{\pi}{4}\right) + \frac{4}{3}\left(x - \frac{\pi}{4}\right)^3 - \frac{4}{15}\left(x - \frac{\pi}{4}\right)^5 + \ldots$ | |

$\approx -0.416147$ cao | M1 A1 | 3 marks [8]

---

Show LaTeX source

5. (a) Find the Taylor expansion of $\cos 2 x$ in ascending powers of $\left( x - \frac { \pi } { 4 } \right)$ up to and including the term in $\left( x - \frac { \pi } { 4 } \right) ^ { 5 }$.\\
(b) Use your answer to (a) to obtain an estimate of $\cos 2$, giving your answer to 6 decimal places.\\
(3)(Total 8 marks)\\

\hfill \mbox{\textit{Edexcel FP2 2006 Q5 [8]}}

This paper (8 questions)

View full paper

Q1 8 Q2 10 Q3 12 Q4 12 Q5 8 Q6 11 Q7 11 Q8 14