| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Combining or manipulating standard series |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring standard manipulation of known series. Part (a) involves substituting the exponential series into cosh x = (e^x + e^{-x})/2 and collecting termsβpurely mechanical. Part (b) requires recognizing that cosh(ix) relates to cos(x) via Osborn's rule or direct substitution, which is a standard result. No novel insight or complex problem-solving is needed, just careful algebraic manipulation of given formulas. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| 10(a) | Uses, or writes, | |
| 1 | 1.2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Ignore errors beyond | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| NMS can score 3/3 4 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 10(b) | π₯π₯ | |
| Substitutes for in their expansion of | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Implied by a correct an4swer of | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| NMS can score 3/3 | 1.2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total | 6 | = cosπ₯π₯ |
| Q | Marking instructions | AO |
Question 10:
--- 10(a) ---
10(a) | Uses, or writes,
1 | 1.2 | B1 | 1 π₯π₯ βπ₯π₯
coshπ₯π₯ = 2(e +e )
2 3 4 2 3 4
1 π₯π₯ π₯π₯ π₯π₯ 1 π₯π₯ π₯π₯ π₯π₯
coshπ₯π₯ = οΏ½1+π₯π₯+ + + + ......οΏ½+ οΏ½1βπ₯π₯+ β + + ......οΏ½
2 2! 3! 4! 2 2! 3! 4!
2 4
π₯π₯ π₯π₯
coshπ₯π₯ = 1+ +
2 24
π₯π₯ βπ₯π₯
Substitutes into thcoesirh π₯π₯ = 2( ethe+ Meacl)aurin
expansions of and up to or beyond.
coshπ₯π₯
Allow one sign eπ₯π₯rror. βπ₯π₯ 3
ππ ππ π₯π₯
Ignore errors beyond | 1.1a | M1
4
Correctly simplifies their expression to
π₯π₯
or equivalent.
2 4
π₯π₯ π₯π₯
Accept 2! for 2 and 4! for 24.
1+ 2 +24
Ignore terms beyond
NMS can score 3/3 4 | 1.1b | A1
--- 10(b) ---
10(b) | π₯π₯
Substitutes for in their expansion of | 1.1a | M1 | 2 4
(iπ₯π₯) (iπ₯π₯)
cosh(iπ₯π₯)= 1+ +
2 24
2 4
β1π₯π₯ 1π₯π₯
= 1+ +
2 24
Correctly simiπ₯π₯plifieπ₯π₯s the powers of i to givceo shπ₯π₯
or equivalent.
2 4
π₯π₯ π₯π₯
Accept 2! for 2 and 4! for 24.
1β 2 +24
Ignore terms beyond
Implied by a correct an4swer of | 1.1b | A1
π₯π₯
Recognises the Maclaurin expansion of
cosπ₯π₯
NMS can score 3/3 | 1.2 | B1
cosπ₯π₯
Total | 6 | = cosπ₯π₯
Q | Marking instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item Using the definition of $\cosh x$ and the Maclaurin series expansion of $e^x$, find the first three non-zero terms in the Maclaurin series expansion of $\cosh x$. [3 marks]
\item Hence find a trigonometric function for which the first three terms of its Maclaurin series are the same as the first three terms of the Maclaurin series for $\cosh(ix)$. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2019 Q10 [6]}}