| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for products/secant |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on Maclaurin series requiring systematic differentiation of sec x (using quotient rule repeatedly), substitution to find coefficients, and then applying series to evaluate a limit. While it involves multiple steps and Further Maths content, the approach is methodical and guided, with part (a) scaffolding part (b). The techniques are standard for FM students who have learned Taylor series. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| \(f''(x) = \sec x(\sec^2 x) + \tan x(\sec x \tan x) = \sec x(\sec^2 x + \tan^2 x)\) | B1 | Differentiates twice correctly |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 5\sec^3 x \tan x + \sec x \tan^3 x\) | M1 | Differentiates three times |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 18\sec^3 x \tan^2 x + 5\sec^5 x + \tan^4 x \sec x\) | A1 | Correct expression for fourth derivative |
| \(f^{(4)}(0) = 0 + 5 + 0 = 5\) | A1 | Obtains required result from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(0)=1,\ f'(0)=0,\ f''(0)=1,\ f^{(3)}(0)=0\) | M1 | Substitutes zero and uses in Maclaurin expansion; condone no division by factorial |
| \[f(x) = 1 + \frac{x^2}{2} + \frac{5x^4}{24} + \cdots\] | A1 | Allow factorial notation |
| Answer | Marks | Guidance |
|---|---|---|
| \[\cosh x = \frac{1}{2}\left(1+x+\frac{x^2}{2!}+\cdots\right) + \frac{1}{2}\left(1-x+\frac{x^2}{2!}+\cdots\right) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!}+\cdots\] | B1 | Deduces correct expansion of \(\cosh x\); OR justifies use of l'Hôpital's rule (must see 0/0 at least once) |
| \[\lim_{x\to 0}\left(\frac{\sec x - \cosh x}{x^4}\right) = \lim_{x\to 0} \frac{\left(1+\frac{x^2}{2!}+\frac{5x^4}{4!}+\cdots\right)-\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)}{x^4}\] | M1 | Substitutes expansions into expression; OR applies l'Hôpital's rule four times |
| \[= \lim_{x\to 0} \frac{\frac{x^4}{6} + \text{higher powers}}{x^4}\] | A1 | Simplifies correctly |
| Answer | Marks | Guidance |
|---|---|---|
| \[\therefore \lim_{x\to 0}\left(\frac{\sec x - \cosh x}{x^4}\right) = \frac{1}{6}\] | R1 | Completes rigorous argument including clear demonstration of limiting process; must show higher powers until limit is taken OR justify l'Hôpital's at each stage |
## Question 8(a)(i):
$f'(x) = \sec x \tan x$
$f''(x) = \sec x(\sec^2 x) + \tan x(\sec x \tan x) = \sec x(\sec^2 x + \tan^2 x)$ | B1 | Differentiates twice correctly
$f^{(3)}(x) = \sec x \tan x(\sec^2 x + \tan^2 x) + \sec x(2\sec x(\sec x \tan x) + 2\tan x \sec^2 x)$
$= 5\sec^3 x \tan x + \sec x \tan^3 x$ | M1 | Differentiates three times
$f^{(4)}(x) = \tan x(15\sec^2 x(\sec x \tan x)) + 5\sec^5 x + \sec x(3\tan^2 x \sec^2 x) + \tan^3 x(\sec x \tan x)$
$= 18\sec^3 x \tan^2 x + 5\sec^5 x + \tan^4 x \sec x$ | A1 | Correct expression for fourth derivative
$f^{(4)}(0) = 0 + 5 + 0 = 5$ | A1 | Obtains required result from correct working
---
## Question 8(a)(ii):
$f(0)=1,\ f'(0)=0,\ f''(0)=1,\ f^{(3)}(0)=0$ | M1 | Substitutes zero and uses in Maclaurin expansion; condone no division by factorial
$$f(x) = 1 + \frac{x^2}{2} + \frac{5x^4}{24} + \cdots$$ | A1 | Allow factorial notation
---
## Question 8(b):
$$\cosh x = \frac{1}{2}\left(1+x+\frac{x^2}{2!}+\cdots\right) + \frac{1}{2}\left(1-x+\frac{x^2}{2!}+\cdots\right) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!}+\cdots$$ | B1 | Deduces correct expansion of $\cosh x$; OR justifies use of l'Hôpital's rule (must see 0/0 at least once)
$$\lim_{x\to 0}\left(\frac{\sec x - \cosh x}{x^4}\right) = \lim_{x\to 0} \frac{\left(1+\frac{x^2}{2!}+\frac{5x^4}{4!}+\cdots\right)-\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)}{x^4}$$ | M1 | Substitutes expansions into expression; OR applies l'Hôpital's rule four times
$$= \lim_{x\to 0} \frac{\frac{x^4}{6} + \text{higher powers}}{x^4}$$ | A1 | Simplifies correctly
$$= \lim_{x\to 0}\left(\frac{1}{6} + \text{higher powers}\right) = \frac{1}{6}$$
$$\therefore \lim_{x\to 0}\left(\frac{\sec x - \cosh x}{x^4}\right) = \frac{1}{6}$$ | R1 | Completes rigorous argument including clear demonstration of limiting process; must show higher powers until limit is taken OR justify l'Hôpital's at each stage8
\begin{enumerate}[label=(\alph*)]
\item The function f is defined as $\mathrm { f } ( x ) = \sec x$
8 (a) (i) Show that $\mathrm { f } ^ { ( 4 ) } ( 0 ) = 5$\\
8 (a) (ii) Hence find the first three non-zero terms of the Maclaurin series for $\mathrm { f } ( x ) = \sec x$\\
8
\item Prove that
$$\lim _ { x \rightarrow 0 } \left( \frac { \sec x - \cosh x } { x ^ { 4 } } \right) = \frac { 1 } { 6 }$$
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2022 Q8 [10]}}