CAIE Further Paper 2 2022 June — Question 2 8 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2022
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTaylor series
TypeFind series for logarithmic function
DifficultyStandard +0.8 This question requires finding a Maclaurin series coefficient through differentiation of a composite function (ln(cos x)), which involves multiple applications of chain rule and product rule. Part (b) then applies this to arc length, requiring integration of √(1 + tan²x). While the techniques are standard Further Maths content, the multi-step differentiation and the need to recognize simplifications make this moderately challenging but not exceptional for FM Pure.
Spec4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n8.06b Arc length and surface area: of revolution, cartesian or parametric

2
  1. Find the coefficient of \(x ^ { 2 }\) in the Maclaurin's series for \(- \ln \cos x\).
  2. Find the length of the arc of the curve with equation \(\mathrm { y } = - \operatorname { Incos } \mathrm { x }\) from the point where \(x = 0\) to the point where \(x = \frac { 1 } { 4 } \pi\).

Question 2(a):
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{d}{dx}(-\ln\cos x) = \tan x\)B1 Finds first derivative
\(\frac{d}{dx}(\tan x) = \sec^2 x\)B1 Finds second derivative
\(\frac{y''(0)}{2!} = \frac{1}{2}\)M1 A1 Evaluates second derivative at zero. Accept \(\frac{1}{2}x^2\)
Total: 4
Question 2(b):
AnswerMarks Guidance
AnswerMarks Guidance
\(\int_0^{\frac{1}{4}\pi} \sqrt{1+\tan^2 x}\, dx\)M1 Forms correct integral
\(= \int_0^{\frac{1}{4}\pi} \sec x\, dx\)A1
\(= \left[\ln\left(\tan\left(\frac{1}{2}x + \frac{1}{4}\pi\right)\right)\right]_0^{\frac{1}{4}\pi}\)M1 Integrates (answer given in List MF19). Accept \(\ln(\sec x + \tan x)\)
\(\ln\tan\left(\frac{3}{8}\pi\right) = 0.881\)A1 Accept \(\ln(\sqrt{2}+1)\)
Total: 4
**Question 2(a):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dx}(-\ln\cos x) = \tan x$ | B1 | Finds first derivative |
| $\frac{d}{dx}(\tan x) = \sec^2 x$ | B1 | Finds second derivative |
| $\frac{y''(0)}{2!} = \frac{1}{2}$ | M1 A1 | Evaluates second derivative at zero. Accept $\frac{1}{2}x^2$ |
| **Total: 4** | | |

## Question 2(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{\frac{1}{4}\pi} \sqrt{1+\tan^2 x}\, dx$ | M1 | Forms correct integral |
| $= \int_0^{\frac{1}{4}\pi} \sec x\, dx$ | A1 | |
| $= \left[\ln\left(\tan\left(\frac{1}{2}x + \frac{1}{4}\pi\right)\right)\right]_0^{\frac{1}{4}\pi}$ | M1 | Integrates (answer given in List MF19). Accept $\ln(\sec x + \tan x)$ |
| $\ln\tan\left(\frac{3}{8}\pi\right) = 0.881$ | A1 | Accept $\ln(\sqrt{2}+1)$ |
| **Total: 4** | | |

---
2
\begin{enumerate}[label=(\alph*)]
\item Find the coefficient of $x ^ { 2 }$ in the Maclaurin's series for $- \ln \cos x$.
\item Find the length of the arc of the curve with equation $\mathrm { y } = - \operatorname { Incos } \mathrm { x }$ from the point where $x = 0$ to the point where $x = \frac { 1 } { 4 } \pi$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q2 [8]}}