CAIE Further Paper 2 2020 June — Question 6 12 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2020
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeMaclaurin series for inverse hyperbolics
DifficultyStandard +0.8 Part (a) is a standard hyperbolic identity proof from definitions (routine for Further Maths). Part (b) requires implicit differentiation with the identity from (a), which is straightforward. Part (c) is more demanding: finding a Maclaurin series requires evaluating the function and derivatives at x=0, involving inverse hyperbolic functions and careful algebraic manipulation. The multi-step nature and integration of several techniques (implicit differentiation, series expansion, exact value calculation) elevates this above average difficulty, though it remains a structured question with clear pathways.
Spec4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions4.08a Maclaurin series: find series for function
  1. Starting from the definitions of \(\tanh\) and \(\sech\) in terms of exponentials, prove that $$1 - \tanh^2 \theta = \sech^2 \theta.$$ [3]
The variables \(x\) and \(y\) are such that \(\tanh y = \cos\left(x + \frac{1}{4}\pi\right)\), for \(-\frac{1}{4}\pi < x < \frac{3}{4}\pi\).
  1. By differentiating the equation \(\tanh y = \cos\left(x + \frac{1}{4}\pi\right)\) with respect to \(x\), show that $$\frac{dy}{dx} = -\operatorname{cosec}\left(x + \frac{1}{4}\pi\right).$$ [4]
  2. Hence find the first three terms in the Maclaurin's series for \(\tanh^{-1}\left(\cos\left(x + \frac{1}{4}\pi\right)\right)\) in the form \(\frac{1}{2}\ln a + bx + cx^2\), giving the exact values of the constants \(a\), \(b\) and \(c\). [5]
Question 6:
AnswerMarks
6Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9231/23 Cambridge International AS & A Level – Mark Scheme May/June 2020
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no “follow through” from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2020 Page 5 of 13
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(a)θ−e −θ
e 2
tanhθ= sechθ=
θ+e −θ θ+e −θ
AnswerMarks
e eB1
( ) ( )
e θ−e −θ 2 e θ+e −θ 2 − e θ−e −θ 2 4
1− = =
AnswerMarks
  e θ+e −θ   ( e θ+e −θ ) 2 ( e θ+e −θ ) 2M1
= sech2θA1
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(b)dy
tanh y = cos(x+ 1 π)  sech2 y = −sin(x+ 1 π)
AnswerMarks
4 dx 4M1 A1
( )dy
1−cos2(x+ 1 π) = −sin(x+ 1 π)
AnswerMarks
4 dx 4M1
dy sin(x+ 1π)
=− 4 =−cosec(x+ 1π)
dx sin2(x+ 1π) 4
AnswerMarks
4A1
4
AnswerMarks
6(c)d2y
=cot(x+1π)cosec(x+1π)
AnswerMarks
dx2 4 4B1
y'(0)=−cosec(1π) y''(0)=
cot(1π)cosec(1π)
AnswerMarks
4 4 4M1
( ) 2+√2
y(0)= tanh−1 1√2 = 1ln 
2 2 2−√2
AnswerMarks
 M1
y= 1ln(3+2√2)−x√2+1 x2√2
AnswerMarks
2 2M1 A1
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9231/23 Cambridge International AS & A Level – Mark Scheme May/June 2020
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no “follow through” from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2020 Page 5 of 13
Question | Answer | Marks
--- 6(a) ---
6(a) | θ−e −θ
e 2
tanhθ= sechθ=
θ+e −θ θ+e −θ
e e | B1
( ) ( )
e θ−e −θ 2 e θ+e −θ 2 − e θ−e −θ 2 4
1− = =
  e θ+e −θ   ( e θ+e −θ ) 2 ( e θ+e −θ ) 2 | M1
= sech2θ | A1
3
Question | Answer | Marks
--- 6(b) ---
6(b) | dy
tanh y = cos(x+ 1 π)  sech2 y = −sin(x+ 1 π)
4 dx 4 | M1 A1
( )dy
1−cos2(x+ 1 π) = −sin(x+ 1 π)
4 dx 4 | M1
dy sin(x+ 1π)
=− 4 =−cosec(x+ 1π)
dx sin2(x+ 1π) 4
4 | A1
4
--- 6(c) ---
6(c) | d2y
=cot(x+1π)cosec(x+1π)
dx2 4 4 | B1
y'(0)=−cosec(1π) y''(0)=
cot(1π)cosec(1π)
4 4 4 | M1
( ) 2+√2
y(0)= tanh−1 1√2 = 1ln 
2 2 2−√2
  | M1
y= 1ln(3+2√2)−x√2+1 x2√2
2 2 | M1 A1
5
Question | Answer | Marks
\begin{enumerate}[label=(\alph*)]
\item Starting from the definitions of $\tanh$ and $\sech$ in terms of exponentials, prove that
$$1 - \tanh^2 \theta = \sech^2 \theta.$$
[3]
\end{enumerate}

The variables $x$ and $y$ are such that $\tanh y = \cos\left(x + \frac{1}{4}\pi\right)$, for $-\frac{1}{4}\pi < x < \frac{3}{4}\pi$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item By differentiating the equation $\tanh y = \cos\left(x + \frac{1}{4}\pi\right)$ with respect to $x$, show that
$$\frac{dy}{dx} = -\operatorname{cosec}\left(x + \frac{1}{4}\pi\right).$$
[4]

\item Hence find the first three terms in the Maclaurin's series for $\tanh^{-1}\left(\cos\left(x + \frac{1}{4}\pi\right)\right)$ in the form
$\frac{1}{2}\ln a + bx + cx^2$, giving the exact values of the constants $a$, $b$ and $c$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q6 [12]}}
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