CAIE Further Paper 2 2023 November — Question 6 14 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2023
SessionNovember
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeProve hyperbolic identity from exponentials
DifficultyStandard +0.8 Part (a) is routine proof from definitions (standard FM exercise). Part (b) requires recognizing the substitution simplifies the integral using the double angle result, which is moderately challenging. Part (c) is a first-order linear ODE requiring integrating factor method with hyperbolic functions—this demands multiple techniques and careful manipulation across 7 marks, placing it above average difficulty but not exceptionally hard for Further Maths students who have practiced these methods.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions4.10c Integrating factor: first order equations
  1. Starting from the definitions of cosh and sinh in terms of exponentials, prove that $$\sinh 2x = 2\sinh x\cosh x.$$ [3]
  2. Using the substitution \(u = \sinh x\), find \(\int \sinh^2 2x\cosh x\,dx\). [4]
  3. Find the particular solution of the differential equation $$\frac{dy}{dx} + y\tanh x = \sinh^2 2x,$$ given that \(y = 4\) when \(x = 0\). Give your answer in the form \(y = f(x)\). [7]
Question 6:
AnswerMarks
6(a)coshx= 1( ex +e−x) sinhx= 1( ex −e−x)
2 2B1
1( ex −e−x)( ex+e−x) = 1( e2x −e−2x) =sinh2x
AnswerMarks Guidance
2 2M1 A1 Expands, AG.
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(b)u=sinhxdu=coshxdx B1
sinh22xcoshxdx=4sinh2xcosh2xdu=4sinh2x ( sinh2x+1 ) duM1 Applies identities to find integral in terms of u.
=4u2( u2 +1 ) duA1
1 1  1 1 
=4 u5 + u3 (+C)=4 sinh5x+ sinh3x (+C)
AnswerMarks
5 3  5 3 A1
4
AnswerMarks Guidance
6(c)tanhxdx
e =elncoshx =coshxM1 A1 Finds integrating factor.
d
(ycoshx)=sinh22xcoshx
AnswerMarks Guidance
dxM1 Correct form on LHS and attempt to integrate RHS.
1 1 
ycoshx=4 sinh5x+ sinh3x +C
AnswerMarks Guidance
5 3 M1 A1 Integrates RHS using their part (b).
4=CM1 Substitutes initial conditions.
1 1 
y=4sechx sinh5x+ sinh3x+1
AnswerMarks
5 3 A1
7
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(a) ---
6(a) | coshx= 1( ex +e−x) sinhx= 1( ex −e−x)
2 2 | B1
1( ex −e−x)( ex+e−x) = 1( e2x −e−2x) =sinh2x
2 2 | M1 A1 | Expands, AG.
3
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | u=sinhxdu=coshxdx | B1
sinh22xcoshxdx=4sinh2xcosh2xdu=4sinh2x ( sinh2x+1 ) du | M1 | Applies identities to find integral in terms of u.
=4u2( u2 +1 ) du | A1
1 1  1 1 
=4 u5 + u3 (+C)=4 sinh5x+ sinh3x (+C)
5 3  5 3  | A1
4
--- 6(c) ---
6(c) | tanhxdx
e =elncoshx =coshx | M1 A1 | Finds integrating factor.
d
(ycoshx)=sinh22xcoshx
dx | M1 | Correct form on LHS and attempt to integrate RHS.
1 1 
ycoshx=4 sinh5x+ sinh3x +C
5 3  | M1 A1 | Integrates RHS using their part (b).
4=C | M1 | Substitutes initial conditions.
1 1 
y=4sechx sinh5x+ sinh3x+1

5 3  | A1
7
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\alph*)]
\item Starting from the definitions of cosh and sinh in terms of exponentials, prove that
$$\sinh 2x = 2\sinh x\cosh x.$$ [3]

\item Using the substitution $u = \sinh x$, find $\int \sinh^2 2x\cosh x\,dx$. [4]

\item Find the particular solution of the differential equation
$$\frac{dy}{dx} + y\tanh x = \sinh^2 2x,$$
given that $y = 4$ when $x = 0$. Give your answer in the form $y = f(x)$. [7]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q6 [14]}}
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