CAIE Further Paper 2 2021 November — Question 8 13 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2021
SessionNovember
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeReduction formulas with hyperbolic integrals
DifficultyChallenging +1.2 This is a structured Further Maths question on hyperbolic functions with clear scaffolding. Part (a) is routine proof from definitions (standard FM exercise). Part (b) is straightforward integration with given substitution. Part (c) requires integration by parts to derive a reduction formula—a standard FM technique, though the algebra needs care. Part (d) applies the formula twice. While this requires multiple techniques and careful manipulation, the question provides significant guidance (substitution hint, derivative formula, clear structure), making it moderately above average but not exceptionally challenging for Further Maths students.
Spec1.08h Integration by substitution4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions8.06a Reduction formulae: establish, use, and evaluate recursively
  1. Starting from the definitions of tanh and sech in terms of exponentials, prove that $$1 - \tanh^2 x = \sech^2 x.$$ [3]
  2. Using the substitution \(u = \tanh x\), or otherwise, find \(\int \sech^2 x \tanh^2 x \, dx\). [2]
  3. It is given that, for \(n \geq 0\), \(I_n = \int_0^{\ln 3} \sech^n x \tanh^2 x \, dx\). Show that, for \(n \geq 2\), $$(n + 1)I_n = \left(\frac{4}{3}\right)^{\frac{3}{n-2}} + (n - 2)I_{n-2}.$$ [You may use the result that \(\frac{d}{dx}(\sech x) = -\tanh x \sech x\).] [5]
  4. Find the value of \(I_4\). [3]
Question 8:
AnswerMarks
8(a)ex −e −x 2
tanhx= sechx=
AnswerMarks Guidance
ex +e −x ex +e −xB1 Writes in exponential form.
( )2 ( )2
ex −e −x  2 ex +e −x − ex −e −x 4
1− = =
AnswerMarks Guidance
 ex +e −x   ( ex +e −x )2 ( ex +e −x )2M1 Writes over common denominator.
sech2 xA1 AG
3
AnswerMarks Guidance
8(b)u2 du= 1tanh3x+C
3M1 A1 Uses substitution correctly. Allow by inspection.
2
AnswerMarks
8(c)I = ln3 sechn−2 xsech2 xtanh2 xdx
n
AnswerMarks Guidance
0B1 Separates into correct structure.
1sechn−2 xtanh3x ln3 +1(n−2) ln3 sechn−2 xtanh4 xdx
AnswerMarks Guidance
3  0 3 0M1 Uses integration by parts correctly.
1 ( 3 )n−2( 4 )3 + 1(n−2) ( I −I )
AnswerMarks Guidance
3 5 5 3 n−2 nM1 A1 Uses 1−sech2x=tanh2x.
( 3+n−2 ) I = ( 3 )n−2( 4 )3 +(n−2)I leading to
n 5 5 n−2
( )n−2( )3
(n+1)I = 3 4 +(n−2)I
AnswerMarks Guidance
n 5 5 n−2A1 AG
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
8(d)( )3
I = 1 4 = 64
AnswerMarks
2 3 5 375B1
( )2( )3 4928
(4+1)I = 3 4 +2I leading to I = =0.105
AnswerMarks Guidance
4 5 5 2 4 46875M1 A1 Applies reduction formula with n=4.
3
Question 8:
--- 8(a) ---
8(a) | ex −e −x 2
tanhx= sechx=
ex +e −x ex +e −x | B1 | Writes in exponential form.
( )2 ( )2
ex −e −x  2 ex +e −x − ex −e −x 4
1− = =
 ex +e −x   ( ex +e −x )2 ( ex +e −x )2 | M1 | Writes over common denominator.
sech2 x | A1 | AG
3
--- 8(b) ---
8(b) | u2 du= 1tanh3x+C
3 | M1 A1 | Uses substitution correctly. Allow by inspection.
2
--- 8(c) ---
8(c) | I = ln3 sechn−2 xsech2 xtanh2 xdx
n
0 | B1 | Separates into correct structure.
1sechn−2 xtanh3x ln3 +1(n−2) ln3 sechn−2 xtanh4 xdx
3  0 3 0 | M1 | Uses integration by parts correctly.
1 ( 3 )n−2( 4 )3 + 1(n−2) ( I −I )
3 5 5 3 n−2 n | M1 A1 | Uses 1−sech2x=tanh2x.
( 3+n−2 ) I = ( 3 )n−2( 4 )3 +(n−2)I leading to
n 5 5 n−2
( )n−2( )3
(n+1)I = 3 4 +(n−2)I
n 5 5 n−2 | A1 | AG
5
Question | Answer | Marks | Guidance
--- 8(d) ---
8(d) | ( )3
I = 1 4 = 64
2 3 5 375 | B1
( )2( )3 4928
(4+1)I = 3 4 +2I leading to I = =0.105
4 5 5 2 4 46875 | M1 A1 | Applies reduction formula with n=4.
3
\begin{enumerate}[label=(\alph*)]
\item Starting from the definitions of tanh and sech in terms of exponentials, prove that

$$1 - \tanh^2 x = \sech^2 x.$$ [3]

\item Using the substitution $u = \tanh x$, or otherwise, find $\int \sech^2 x \tanh^2 x \, dx$. [2]

\item It is given that, for $n \geq 0$, $I_n = \int_0^{\ln 3} \sech^n x \tanh^2 x \, dx$.

Show that, for $n \geq 2$,

$$(n + 1)I_n = \left(\frac{4}{3}\right)^{\frac{3}{n-2}} + (n - 2)I_{n-2}.$$

[You may use the result that $\frac{d}{dx}(\sech x) = -\tanh x \sech x$.] [5]

\item Find the value of $I_4$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q8 [13]}}
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