Question 5 - A-Level Maths

Edexcel F3 2024 January — Question 5 11 marks

Exam Board	Edexcel
Module	F3 (Further Pure Mathematics 3)
Year	2024
Session	January
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Hyperbolic functions
Type	Reduction formulas with hyperbolic integrals
Difficulty	Challenging +1.3 This is a structured Further Maths question on reduction formulas with hyperbolic functions. Part (a) is routine identity verification using exponential definitions. Part (b) requires applying the identity to derive a reduction formula—a standard technique in F3. Part (c) involves iterating the formula twice and evaluating definite integrals. While requiring multiple steps and careful algebra, the question follows a well-signposted path with no novel insights needed, making it moderately above average difficulty for Further Maths content.
Spec	4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials 4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 8.06a Reduction formulae: establish, use, and evaluate recursively

(a) Use the definitions of hyperbolic functions in terms of exponentials to prove that

$$\begin{gathered} 1 - \operatorname { sech } ^ { 2 } x \equiv \tanh ^ { 2 } x \\ I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 3 } \ln 2 } \tanh ^ { n } 3 x \mathrm {~d} x \quad n \in \mathbb { Z } \quad n \geqslant 0 \end{gathered}$$ (b) Show that $$I _ { n } = I _ { n - 2 } - \frac { p ^ { n - 1 } } { 3 ( n - 1 ) } \quad n \geqslant 2$$ where $p$ is a rational number to be determined.
(c) Hence determine the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 3 } \ln 2 } \tanh ^ { 5 } 3 x \mathrm {~d} x$$ giving your answer in the form $a \ln b + c$ where $a , b$ and $c$ are rational numbers to be found.

Show mark scheme Show mark scheme source

Question 5(a):

Way 1 (From LHS):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$(1-\text{sech}^2 x =)\ 1-\left(\frac{2}{e^x+e^{-x}}\right)^2$	B1	Replaces $\text{sech}\, x$ with correct expression in terms of exponentials
$= \frac{(e^x+e^{-x})^2-4}{(e^x+e^{-x})^2} = \frac{e^{2x}+2+e^{-2x}-4}{(e^x+e^{-x})^2}$	M1	Expresses as single fraction and expands numerator
$= \frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2} = \tanh^2 x$	A1*	Fully correct proof

Way 2 (Difference of squares):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$1-\text{sech}^2x = (1+\text{sech}\,x)(1-\text{sech}\,x) = \left(1+\frac{2}{e^x+e^{-x}}\right)\left(1-\frac{2}{e^x+e^{-x}}\right)$	B1	Uses difference of two squares and replaces $\text{sech}\,x$ correctly
$= \frac{(e^x+e^{-x}+2)(e^x+e^{-x}-2)}{(e^x+e^{-x})^2}$ expanded	M1	Expresses as single fraction and expands numerator
$= \frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2} = \tanh^2 x$	A1*	Fully correct proof

Way 3 (From RHS):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\tanh^2 x = \frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}$	B1	Replaces $\tanh x$ with correct expression in exponentials
$= \frac{e^{2x}+2+e^{-2x}}{(e^x+e^{-x})^2} - \frac{4}{(e^x+e^{-x})^2}$	M1	Expands numerator and splits into two fractions
$= 1 - \left(\frac{2}{e^x+e^{-x}}\right)^2 = 1-\text{sech}^2 x$	A1*	Fully correct proof

Question (b): Reduction Formula

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\int \tanh^n 3x \, dx = \int \tanh^{n-2} 3x \tanh^2 3x \, dx = \int \tanh^{n-2} 3x(1-\text{sech}^2 3x) \, dx$	M1	Splits $\tanh^n 3x$ into $\tanh^{n-2} 3x \tanh^2 3x$ and applies $\tanh^2 3x = 1 - \text{sech}^2 3x$
$\int \tanh^{n-2} 3x \, \text{sech}^2 3x \, dx = \frac{1}{3(n-1)} \tanh^{n-1} 3x$	dM1	Expands and integrates $\tanh^{n-2} 3x \, \text{sech}^2 3x$ to obtain $\alpha \tanh^{n-1} 3x$. Integration by parts also acceptable. Must be complete method leading to $\alpha \tanh^{n-1} 3x$
$I_n = I_{n-2} - \frac{1}{3(n-1)}\left[\tanh^{n-1} 3x\right]_0^{\frac{1}{3}\ln 2} = I_{n-2} - \frac{1}{3(n-1)}\left(\frac{e^{2\ln 2}-1}{e^{2\ln 2}+1}\right)^{n-1}$	ddM1	Introduces $I_{n-2}$ and applies $x = \frac{1}{3}\ln 2$ using correct exponential definition of tanh, or accept calculator use e.g. $\tanh(\ln 2) = \frac{3}{5}$
$I_n = I_{n-2} - \dfrac{\left(\frac{3}{5}\right)^{n-1}}{3(n-1)}$	A1	Fully correct proof. Allow recovery from slips e.g. tanh→tan→tanh. Clear unrecovered errors score A0.

Question (c): Evaluating $I_5$

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$I_5 = I_3 - \dfrac{\left(\frac{3}{5}\right)^{5-1}}{3(5-1)} = I_1 - \dfrac{\left(\frac{3}{5}\right)^{3-1}}{3(3-1)} - \dfrac{\left(\frac{3}{5}\right)^{5-1}}{3(5-1)}$	M1	Uses reduction formula to obtain $I_5$ in terms of $I_1$. Condone use of letter $p$ for $\frac{3}{5}$
$\int \tanh 3x \, dx = \frac{1}{3}\ln(\cosh 3x)$	M1	Integrates to obtain $q\ln(\cosh rx)$ or e.g. $q\ln(\text{sech}\, rx)$. Condone $q$ and/or $r=1$
$I_5 = \frac{1}{3}\ln\!\left(\dfrac{e^{\ln 2}+e^{-\ln 2}}{2}\right) - \dfrac{\left(\frac{9}{25}\right)}{6} - \dfrac{\left(\frac{81}{625}\right)}{12}$	ddM1	Applies $x = \frac{1}{3}\ln 2$ using correct exponential definition of cosh, or calculator e.g. $\cosh(\ln 2) = \frac{5}{4}$. Must not be in terms of $p$; must use value of $p$ from part (b)
$\dfrac{1}{3}\ln\dfrac{5}{4} - \dfrac{177}{2500}$	A1	Correct answer in correct form (allow $a=\ldots,\, b=\ldots,\, c=\ldots$). Allow $-0.0708$ for $c$

## Question 5(a):

**Way 1 (From LHS):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1-\text{sech}^2 x =)\ 1-\left(\frac{2}{e^x+e^{-x}}\right)^2$ | B1 | Replaces $\text{sech}\, x$ with correct expression in terms of exponentials |
| $= \frac{(e^x+e^{-x})^2-4}{(e^x+e^{-x})^2} = \frac{e^{2x}+2+e^{-2x}-4}{(e^x+e^{-x})^2}$ | M1 | Expresses as single fraction and expands numerator |
| $= \frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2} = \tanh^2 x$ | A1* | Fully correct proof |

**Way 2 (Difference of squares):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $1-\text{sech}^2x = (1+\text{sech}\,x)(1-\text{sech}\,x) = \left(1+\frac{2}{e^x+e^{-x}}\right)\left(1-\frac{2}{e^x+e^{-x}}\right)$ | B1 | Uses difference of two squares and replaces $\text{sech}\,x$ correctly |
| $= \frac{(e^x+e^{-x}+2)(e^x+e^{-x}-2)}{(e^x+e^{-x})^2}$ expanded | M1 | Expresses as single fraction and expands numerator |
| $= \frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2} = \tanh^2 x$ | A1* | Fully correct proof |

**Way 3 (From RHS):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tanh^2 x = \frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}$ | B1 | Replaces $\tanh x$ with correct expression in exponentials |
| $= \frac{e^{2x}+2+e^{-2x}}{(e^x+e^{-x})^2} - \frac{4}{(e^x+e^{-x})^2}$ | M1 | Expands numerator and splits into two fractions |
| $= 1 - \left(\frac{2}{e^x+e^{-x}}\right)^2 = 1-\text{sech}^2 x$ | A1* | Fully correct proof |

# Question (b): Reduction Formula

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \tanh^n 3x \, dx = \int \tanh^{n-2} 3x \tanh^2 3x \, dx = \int \tanh^{n-2} 3x(1-\text{sech}^2 3x) \, dx$ | M1 | Splits $\tanh^n 3x$ into $\tanh^{n-2} 3x \tanh^2 3x$ and applies $\tanh^2 3x = 1 - \text{sech}^2 3x$ |
| $\int \tanh^{n-2} 3x \, \text{sech}^2 3x \, dx = \frac{1}{3(n-1)} \tanh^{n-1} 3x$ | dM1 | Expands and integrates $\tanh^{n-2} 3x \, \text{sech}^2 3x$ to obtain $\alpha \tanh^{n-1} 3x$. Integration by parts also acceptable. Must be complete method leading to $\alpha \tanh^{n-1} 3x$ |
| $I_n = I_{n-2} - \frac{1}{3(n-1)}\left[\tanh^{n-1} 3x\right]_0^{\frac{1}{3}\ln 2} = I_{n-2} - \frac{1}{3(n-1)}\left(\frac{e^{2\ln 2}-1}{e^{2\ln 2}+1}\right)^{n-1}$ | ddM1 | Introduces $I_{n-2}$ and applies $x = \frac{1}{3}\ln 2$ using correct exponential definition of tanh, or accept calculator use e.g. $\tanh(\ln 2) = \frac{3}{5}$ |
| $I_n = I_{n-2} - \dfrac{\left(\frac{3}{5}\right)^{n-1}}{3(n-1)}$ | A1 | Fully correct proof. Allow recovery from slips e.g. tanh→tan→tanh. Clear unrecovered errors score A0. |

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# Question (c): Evaluating $I_5$

| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_5 = I_3 - \dfrac{\left(\frac{3}{5}\right)^{5-1}}{3(5-1)} = I_1 - \dfrac{\left(\frac{3}{5}\right)^{3-1}}{3(3-1)} - \dfrac{\left(\frac{3}{5}\right)^{5-1}}{3(5-1)}$ | M1 | Uses reduction formula to obtain $I_5$ in terms of $I_1$. Condone use of letter $p$ for $\frac{3}{5}$ |
| $\int \tanh 3x \, dx = \frac{1}{3}\ln(\cosh 3x)$ | M1 | Integrates to obtain $q\ln(\cosh rx)$ or e.g. $q\ln(\text{sech}\, rx)$. Condone $q$ and/or $r=1$ |
| $I_5 = \frac{1}{3}\ln\!\left(\dfrac{e^{\ln 2}+e^{-\ln 2}}{2}\right) - \dfrac{\left(\frac{9}{25}\right)}{6} - \dfrac{\left(\frac{81}{625}\right)}{12}$ | ddM1 | Applies $x = \frac{1}{3}\ln 2$ using correct exponential definition of cosh, or calculator e.g. $\cosh(\ln 2) = \frac{5}{4}$. Must not be in terms of $p$; must use value of $p$ from part (b) |
| $\dfrac{1}{3}\ln\dfrac{5}{4} - \dfrac{177}{2500}$ | A1 | Correct answer in correct form (allow $a=\ldots,\, b=\ldots,\, c=\ldots$). Allow $-0.0708$ for $c$ |

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Show LaTeX source

\begin{enumerate}
  \item (a) Use the definitions of hyperbolic functions in terms of exponentials to prove that
\end{enumerate}

$$\begin{gathered}
1 - \operatorname { sech } ^ { 2 } x \equiv \tanh ^ { 2 } x \\
I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 3 } \ln 2 } \tanh ^ { n } 3 x \mathrm {~d} x \quad n \in \mathbb { Z } \quad n \geqslant 0
\end{gathered}$$

(b) Show that

$$I _ { n } = I _ { n - 2 } - \frac { p ^ { n - 1 } } { 3 ( n - 1 ) } \quad n \geqslant 2$$

where $p$ is a rational number to be determined.\\
(c) Hence determine the exact value of

$$\int _ { 0 } ^ { \frac { 1 } { 3 } \ln 2 } \tanh ^ { 5 } 3 x \mathrm {~d} x$$

giving your answer in the form $a \ln b + c$ where $a , b$ and $c$ are rational numbers to be found.

\hfill \mbox{\textit{Edexcel F3 2024 Q5 [11]}}

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