Pre-U Pre-U 9795/1 2013 June — Question 13 4 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2013
SessionJune
Marks4
TopicHyperbolic functions
TypeReduction formulas with hyperbolic integrals
DifficultyChallenging +1.8 This is a substantial Further Maths question requiring multiple techniques: proving hyperbolic identities from definitions, deriving a reduction formula, evaluating definite integrals with hyperbolic functions, applying method of differences, and deducing an infinite series sum. While each individual step is methodical, the multi-part structure, combination of proof and calculation, and the final series summation requiring careful algebraic manipulation place this well above average difficulty but within reach of strong Further Maths students.
Spec4.06b Method of differences: telescoping series4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions
13
  1. Use the definitions \(\tanh \theta = \frac { \mathrm { e } ^ { \theta } - \mathrm { e } ^ { - \theta } } { \mathrm { e } ^ { \theta } + \mathrm { e } ^ { - \theta } }\) and \(\operatorname { sech } \theta = \frac { 2 } { \mathrm { e } ^ { \theta } + \mathrm { e } ^ { - \theta } }\) to prove the results
    1. \(\tanh ^ { 2 } \theta \equiv 1 - \operatorname { sech } ^ { 2 } \theta\),
    2. \(\frac { \mathrm { d } } { \mathrm { d } \theta } ( \tanh \theta ) = \operatorname { sech } ^ { 2 } \theta\).
    3. Let \(I _ { n } = \int _ { 0 } ^ { \alpha } \tanh ^ { 2 n } \theta \mathrm {~d} \theta\) for \(n \geqslant 0\), where \(\alpha > 0\).
      (a) Show that \(I _ { n - 1 } - I _ { n } = \frac { \tanh ^ { 2 n - 1 } \alpha } { 2 n - 1 }\) for \(n \geqslant 1\). Given that \(\alpha = \frac { 1 } { 2 } \ln 3\),
      (b) evaluate \(I _ { 0 }\),
    4. use the method of differences to show that \(I _ { n } = \frac { 1 } { 2 } \ln 3 - \sum _ { r = 1 } ^ { n } \frac { \left( \frac { 1 } { 2 } \right) ^ { 2 r - 1 } } { 2 r - 1 }\) and deduce the sum of the infinite series \(\sum _ { r = 0 } ^ { \infty } \frac { 1 } { ( 2 r + 1 ) 4 ^ { r } }\).
(i)(a) \(1 - \mathrm{sech}^2\theta \equiv \dfrac{(e^\theta+e^{-\theta})^2 - 4}{(e^\theta+e^{-\theta})^2} \equiv \dfrac{(e^\theta-e^{-\theta})^2}{(e^\theta+e^{-\theta})^2} \equiv \tanh^2\theta\) shown legitimately M1 A1
(b) \(\dfrac{\mathrm{d}}{\mathrm{d}\theta}(\tanh\theta) \equiv \dfrac{(e^\theta+e^{-\theta})(e^\theta+e^{-\theta}) - (e^\theta-e^{-\theta})(e^\theta-e^{-\theta})}{(e^\theta+e^{-\theta})^2} \equiv \mathrm{sech}^2\theta\) from (a) M1 A1
[4]
(ii)(a) \(I_n = \displaystyle\int_0^\alpha \tanh^{2n-2}\theta\cdot\tanh^2\theta\,\mathrm{d}\theta = \int_0^\alpha \tanh^{2n-2}\theta(1-\mathrm{sech}^2\theta)\,\mathrm{d}\theta\) M1 M1
\(= I_{n-1} - \left[\dfrac{\tanh^{2n-1}\theta}{2n-1}\right]_0^\alpha \Rightarrow I_{n-1} - I_n = \dfrac{(\tanh\alpha)^{2n-1}}{2n-1}\) M1 A1
ALT: \(I_{n-1} - I_n = \displaystyle\int_0^\alpha \tanh^{2n-2}\theta(1-\tanh^2\theta)\,\mathrm{d}\theta = \int_0^\alpha \tanh^{2n-2}\theta\cdot\mathrm{sech}^2\theta\,\mathrm{d}\theta\) M1 M1
\(= \left[\dfrac{\tanh^{2n-1}\theta}{2n-1}\right]_0^\alpha = \dfrac{(\tanh\alpha)^{2n-1}}{2n-1}\) M1 A1
[4]
(b) \(I_0 = \displaystyle\int_0^\alpha 1\,\mathrm{d}\theta = \alpha = \frac{1}{2}\ln 3\) B1
[1]
(c) \((I_{n-1}-I_n)+(I_{n-2}-I_{n-1})+(I_{n-3}-I_{n-2})+\ldots+(I_2-I_3)+(I_1-I_2)+(I_0-I_1)\)
Use of the method of differences M1
\(= \displaystyle\sum_{r=1}^{n}\dfrac{(\tanh\alpha)^{2r-1}}{2r-1} = \sum_{r=1}^{n}\dfrac{(\frac{1}{2})^{2r-1}}{2r-1}\) when \(\alpha = \frac{1}{2}\ln 3\) A1
\(\Rightarrow I_0 - I_n = \displaystyle\sum_{r=1}^{n}\dfrac{(\frac{1}{2})^{2r-1}}{2r-1}\) Cancellation of terms in the summation M1
\(\Rightarrow I_n = \frac{1}{2}\ln 3 - \displaystyle\sum_{r=1}^{n}\dfrac{(\frac{1}{2})^{2r-1}}{2r-1}\) AG A1
AnswerMarks Guidance
As \(n\to\infty\), \(I_n \to 0\) since \(\tanh < 1\) E1
\(\Rightarrow \frac{1}{2}\ln 3 = \displaystyle\sum_{r=1}^{n}\dfrac{(\frac{1}{2})^{2r-1}}{2r-1} = \dfrac{\frac{1}{2}}{1} + \dfrac{(\frac{1}{2})^3}{3} + \dfrac{(\frac{1}{2})^5}{5} + \dfrac{(\frac{1}{2})^7}{7} + \ldots\) M1
\(\Rightarrow \ln 3 = 1 + \dfrac{1}{3\cdot4} + \dfrac{1}{5\cdot4^2} + \dfrac{1}{7\cdot4^3} + \ldots = \displaystyle\sum_{r=0}^{\infty}\dfrac{1}{(2r+1)4^r}\) A1
Ignoring "method of differences", but opting for a direct iterative approach scores max 3/4 … M0 M1 A1 A1
[7]
**(i)(a)** $1 - \mathrm{sech}^2\theta \equiv \dfrac{(e^\theta+e^{-\theta})^2 - 4}{(e^\theta+e^{-\theta})^2} \equiv \dfrac{(e^\theta-e^{-\theta})^2}{(e^\theta+e^{-\theta})^2} \equiv \tanh^2\theta$ shown legitimately **M1 A1**

**(b)** $\dfrac{\mathrm{d}}{\mathrm{d}\theta}(\tanh\theta) \equiv \dfrac{(e^\theta+e^{-\theta})(e^\theta+e^{-\theta}) - (e^\theta-e^{-\theta})(e^\theta-e^{-\theta})}{(e^\theta+e^{-\theta})^2} \equiv \mathrm{sech}^2\theta$ from (a) **M1 A1**

**[4]**

**(ii)(a)** $I_n = \displaystyle\int_0^\alpha \tanh^{2n-2}\theta\cdot\tanh^2\theta\,\mathrm{d}\theta = \int_0^\alpha \tanh^{2n-2}\theta(1-\mathrm{sech}^2\theta)\,\mathrm{d}\theta$ **M1 M1**

$= I_{n-1} - \left[\dfrac{\tanh^{2n-1}\theta}{2n-1}\right]_0^\alpha \Rightarrow I_{n-1} - I_n = \dfrac{(\tanh\alpha)^{2n-1}}{2n-1}$ **M1 A1**

**ALT:** $I_{n-1} - I_n = \displaystyle\int_0^\alpha \tanh^{2n-2}\theta(1-\tanh^2\theta)\,\mathrm{d}\theta = \int_0^\alpha \tanh^{2n-2}\theta\cdot\mathrm{sech}^2\theta\,\mathrm{d}\theta$ **M1 M1**

$= \left[\dfrac{\tanh^{2n-1}\theta}{2n-1}\right]_0^\alpha = \dfrac{(\tanh\alpha)^{2n-1}}{2n-1}$ **M1 A1**

**[4]**

**(b)** $I_0 = \displaystyle\int_0^\alpha 1\,\mathrm{d}\theta = \alpha = \frac{1}{2}\ln 3$ **B1**

**[1]**

**(c)** $(I_{n-1}-I_n)+(I_{n-2}-I_{n-1})+(I_{n-3}-I_{n-2})+\ldots+(I_2-I_3)+(I_1-I_2)+(I_0-I_1)$

Use of the method of differences **M1**

$= \displaystyle\sum_{r=1}^{n}\dfrac{(\tanh\alpha)^{2r-1}}{2r-1} = \sum_{r=1}^{n}\dfrac{(\frac{1}{2})^{2r-1}}{2r-1}$ when $\alpha = \frac{1}{2}\ln 3$ **A1**

$\Rightarrow I_0 - I_n = \displaystyle\sum_{r=1}^{n}\dfrac{(\frac{1}{2})^{2r-1}}{2r-1}$ Cancellation of terms in the summation **M1**

$\Rightarrow I_n = \frac{1}{2}\ln 3 - \displaystyle\sum_{r=1}^{n}\dfrac{(\frac{1}{2})^{2r-1}}{2r-1}$ **AG** **A1**

As $n\to\infty$, $I_n \to 0$ since $|\tanh| < 1$ **E1**

$\Rightarrow \frac{1}{2}\ln 3 = \displaystyle\sum_{r=1}^{n}\dfrac{(\frac{1}{2})^{2r-1}}{2r-1} = \dfrac{\frac{1}{2}}{1} + \dfrac{(\frac{1}{2})^3}{3} + \dfrac{(\frac{1}{2})^5}{5} + \dfrac{(\frac{1}{2})^7}{7} + \ldots$ **M1**

$\Rightarrow \ln 3 = 1 + \dfrac{1}{3\cdot4} + \dfrac{1}{5\cdot4^2} + \dfrac{1}{7\cdot4^3} + \ldots = \displaystyle\sum_{r=0}^{\infty}\dfrac{1}{(2r+1)4^r}$ **A1**

Ignoring "method of differences", but opting for a direct iterative approach scores max 3/4 … **M0 M1 A1 A1**

**[7]**
13 (i) Use the definitions $\tanh \theta = \frac { \mathrm { e } ^ { \theta } - \mathrm { e } ^ { - \theta } } { \mathrm { e } ^ { \theta } + \mathrm { e } ^ { - \theta } }$ and $\operatorname { sech } \theta = \frac { 2 } { \mathrm { e } ^ { \theta } + \mathrm { e } ^ { - \theta } }$ to prove the results
\begin{enumerate}[label=(\alph*)]
\item $\tanh ^ { 2 } \theta \equiv 1 - \operatorname { sech } ^ { 2 } \theta$,
\item $\frac { \mathrm { d } } { \mathrm { d } \theta } ( \tanh \theta ) = \operatorname { sech } ^ { 2 } \theta$.\\
(ii) Let $I _ { n } = \int _ { 0 } ^ { \alpha } \tanh ^ { 2 n } \theta \mathrm {~d} \theta$ for $n \geqslant 0$, where $\alpha > 0$.\\
(a) Show that $I _ { n - 1 } - I _ { n } = \frac { \tanh ^ { 2 n - 1 } \alpha } { 2 n - 1 }$ for $n \geqslant 1$.

Given that $\alpha = \frac { 1 } { 2 } \ln 3$,\\
(b) evaluate $I _ { 0 }$,
\item use the method of differences to show that $I _ { n } = \frac { 1 } { 2 } \ln 3 - \sum _ { r = 1 } ^ { n } \frac { \left( \frac { 1 } { 2 } \right) ^ { 2 r - 1 } } { 2 r - 1 }$ and deduce the sum of the infinite series $\sum _ { r = 0 } ^ { \infty } \frac { 1 } { ( 2 r + 1 ) 4 ^ { r } }$.
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2013 Q13 [4]}}
This paper (13 questions)
View full paper
Q1 4 Q2 4 Q3 2 Q4 2 Q5 5 Q6 8 Q7 7 Q8 8 Q9 8 Q10 18 Q11 13 Q12 11 Q13 4