Pre-U Pre-U 9795/1 2017 June — Question 5 8 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2017
SessionJune
Marks8
TopicHyperbolic functions
TypeProve inverse hyperbolic logarithmic form
DifficultyStandard +0.3 Part (i) is a standard derivation of the inverse hyperbolic tangent logarithmic form requiring algebraic manipulation of exponentials—a common textbook exercise. Part (ii) requires substituting definitions, solving a quadratic in e^(2x), and expressing the answer in logarithmic form, which is routine for Further Maths students familiar with hyperbolic functions. This is slightly above average difficulty due to the two-part structure and algebraic manipulation required, but remains a standard Further Maths question without requiring novel insight.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07e Inverse hyperbolic: definitions, domains, ranges
5
  1. Use the definition \(\tanh y = \frac { \mathrm { e } ^ { 2 y } - 1 } { \mathrm { e } ^ { 2 y } + 1 }\) to show that \(\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\) for \(| x | < 1\).
  2. Solve the equation \(\tanh x + \operatorname { coth } x = 4\), giving your answer in the form \(p \ln m\), where \(p\) is a positive rational number and \(m\) is a positive integer.
Question 5(i)
\(y = \tanh^{-1}x \Leftrightarrow \tanh y = x = \dfrac{e^{2y}-1}{e^{2y}+1}\) M1
\(xe^{2y} + x = e^{2y} - 1 \Leftrightarrow 1+x = e^{2y}(1-x)\) M1 Identifying \(e^{2y}\)
\(y = \tanh^{-1}x = \dfrac{1}{2}\ln\!\left(\dfrac{1+x}{1-x}\right)\) A1 Legitimately obtained by taking logs. Allow verification by substitution of given result.
Question 5(ii) Method I
\(t + \dfrac{1}{t} = 4 \Rightarrow t^2 - 4t + 1 = 0\) M1 Creating a quadratic in \(\tanh x\)
\(\Rightarrow t = 2 \pm \sqrt{3}\) M1 Solving
AnswerMarks Guidance
Using \(\dfrac{1}{2}\ln\!\left(\dfrac{1+t}{1-t}\right)\) with \(t = 2-\sqrt{3}\) and/or \(2+\sqrt{3}\) M1 (NB since \(\tanh x < 1\), it must be \(t = 2-\sqrt{3}\))
\(x = \dfrac{1}{2}\ln\!\left(\dfrac{3-\sqrt{3}}{-1+\sqrt{3}}\times\dfrac{1+\sqrt{3}}{1+\sqrt{3}}\right) = \dfrac{1}{2}\ln(\sqrt{3})\) M1 By rationalising denominator or direct observation
\(= \dfrac{1}{4}\ln(3)\) A1 Must be in this form
Method II
\(\dfrac{\text{sh}}{\text{ch}} + \dfrac{\text{ch}}{\text{sh}} = 4\) M1
\(\Rightarrow \text{ch}^2 + \text{sh}^2 = 4\,\text{sh.ch} \Rightarrow \cosh(2x) = 2\sinh(2x)\) M1 Conversion to double-"angles"
\(\Rightarrow \tanh(2x) = \dfrac{1}{2}\) A1
\(\Rightarrow 2x = \dfrac{1}{2}\ln\!\left(\dfrac{3/2}{1/2}\right)\) M1 Use of \(\tanh^{-1}x\) formula from (i)
\(\Rightarrow x = \dfrac{1}{4}\ln(3)\) A1 Must be in this form
Method III
\(\dfrac{e^{2x}-1}{e^{2x}+1} + \dfrac{e^{2x}+1}{e^{2x}-1} = 4\) M1
\(\Rightarrow (e^{2x}-1)^2 + (e^{2x}+1)^2 = 4(e^{2x}-1)(e^{2x}+1)\) M1
\(\Rightarrow e^{4x} - 2e^{2x}+1 + e^{4x}+2e^{2x}+1 = 4(e^{4x}-1)\) A2 (A1 LHS, A1 RHS)
\(\Rightarrow 6 = 2e^{4x} \Rightarrow x = \dfrac{1}{4}\ln(3)\) A1 Must be in this form
Total: 8 marks
**Question 5(i)**

$y = \tanh^{-1}x \Leftrightarrow \tanh y = x = \dfrac{e^{2y}-1}{e^{2y}+1}$ **M1**

$xe^{2y} + x = e^{2y} - 1 \Leftrightarrow 1+x = e^{2y}(1-x)$ **M1** Identifying $e^{2y}$

$y = \tanh^{-1}x = \dfrac{1}{2}\ln\!\left(\dfrac{1+x}{1-x}\right)$ **A1** Legitimately obtained by taking logs. Allow verification by substitution of given result.

**Question 5(ii) Method I**

$t + \dfrac{1}{t} = 4 \Rightarrow t^2 - 4t + 1 = 0$ **M1** Creating a quadratic in $\tanh x$

$\Rightarrow t = 2 \pm \sqrt{3}$ **M1** Solving

Using $\dfrac{1}{2}\ln\!\left(\dfrac{1+t}{1-t}\right)$ with $t = 2-\sqrt{3}$ and/or $2+\sqrt{3}$ **M1** (NB since $|\tanh x| < 1$, it must be $t = 2-\sqrt{3}$)

$x = \dfrac{1}{2}\ln\!\left(\dfrac{3-\sqrt{3}}{-1+\sqrt{3}}\times\dfrac{1+\sqrt{3}}{1+\sqrt{3}}\right) = \dfrac{1}{2}\ln(\sqrt{3})$ **M1** By rationalising denominator or direct observation

$= \dfrac{1}{4}\ln(3)$ **A1** Must be in this form

**Method II**

$\dfrac{\text{sh}}{\text{ch}} + \dfrac{\text{ch}}{\text{sh}} = 4$ **M1**

$\Rightarrow \text{ch}^2 + \text{sh}^2 = 4\,\text{sh.ch} \Rightarrow \cosh(2x) = 2\sinh(2x)$ **M1** Conversion to double-"angles"

$\Rightarrow \tanh(2x) = \dfrac{1}{2}$ **A1**

$\Rightarrow 2x = \dfrac{1}{2}\ln\!\left(\dfrac{3/2}{1/2}\right)$ **M1** Use of $\tanh^{-1}x$ formula from **(i)**

$\Rightarrow x = \dfrac{1}{4}\ln(3)$ **A1** Must be in this form

**Method III**

$\dfrac{e^{2x}-1}{e^{2x}+1} + \dfrac{e^{2x}+1}{e^{2x}-1} = 4$ **M1**

$\Rightarrow (e^{2x}-1)^2 + (e^{2x}+1)^2 = 4(e^{2x}-1)(e^{2x}+1)$ **M1**

$\Rightarrow e^{4x} - 2e^{2x}+1 + e^{4x}+2e^{2x}+1 = 4(e^{4x}-1)$ **A2** (A1 LHS, A1 RHS)

$\Rightarrow 6 = 2e^{4x} \Rightarrow x = \dfrac{1}{4}\ln(3)$ **A1** Must be in this form

**Total: 8 marks**
5\\
(i) Use the definition $\tanh y = \frac { \mathrm { e } ^ { 2 y } - 1 } { \mathrm { e } ^ { 2 y } + 1 }$ to show that $\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)$ for $| x | < 1$.\\
(ii) Solve the equation $\tanh x + \operatorname { coth } x = 4$, giving your answer in the form $p \ln m$, where $p$ is a positive rational number and $m$ is a positive integer.

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2017 Q5 [8]}}
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