Pre-U Pre-U 9795/1 2010 June — Question 4 5 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2010
SessionJune
Marks5
TopicHyperbolic functions
TypeSolve mixed sinh/cosh linear combinations
DifficultyStandard +0.3 Part (i) is a standard bookwork proof using exponential definitions that most students memorize. Part (ii) requires applying the identity to rewrite the equation, then solving a quadratic in e^x, which is a routine technique for hyperbolic equations. The final form requires some algebraic manipulation but follows a well-practiced method. This is slightly easier than average due to being a standard textbook exercise with clear steps.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1

4
  1. Using the definitions of sinh and cosh in terms of exponentials, prove that $$\cosh A \cosh B + \sinh A \sinh B \equiv \cosh ( A + B )$$
  2. Solve the equation \(5 \cosh x + 3 \sinh x = 12\), giving your answers in the form \(\ln ( p \pm q \sqrt { 2 } )\) for rational numbers \(p\) and \(q\) to be determined.

(i)
\(\cosh A\cosh B + \sinh A\sinh B \equiv \frac{1}{2}(e^A+e^{-A})\frac{1}{2}(e^B+e^{-B}) + \frac{1}{2}(e^A-e^{-A})\frac{1}{2}(e^B-e^{-B})\)
\(\equiv \frac{1}{4}(e^{A+B}+e^{B-A}+e^{A-B}+e^{-A-B}) + \frac{1}{4}(e^{A+B}-e^{B-A}-e^{A-B}+e^{-A-B})\)
\(\equiv \frac{1}{2}(e^{A+B}+e^{-(A+B)})\)
\(\equiv \cosh(A+B)\) B1 Shown legitimately [1]
(ii) METHOD I \(5\cdot\frac{1}{2}(e^x+e^{-x})+3\cdot\frac{1}{2}(e^x-e^{-x}) = 4e^x+e^{-x}\) M1 Use of exponential forms
\(4(e^x)^2 - 12(e^x) + 1 = 0\) M1 Writing as quadratic in \(e^x\); A1 correct
\(e^x = \frac{12\pm\sqrt{144-16}}{8}\) M1 \(e^x = \frac{3}{2}\pm\sqrt{2}\)
A1 for \(\ln(1.5\pm\sqrt{2})\)
METHOD II \(4\cosh\left(x+\tanh^{-1}\frac{3}{5}\right)=12\) M1A1
M1 \(x = (\pm)\cosh^{-1}3 - \tanh^{-1}\frac{3}{5}\)
M1 Use of \(\cosh^{-1}x = \ln\left(x\pm\sqrt{x^2-1}\right)\), \(\tanh^{-1}x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\) ignore \(\pm\)
A1 \(\ln(1.5\pm\sqrt{2})\) [5]
**(i)**
$\cosh A\cosh B + \sinh A\sinh B \equiv \frac{1}{2}(e^A+e^{-A})\frac{1}{2}(e^B+e^{-B}) + \frac{1}{2}(e^A-e^{-A})\frac{1}{2}(e^B-e^{-B})$

$\equiv \frac{1}{4}(e^{A+B}+e^{B-A}+e^{A-B}+e^{-A-B}) + \frac{1}{4}(e^{A+B}-e^{B-A}-e^{A-B}+e^{-A-B})$

$\equiv \frac{1}{2}(e^{A+B}+e^{-(A+B)})$

$\equiv \cosh(A+B)$ **B1** Shown legitimately [1]

**(ii)** METHOD I $5\cdot\frac{1}{2}(e^x+e^{-x})+3\cdot\frac{1}{2}(e^x-e^{-x}) = 4e^x+e^{-x}$ **M1** Use of exponential forms

$4(e^x)^2 - 12(e^x) + 1 = 0$ **M1** Writing as quadratic in $e^x$; **A1** correct

$e^x = \frac{12\pm\sqrt{144-16}}{8}$ **M1** $e^x = \frac{3}{2}\pm\sqrt{2}$

**A1** for $\ln(1.5\pm\sqrt{2})$

METHOD II $4\cosh\left(x+\tanh^{-1}\frac{3}{5}\right)=12$ **M1A1**

**M1** $x = (\pm)\cosh^{-1}3 - \tanh^{-1}\frac{3}{5}$

**M1** Use of $\cosh^{-1}x = \ln\left(x\pm\sqrt{x^2-1}\right)$, $\tanh^{-1}x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ ignore $\pm$

**A1** $\ln(1.5\pm\sqrt{2})$ **[5]**
4 (i) Using the definitions of sinh and cosh in terms of exponentials, prove that

$$\cosh A \cosh B + \sinh A \sinh B \equiv \cosh ( A + B )$$

(ii) Solve the equation $5 \cosh x + 3 \sinh x = 12$, giving your answers in the form $\ln ( p \pm q \sqrt { 2 } )$ for rational numbers $p$ and $q$ to be determined.

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2010 Q4 [5]}}