Pre-U Pre-U 9795/1 2011 June — Question 4 8 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2011
SessionJune
Marks8
TopicHyperbolic functions
TypeSketch graphs of hyperbolic functions
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring sketching hyperbolic functions (less familiar than trig), analyzing intersections, verifying bounds, and applying Newton-Raphson with derivatives of hyperbolic functions. While the individual techniques are standard, the combination of hyperbolic function manipulation and numerical methods places it moderately above average difficulty.
Spec1.09d Newton-Raphson method4.07b Hyperbolic graphs: sketch and properties4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1

  1. On a single diagram, sketch the graphs of \(y = \tanh x\) and \(y = \cosh x - 1\), and use your diagram to explain why the equation \(\text{f}(x) = 0\) has exactly two roots, where $$\text{f}(x) = 1 + \tanh x - \cosh x.$$ [3]
  2. The non-zero root of \(\text{f}(x) = 0\) is \(\alpha\).
    1. Verify that \(1 < \alpha < 1.5\). [1]
    2. Taking \(x_1 = 1.25\) as an initial approximation to \(\alpha\), use the Newton-Raphson iterative method to find \(x_3\), giving your answer to 5 decimal places. [4]

AnswerMarks
(i) [Graph showing \(y = \tanh x\) and \(y = \cosh x - 1\) with curves crossing twice]B1
Grad. of \(\tanh x\) should be 1B1
Curves cross twice (at \(x = 0\) and \(x = \alpha\)) so there are 2 roots to \(\tanh x = \cosh x - 1\) i.e. \(1 + \tanh x - \cosh x = 0\)B1
[3]
(ii) (a) \(f(1)f(1.5) = 0.22... \times (-0.45...) < 0\) \(\Rightarrow 1 < \alpha < 1.5\) by the "Change-of-Sign" RuleB1
[1]
(b) \(f'(x) = \text{sech}^2 x - \sinh x\)B1, B1
Use of \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) at least onceM1
\(x_1 = 1.25\), \(x_2 = 1.219\ 625\ 3\), \(x_3 = 1.218\ 76\) to 5 d.p.A1
[4]
**(i)** [Graph showing $y = \tanh x$ and $y = \cosh x - 1$ with curves crossing twice] | B1 |
Grad. of $\tanh x$ should be 1 | B1 |
Curves cross twice (at $x = 0$ and $x = \alpha$) so there are 2 roots to $\tanh x = \cosh x - 1$ i.e. $1 + \tanh x - \cosh x = 0$ | B1 |
| [3] |

**(ii)** **(a)** $f(1)f(1.5) = 0.22... \times (-0.45...) < 0$ $\Rightarrow 1 < \alpha < 1.5$ by the "Change-of-Sign" Rule | B1 |
| [1] |

**(b)** $f'(x) = \text{sech}^2 x - \sinh x$ | B1, B1 |
Use of $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ at least once | M1 |
$x_1 = 1.25$, $x_2 = 1.219\ 625\ 3$, $x_3 = 1.218\ 76$ to 5 d.p. | A1 |
| [4] |

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\begin{enumerate}[label=(\roman*)]
\item On a single diagram, sketch the graphs of $y = \tanh x$ and $y = \cosh x - 1$, and use your diagram to explain why the equation $\text{f}(x) = 0$ has exactly two roots, where
$$\text{f}(x) = 1 + \tanh x - \cosh x.$$ [3]
\item The non-zero root of $\text{f}(x) = 0$ is $\alpha$.
\begin{enumerate}[label=(\alph*)]
\item Verify that $1 < \alpha < 1.5$. [1]
\item Taking $x_1 = 1.25$ as an initial approximation to $\alpha$, use the Newton-Raphson iterative method to find $x_3$, giving your answer to 5 decimal places. [4]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2011 Q4 [8]}}