1988 Paper 2 Q6

Year: 1988
Paper: 2
Question Number: 6

Course: UFM Pure
Section: Hyperbolic functions

Difficulty: 1600.0 Banger: 1500.0

Problem

Show that the following functions are positive when \(x\) is positive:
  1. [ \(x-\tanh x\)
  2. \(x\sinh x-2\cosh x+2\)
  3. \(2x\cosh2x-3\sinh2x+4x\).
The function \(\mathrm{f}\) is defined for \(x>0\) by \[ \mathrm{f}(x)=\frac{x(\cosh x)^{\frac{1}{3}}}{\sinh x}. \] Show that \(\mathrm{f}(x)\) has no turning points when \(x>0,\) and sketch \(\mathrm{f}(x)\) for \(x>0.\)

Solution

  1. Notice that \(f(x) = x - \tanh x\) has \(f'(x) = 1-\textrm{sech}^2 x = \tanh^2 x > 0\) so \(f(x)\) is strictly increasing on \((0, \infty)\) and \(f(0) = 0\) therefore \(f(x)\) is positive for all \(x\) positive
  2. Let \(f(x) = x\sinh x-2\cosh x+2\) then \(f'(x) = \sinh x +x \cosh x - 2 \sinh x = x \cosh x -\sinh x = \cosh x ( x - \tanh x) > 0\) by the first part. \(f(0) = 0\) so \(f(x)\) is positive for all \(x\) positive.
  3. Let \(f(x) = 2x\cosh2x-3\sinh2x+4x\) then \begin{align*} f'(x) &= 2\cosh 2x +4x\sinh 2x - 6 \cosh 2x + 4 \\ &= 4( x\sinh 2x-\cosh 2x +1) \\ &= 4(x2\cosh x \sinh x -2\cosh^2x ) \\ &= 8 \cosh^2 x (x - \tanh x) \end{align*} Which is always positive when \(x\) > 0, \(f(0) = 0\) so \(f(x) > 0\) for all positive \(x\).
Let \(f(x) = \frac{x(\cosh x)^{\frac{1}{3}}}{\sinh x}\) then \begin{align*} f'(x) &= \frac{(\cosh x)^{\frac13}\sinh x+\frac13 x \cosh^{-\frac23} x \sinh^2 x - x(\cosh x)^{\frac13} \cosh x}{\sinh^2 x} \\ &= \frac{\cosh x \sinh x + \frac13 x \sinh^2 x - x \cosh^2 x}{\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{3\cosh x \sinh x + x( \sinh^2 x - 3 \cosh^2 x)}{3\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{\frac32 \sinh 2x + x( -2\cosh 2x - 2)}{3\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{3 \sinh 2x -4x\cosh 2x - 4x}{6\cosh x^{\frac23} x \sinh^2 x} \\ \end{align*} which from the earlier part is always negative.
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Difficulty Rating: 1600.0

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Problem source
Show that the following functions are positive when $x$ is positive: 
\begin{questionparts}
\item[ $x-\tanh x$
\item $x\sinh x-2\cosh x+2$
\item $2x\cosh2x-3\sinh2x+4x$. 
\end{questionparts}
The function $\mathrm{f}$ is defined for $x>0$ by 
\[
\mathrm{f}(x)=\frac{x(\cosh x)^{\frac{1}{3}}}{\sinh x}.
\]
Show that $\mathrm{f}(x)$ has no turning points when $x>0,$ and
sketch $\mathrm{f}(x)$ for $x>0.$
Solution source
\begin{questionparts}
\item Notice that $f(x) = x - \tanh x$ has $f'(x) = 1-\textrm{sech}^2 x =  \tanh^2 x > 0$ so $f(x)$ is strictly increasing on $(0, \infty)$ and $f(0) = 0$ therefore $f(x)$ is positive for all $x$ positive
\item Let $f(x) = x\sinh x-2\cosh x+2$ then $f'(x) = \sinh x +x \cosh x - 2 \sinh x  = x \cosh x -\sinh x = \cosh x ( x - \tanh x) > 0$ by the first part. $f(0) = 0$ so $f(x)$ is positive for all $x$ positive.
\item Let $f(x) = 2x\cosh2x-3\sinh2x+4x$ then 
\begin{align*}
f'(x) &= 2\cosh 2x +4x\sinh 2x - 6 \cosh 2x + 4 \\
&= 4( x\sinh 2x-\cosh 2x +1) \\
&= 4(x2\cosh x \sinh x -2\cosh^2x ) \\
&= 8 \cosh^2 x (x - \tanh x) 
\end{align*}

Which is always positive when $x$ > 0, $f(0) = 0$ so $f(x) > 0$ for all positive $x$.
\end{questionparts}

Let $f(x) = \frac{x(\cosh x)^{\frac{1}{3}}}{\sinh x}$ then

\begin{align*}
f'(x) &= \frac{(\cosh x)^{\frac13}\sinh x+\frac13 x \cosh^{-\frac23} x \sinh^2 x - x(\cosh x)^{\frac13} \cosh x}{\sinh^2 x} \\
&= \frac{\cosh x \sinh x + \frac13 x \sinh^2 x - x \cosh^2 x}{\cosh x^{\frac23} x \sinh^2 x} \\
&= \frac{3\cosh x \sinh x + x( \sinh^2 x - 3 \cosh^2 x)}{3\cosh x^{\frac23} x \sinh^2 x} \\
&= \frac{\frac32 \sinh 2x + x( -2\cosh 2x - 2)}{3\cosh x^{\frac23} x \sinh^2 x} \\
&= \frac{3 \sinh 2x -4x\cosh 2x - 4x}{6\cosh x^{\frac23} x \sinh^2 x} \\
\end{align*}

which from the earlier part is always negative.

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