4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials

241 questions

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OCR MEI FP2 2012 January Q4
18 marks Standard +0.8
4
  1. Define tanh \(t\) in terms of exponential functions. Sketch the graph of \(\tanh t\).
  2. Show that \(\operatorname { artanh } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\). State the set of values of \(x\) for which this equation is valid.
  3. Differentiate the equation \(\tanh y = x\) with respect to \(x\) and hence show that the derivative of \(\operatorname { artanh } x\) is \(\frac { 1 } { 1 - x ^ { 2 } }\). Show that this result may also be obtained by differentiating the equation in part (ii).
  4. By considering \(\operatorname { artanh } x\) as \(1 \times \operatorname { artanh } x\) and using integration by parts, show that $$\int _ { 0 } ^ { \frac { 1 } { 2 } } \operatorname { artanh } x \mathrm {~d} x = \frac { 1 } { 4 } \ln \frac { 27 } { 16 }$$
OCR MEI FP2 2013 January Q4
18 marks Challenging +1.8
4
  1. Show that the curve with equation $$y = 3 \sinh x - 2 \cosh x$$ has no turning points.
    Show that the curve crosses the \(x\)-axis at \(x = \frac { 1 } { 2 } \ln 5\). Show that this is also the point at which the gradient of the curve has a stationary value.
  2. Sketch the curve.
  3. Express \(( 3 \sinh x - 2 \cosh x ) ^ { 2 }\) in terms of \(\sinh 2 x\) and \(\cosh 2 x\). Hence or otherwise, show that the volume of the solid of revolution formed by rotating the region bounded by the curve and the axes through \(360 ^ { \circ }\) about the \(x\)-axis is $$\pi \left( 3 - \frac { 5 } { 4 } \ln 5 \right) .$$ Option 2: Investigation of curves \section*{This question requires the use of a graphical calculator.}
OCR MEI FP2 2015 June Q4
18 marks Standard +0.8
4
  1. Starting with the relationship \(\cosh ^ { 2 } t - \sinh ^ { 2 } t = 1\), deduce a relationship between \(\tanh ^ { 2 } t\) and \(\operatorname { sech } ^ { 2 } t\). You are given that \(y = \operatorname { artanh } x\).
  2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 - x ^ { 2 } }\).
  3. Show, by integrating the result in part (ii), that \(y = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\).
  4. Show that \(\int _ { 0 } ^ { \frac { \sqrt { 3 } } { 6 } } \frac { 1 } { 1 - 3 x ^ { 2 } } \mathrm {~d} x = \frac { 1 } { \sqrt { 3 } } \operatorname { artanh } \frac { 1 } { 2 }\). Express this answer in logarithmic form.
  5. Use integration by parts to find \(\int \operatorname { artanh } x \mathrm {~d} x\), giving your answer in terms of logarithms. \section*{END OF QUESTION PAPER}
OCR MEI FP2 2012 June Q4
18 marks Challenging +1.2
4
  1. Prove, from definitions involving exponential functions, that $$\cosh 2 u = 2 \sinh ^ { 2 } u + 1$$
  2. Prove that, if \(y \geqslant 0\) and \(\cosh y = u\), then \(y = \ln \left( u + \sqrt { } \left( u ^ { 2 } - 1 \right) \right)\).
  3. Using the substitution \(2 x = \cosh u\), show that $$\int \sqrt { 4 x ^ { 2 } - 1 } \mathrm {~d} x = a x \sqrt { 4 x ^ { 2 } - 1 } - b \operatorname { arcosh } 2 x + c$$ where \(a\) and \(b\) are constants to be determined and \(c\) is an arbitrary constant.
  4. Find \(\int _ { \frac { 1 } { 2 } } ^ { 1 } \sqrt { 4 x ^ { 2 } - 1 } \mathrm {~d} x\), expressing your answer in an exact form involving logarithms.
OCR MEI FP2 2013 June Q4
18 marks Challenging +1.2
4
  1. Prove, using exponential functions, that \(\cosh ^ { 2 } u - \sinh ^ { 2 } u = 1\).
  2. Given that \(y = \operatorname { arsinh } x\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 1 + x ^ { 2 } } }$$ and that $$y = \ln \left( x + \sqrt { 1 + x ^ { 2 } } \right)$$
  3. Show that $$\int _ { 0 } ^ { 2 } \frac { 1 } { \sqrt { 4 + 9 x ^ { 2 } } } \mathrm {~d} x = \frac { 1 } { 3 } \ln ( 3 + \sqrt { 10 } )$$
  4. Find, in exact logarithmic form, $$\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 + x ^ { 2 } } } \operatorname { arsinh } x \mathrm {~d} x$$
OCR MEI FP2 2011 June Q4
18 marks Challenging +1.2
4
  1. Given that \(\cosh y = x\), show that \(y = \pm \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)\) and that \(\operatorname { arcosh } x = \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)\).
  2. Find \(\int _ { \frac { 4 } { 5 } } ^ { 1 } \frac { 1 } { \sqrt { 25 x ^ { 2 } - 16 } } \mathrm {~d} x\), expressing your answer in an exact logarithmic form.
  3. Solve the equation $$5 \cosh x - \cosh 2 x = 3$$ giving your answers in an exact logarithmic form.
OCR FP2 2011 January Q4
9 marks Standard +0.8
4
  1. Use the definitions of hyperbolic functions in terms of exponentials to prove that $$8 \sinh ^ { 4 } x \equiv \cosh 4 x - 4 \cosh 2 x + 3$$
  2. Solve the equation $$\cosh 4 x - 3 \cosh 2 x + 1 = 0$$ giving your answer(s) in logarithmic form.
OCR FP2 2013 January Q3
6 marks Standard +0.3
3 By first expressing \(\cosh x\) and \(\sinh x\) in terms of exponentials, solve the equation $$3 \cosh x - 4 \sinh x = 7$$ giving your answer in an exact logarithmic form.
OCR FP2 2009 June Q7
10 marks Standard +0.3
7
  1. Sketch the graph of \(y = \operatorname { coth } x\), and give the equations of any asymptotes.
  2. It is given that \(\mathrm { f } ( x ) = x \tanh x - 2\). Use the Newton-Raphson method, with a first approximation \(x _ { 1 } = 2\), to find the next three approximations \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\) to a root of \(\mathrm { f } ( x ) = 0\). Give the answers correct to 4 decimal places.
  3. If \(\mathrm { f } ( x ) = 0\), show that \(\operatorname { coth } x = \frac { 1 } { 2 } x\). Hence write down the roots of \(\mathrm { f } ( x ) = 0\), correct to 4 decimal places.
OCR FP2 2009 June Q8
14 marks Standard +0.3
8
  1. Using the definitions of \(\sinh x\) and \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that
    1. \(\cosh ( \ln a ) \equiv \frac { a ^ { 2 } + 1 } { 2 a }\), where \(a > 0\),
    2. \(\cosh x \cosh y - \sinh x \sinh y \equiv \cosh ( x - y )\).
    3. Use part (i)(b) to show that \(\cosh ^ { 2 } x - \sinh ^ { 2 } x \equiv 1\).
    4. Given that \(R > 0\) and \(a > 1\), find \(R\) and \(a\) such that $$13 \cosh x - 5 \sinh x \equiv R \cosh ( x - \ln a )$$
    5. Hence write down the coordinates of the minimum point on the curve with equation \(y = 13 \cosh x - 5 \sinh x\).
OCR FP2 2010 June Q6
7 marks Standard +0.8
6
  1. Show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sinh ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 2 } + 1 } }\).
  2. Given that \(y = \cosh \left( a \sinh ^ { - 1 } x \right)\), where \(a\) is a constant, show that $$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + x \frac { \mathrm {~d} y } { \mathrm {~d} x } - a ^ { 2 } y = 0$$
OCR FP2 2010 June Q8
10 marks Challenging +1.2
8
  1. Using the definition of \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that $$4 \cosh ^ { 3 } x - 3 \cosh x \equiv \cosh 3 x$$
  2. Use the substitution \(u = \cosh x\) to find, in terms of \(5 ^ { \frac { 1 } { 3 } }\), the real root of the equation $$20 u ^ { 3 } - 15 u - 13 = 0 .$$
OCR FP2 2012 June Q1
5 marks Standard +0.3
1 Express sech \(2 x\) in terms of exponentials and hence, by using the substitution \(u = e ^ { 2 x }\), find \(\int \operatorname { sech } 2 x \mathrm {~d} x\).
OCR FP2 2012 June Q3
8 marks Standard +0.3
3
  1. By quoting results given in the List of Formulae (MF1), prove that \(\tanh 2 x \equiv \frac { 2 \tanh x } { 1 + \tanh ^ { 2 } x }\).
  2. Solve the equation \(5 \tanh 2 x = 1 + 6 \tanh x\), giving your answers in logarithmic form.
OCR FP2 2013 June Q2
8 marks Standard +0.3
2
  1. Using the definitions for \(\cosh x\) and \(\sinh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that \(\cosh ^ { 2 } x - \sinh ^ { 2 } x \equiv 1\).
  2. Hence solve the equation \(\sinh ^ { 2 } x = 5 \cosh x - 7\), giving your answers in logarithmic form.
OCR FP2 2015 June Q1
3 marks Standard +0.8
1 By first expressing \(\tanh y\) in terms of exponentials, prove that \(\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\).
OCR FP2 2015 June Q8
9 marks Standard +0.3
8 It is given that \(\mathrm { f } ( x ) = 2 \sinh x + 3 \cosh x\).
  1. Show that the curve \(y = \mathrm { f } ( x )\) has a stationary point at \(x = - \frac { 1 } { 2 } \ln 5\) and find the value of \(y\) at this point.
  2. Solve the equation \(\mathrm { f } ( x ) = 5\), giving your answers exactly. \section*{Question 9 begins on page 4.}
CAIE FP1 2017 June Q12 EITHER
Challenging +1.2
The curve \(C\) has equation \(y = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right)\) for \(0 \leqslant x \leqslant 4\).
  1. The region \(R\) is bounded by \(C\), the \(x\)-axis, the \(y\)-axis and the line \(x = 4\). Find, in terms of e, the coordinates of the centroid of the region \(R\).
  2. Show that \(\frac { \mathrm { d } s } { \mathrm {~d} x } = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right)\), where \(s\) denotes the arc length of \(C\), and find the surface area generated when \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
CAIE FP1 2019 June Q5
8 marks Challenging +1.2
5 A curve \(C\) is defined parametrically by $$x = \frac { 2 } { \mathrm { e } ^ { t } + \mathrm { e } ^ { - t } } \quad \text { and } \quad y = \frac { \mathrm { e } ^ { t } - \mathrm { e } ^ { - t } } { \mathrm { e } ^ { t } + \mathrm { e } ^ { - t } }$$ for \(0 \leqslant t \leqslant 1\). The area of the surface generated when \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis is denoted by \(S\).
  1. Show that \(S = 4 \pi \int _ { 0 } ^ { 1 } \frac { \mathrm { e } ^ { t } - \mathrm { e } ^ { - t } } { \left( \mathrm { e } ^ { t } + \mathrm { e } ^ { - t } \right) ^ { 2 } } \mathrm {~d} t\).
  2. Using the substitution \(u = \mathrm { e } ^ { t } + \mathrm { e } ^ { - t }\), or otherwise, find \(S\) in terms of \(\pi\) and e .
CAIE FP1 2010 November Q1
4 marks Challenging +1.2
1 The curve \(C\) has equation \(y = \frac { 1 } { 4 } \left( \mathrm { e } ^ { 2 x } + \mathrm { e } ^ { - 2 x } \right)\). Show that the length of the \(\operatorname { arc }\) of \(C\) from the point where \(x = 0\) to the point where \(x = \frac { 1 } { 2 }\) is \(\frac { \mathrm { e } ^ { 2 } - 1 } { 4 \mathrm { e } }\).
AQA Further AS Paper 1 2023 June Q1
1 marks Easy -2.0
1 Which expression below is equivalent to \(\tanh x\) ? Circle your answer. \(\sinh x \cosh x\) \(\frac { \sinh x } { \cosh x }\) \(\frac { \cosh x } { \sinh x }\) \(\sinh x + \cosh x\)
OCR Further Pure Core 1 2019 June Q5
7 marks Standard +0.3
5 The diagram shows part of the curve \(y = 5 \cosh x + 3 \sinh x\). \includegraphics[max width=\textwidth, alt={}, center]{a6d9b3ec-5170-4f06-a8a3-b854efe36f07-3_496_771_315_246}
  1. Solve the equation \(5 \cosh x + 3 \sinh x = 4\) giving your solution in exact form.
  2. In this question you must show detailed reasoning. Find \(\int _ { - 1 } ^ { 1 } ( 5 \cosh x + 3 \sinh x ) \mathrm { d } x\) giving your answer in the form \(a \mathrm { e } + \frac { b } { \mathrm { e } }\) where \(a\) and \(b\) are integers to be determined.
OCR Further Pure Core 1 2019 June Q7
6 marks Standard +0.8
7 The function sech \(x\) is defined by \(\operatorname { sech } x = \frac { 1 } { \cosh x }\).
  1. Show that \(\operatorname { sech } x = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 }\).
  2. Using a suitable substitution, find \(\int \operatorname { sech } x \mathrm {~d} x\).
OCR Further Pure Core 1 2023 June Q7
11 marks Challenging +1.2
7 An engineer is modelling the motion of a particle \(P\) of mass 0.5 kg in a wind tunnel. \(P\) is modelled as travelling in a straight line. The point \(O\) is a fixed point within the wind tunnel. The displacement of \(P\) from \(O\) at time \(t\) seconds is \(x\) metres, for \(t \geqslant 0\). You are given that \(x \geqslant 0\) for all \(t \geqslant 0\) and that \(P\) does not reach the end of the wind tunnel.
If \(t \geqslant 0\), then \(P\) is subject to three forces which are modelled in the following way.
  • The first force has a magnitude of \(5 ( t + 1 ) \cosh t \mathrm {~N}\) and acts in the positive \(x\)-direction.
  • The second force has a magnitude of \(0.5 x \mathrm {~N}\) and acts towards \(O\).
  • The third force has a magnitude of \(\left| \frac { d x } { d t } \right| \mathrm { N }\) and acts in the direction of motion of the particle.
      1. Show that the Maclaurin series for \(\mathrm { f } ( t )\) up to and including the term in \(t\) is \(6 - 5 t\).
      2. Use your answer to part (a)(ii) to show that the term in \(t ^ { 2 }\) in the Maclaurin series for \(\mathrm { f } ( t )\) is \(- 3 t ^ { 2 }\).
      3. By differentiating the differential equation in part (a) with respect to \(t\), show that the term in \(t ^ { 3 }\) in the Maclaurin series for \(\mathrm { f } ( t )\) is \(0.5 t ^ { 3 }\). You are given that the complete Maclaurin series for the function f is valid for all values of \(t \geqslant 0\).
        After 0.25 seconds \(P\) has travelled 1.43 m towards the origin.
      1. By using the Maclaurin series for \(\mathrm { f } ( t )\) up to and including the term in \(t ^ { 3 }\), evaluate the suitability of the model for determining the displacement of \(P\) from \(O\) when \(t = 0.25\).
      2. Explain why it might not be sensible to use the Maclaurin series for \(\mathrm { f } ( t )\) up to and including the term in \(t ^ { 3 }\) to evaluate the suitability of the model for determining the displacement of \(P\) from \(O\) when \(t = 10\).
OCR Further Pure Core 1 2024 June Q7
9 marks Standard +0.8
7
  1. By using the definitions of \(\cosh u\) and \(\sinh u\) in terms of \(\mathrm { e } ^ { u }\) and \(\mathrm { e } ^ { - u }\), show that \(\sinh 2 u \equiv 2 \sinh u \cosh u\). The equation of a curve, \(C\), is \(\mathrm { y } = 16 \cosh \mathrm { x } - \sinh 2 \mathrm { x }\).
  2. Show that there is only one solution to the equation \(\frac { d ^ { 2 } y } { d x ^ { 2 } } = 0\) You are now given that \(C\) has exactly one point of inflection.
  3. Use your answer to part (b) to determine the exact coordinates of this point of inflection. Give your answer in a logarithmic form where appropriate.