Edexel_2017_3

Edexel_2017_3

(a)

We have:

\displaystyle \cosh x = \frac {e^x + e^{-x}} 2

so:

\begin{align*}2 \cosh^2 x - 1 & = 2 \left(\frac {e^x + e^{-x}} 2\right)^2 - 1 \\ & = 2 \times \frac {(e^x + e^{-x})^2} 4 - 1 \\ & = \frac {e^{2x} + 2 + e^{-2x}} 2 -1 \\ & = \frac {e^{2x} + e^{-2x}} 2 + \frac 2 2 - 1 \\ & = \cosh 2x\end{align*}

(b)

Using part (a) we have:

29 \cosh x - 3 \cosh 2x = 29 \cosh x - 3(2 \cosh^2 x - 1) = 29 \cosh x - 6 \cosh^2 x + 3 = 38

so:

-6 \cosh^2 x + 29 \cosh x - 35 = 0

that is:

6 \cosh^2 x - 29 \cosh x + 35 = 0

We can write this as:

(2 \cosh x - 5)(3 \cosh x - 7) = 0

so either:

\cosh x = \dfrac 5 2

giving:

\begin{align*}x &= \pm \operatorname {arcosh} \frac 5 2 \\ & = \pm \ln \left(\frac 5 2 + \sqrt {\left(\frac 5 2\right)^2 - 1}\right) \\ & = \pm \ln \left(\frac {5 + \sqrt {21}} 3\right)\end{align*}

or:

\cosh x = \dfrac 7 3

giving:

\begin{align*} x &= \pm \operatorname {arcosh} \frac 7 3 \\ &= \pm \ln \left(\frac 7 3 + \sqrt {\left(\frac 7 3\right)^2 - 1}\right) \\ & = \pm \ln \left(\frac {7 + \sqrt {40}} 3\right)\end{align*}

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Nice :sunglasses: