题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Solve the equation

2\sinh^2 x + 3\cosh x = 7

Give your answers as simplified natural logarithms.

(6)

题目中文翻译

在本题中，你必须写出所有解题步骤。

完全依赖计算器技术求解是不可以的。

求方程

2\sinh^2 x + 3\cosh x = 7

的解，并将答案写成最简自然对数形式。

(6)

解答

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先用恒等式 $\sinh^2x=\cosh^2x-1$ ，把原方程化成关于 $\cosh x$ 的二次方程。舍去不可能的负值后，再用指数定义把 $\cosh x=\frac32$ 化成关于 $e^x$ 的二次方程。

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Using $\sinh^2x=\cosh^2x-1$ ,

\begin{align*} 2(\cosh^2x-1)+3\cosh x=&\,7,\\ 2\cosh^2x+3\cosh x-9=&\,0,\\ (2\cosh x-3)(\cosh x+3)=&\,0. \end{align*}

Thus

\cosh x=\frac32 \quad\text{or}\quad \cosh x=-3.

For real $x$ , $\cosh x\geqslant1$ , so $\cosh x=-3$ is rejected. Using the exponential definition,

\frac{e^x+e^{-x}}2=\frac32.

Let $u=e^x$ , where $u>0$ . Then

\begin{align*} u+\frac1u=&\,3,\\ u^2-3u+1=&\,0,\\ \big(u-\tfrac32\big)^2=&\,\tfrac54,\\ u=&\,\frac{3\pm\sqrt5}{2}. \end{align*}

Both values of $u$ are positive. Therefore,

\boxed{ \begin{aligned} x=&\,\ln\big(\frac{3+\sqrt5}{2}\big),\\ \text{or}\quad x=&\,\ln\big(\frac{3-\sqrt5}{2}\big). \end{aligned}}

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把 $\sinh x$ 、 $\cosh x$ 都直接写成指数形式。令 $u=e^x>0$ 后会得到一个可分解为两个二次式的四次方程；最后必须利用 $u>0$ 排除负根。

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Let $u=e^x$ , so $u>0$ . Then

\sinh x=\frac{u-u^{-1}}2 \quad\text{and}\quad \cosh x=\frac{u+u^{-1}}2.

Substituting into the equation gives

2\big(\frac{u-u^{-1}}2\big)^2 +3\big(\frac{u+u^{-1}}2\big)=7.

Multiplying by $2u^2$ ,

\begin{align*} (u^2-1)^2+3u(u^2+1)=&\,14u^2,\\ u^4+3u^3-16u^2+3u+1=&\,0,\\ (u^2+6u+1)(u^2-3u+1)=&\,0. \end{align*}

The first factor gives

\begin{align*} u^2+6u+1=&\,0,\\ (u+3)^2=&\,8,\\ u=&\,-3\pm2\sqrt2. \end{align*}

and both values are negative, so they are invalid. The second factor gives

\begin{align*} u^2-3u+1=&\,0,\\ \big(u-\tfrac32\big)^2=&\,\tfrac54,\\ u=&\,\frac{3\pm\sqrt5}{2}. \end{align*}

with both values positive. Hence

\boxed{ \begin{aligned} x=&\,\ln\big(\frac{3+\sqrt5}{2}\big),\\ \text{or}\quad x=&\,\ln\big(\frac{3-\sqrt5}{2}\big). \end{aligned}}