参数方程 $\begin{cases} x = a\left( {t - \sin t} \right) ,\\ y = a\left( {1 - \cos t} \right),\end{cases}(a > 0),$ 所表示的函数 $y = f\left( x \right)$ 是 \((\qquad)\)
【难度】
【出处】
2010年复旦大学优秀高中生文化水平选拔测试
【标注】
【答案】
C
【解析】
设 $\varphi\left( t \right) = a\left( {t - \sin t} \right)$，则 $\varphi\left( t \right)$ 是奇函数．
因为$$\varphi'\left( t \right) = a\left( {1 - \cos t} \right) > 0,$$所以 $\varphi\left( t \right)$ 是单调递增函数，于是 $\varphi\left( t \right)$ 存在反函数，设为 ${\varphi^{ - 1}}$．
显然函数 ${\varphi^{ - 1}}$ 也为单调递增的奇函数．
所以$$f\left( x \right) = a\left[ {1 - \cos {\varphi ^{ - 1}}\left( x \right)} \right].$$对于选项A、B．
因为\[\begin{split}f\left( { - x} \right) &= a\left[ {1 - \cos {\varphi ^{ - 1}}\left( { - x} \right)} \right] \\&= a\left[ {1 - \cos \left( { - {\varphi ^{ - 1}}\left( x \right)} \right)} \right] \\&= a\left[ {1 - \cos {\varphi ^{ - 1}}\left( x \right)} \right] = f\left( x \right).\end{split}\]所以 $f\left( x \right)$ 是偶函数，图象关于 $y$ 轴对称．
对于选项C、D．
因为$$\varphi \left( {t + 2{\mathrm{\pi }}} \right) - \varphi \left( t \right) = 2a{\mathrm{\pi }},$$所以$$\varphi \left( {t + 2{\mathrm{\pi }}} \right) = 2a{\mathrm{\pi }} + \varphi \left( t \right),t + 2{\mathrm{\pi }} = {\varphi ^{ - 1}}\left[ {2a{\mathrm{\pi }} + \varphi \left( t \right)} \right].$$取 $t = {\varphi ^{ - 1}}\left( x \right)$，有$${\varphi ^{ - 1}}\left( {2a{\mathrm{\pi }} + x} \right) = 2{\mathrm{\pi }} + {\varphi ^{ - 1}}\left( x \right).$$因此\[\begin{split}f\left( {x + 2a{\mathrm{\pi }}} \right) &= a\left[ {1 - \cos {\varphi ^{ - 1}}\left( {x + 2a{\mathrm{\pi }}} \right)} \right] \\&= a\left[ {1 - \cos {\varphi ^{ - 1}}\left( x \right)} \right] = f\left( x \right).\end{split}\]综上，选项C正确．
因为$$\varphi'\left( t \right) = a\left( {1 - \cos t} \right) > 0,$$所以 $\varphi\left( t \right)$ 是单调递增函数，于是 $\varphi\left( t \right)$ 存在反函数，设为 ${\varphi^{ - 1}}$．
显然函数 ${\varphi^{ - 1}}$ 也为单调递增的奇函数．
所以$$f\left( x \right) = a\left[ {1 - \cos {\varphi ^{ - 1}}\left( x \right)} \right].$$对于选项A、B．
因为\[\begin{split}f\left( { - x} \right) &= a\left[ {1 - \cos {\varphi ^{ - 1}}\left( { - x} \right)} \right] \\&= a\left[ {1 - \cos \left( { - {\varphi ^{ - 1}}\left( x \right)} \right)} \right] \\&= a\left[ {1 - \cos {\varphi ^{ - 1}}\left( x \right)} \right] = f\left( x \right).\end{split}\]所以 $f\left( x \right)$ 是偶函数，图象关于 $y$ 轴对称．
对于选项C、D．
因为$$\varphi \left( {t + 2{\mathrm{\pi }}} \right) - \varphi \left( t \right) = 2a{\mathrm{\pi }},$$所以$$\varphi \left( {t + 2{\mathrm{\pi }}} \right) = 2a{\mathrm{\pi }} + \varphi \left( t \right),t + 2{\mathrm{\pi }} = {\varphi ^{ - 1}}\left[ {2a{\mathrm{\pi }} + \varphi \left( t \right)} \right].$$取 $t = {\varphi ^{ - 1}}\left( x \right)$，有$${\varphi ^{ - 1}}\left( {2a{\mathrm{\pi }} + x} \right) = 2{\mathrm{\pi }} + {\varphi ^{ - 1}}\left( x \right).$$因此\[\begin{split}f\left( {x + 2a{\mathrm{\pi }}} \right) &= a\left[ {1 - \cos {\varphi ^{ - 1}}\left( {x + 2a{\mathrm{\pi }}} \right)} \right] \\&= a\left[ {1 - \cos {\varphi ^{ - 1}}\left( x \right)} \right] = f\left( x \right).\end{split}\]综上，选项C正确．
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