高中数学选择性必修 第二册（381题）

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知识点：第五章 一元函数的导数及其应用

参考答案：见解析

解析：

由对称性，不妨设 \( 0\le {x}_{1}<{x}_{2}\le \pi \) ，则 \( \frac{f\left({x}_{1}\right)-f\left({x}_{2}\right)}{{x}_{1}^{2}-{x}_{2}^{2}}+a>0\) 即为 \( f\left({x}_{2}\right)+a{x}_{2}^{2}>f\left({x}_{1}\right)+a{x}_{1}^{2}\).

设 \( g\left(x\right)=f\left(x\right)+a{x}^{2}\) ，则\( g\left(x\right)\)在 \(\left [ {0，\pi } \right ]\) 上单调递增，

故\(g'\left ( {x} \right )={e}^{x}\left ( {\sin {x+\cos {x}}} \right )+2ax\geq 0\)​ ​ 在 \(\left [ {0，\pi } \right ]\) 上恒成立.

方法一：（含参讨论）

设\(h\left ( {x} \right )=g'\left ( {x} \right )={e}^{x}\left ( {\sin {x+\cos {x}}} \right )+2ax\geq 0\)​ ​，

则 \( h\left(0\right)=1>0\) ， \( h\left(\pi \right)=-{e}^{\pi }+2a\pi \ge 0\)，解得 \( a\ge \frac{{e}^{\pi }}{2\pi }\).

\(h'\left ( {x} \right )=2\left ( {{e}^{x}\cos {x+a}} \right )\) ， \({h}^{'}\left ( {0} \right )=2\left ( {a+1} \right )>0\) ， \({h}^{'}\left ( {\pi } \right )=2\left ( {a-{e}^{\pi }} \right )\).

①当  \( a\ge {e}^{\pi }\) 时，\(\left [ {h'\left ( {x} \right )} \right ]'={2e}^{x}\left ( {\cos {x-}\sin {x}} \right )\) ​，

故，当 \(x\in \left [ {0，\frac {\pi } {4}} \right ]\) 时 ， \(\left [ {h'\left ( {x} \right )} \right ]'={2e}^{x}\left ( {\cos {x-}\sin {x}} \right )\geq 0\)  ， \({h}^{'}\left ( {x} \right )\) 递增；

当 \(x\in \left [ {\frac {\pi } {4}，\pi } \right ]\) 时， \(\left [ {h'\left ( {x} \right )} \right ]'={2e}^{x}\left ( {\cos {x-}\sin {x}} \right )\leq 0\)，\({h}^{'}\left ( {x} \right )\)递减；

此时，\(h'\left ( {x} \right )\geq \text{min}\left \{ {h'\left ( {0} \right )，h'\left ( {\pi } \right )} \right \} =h'\left ( {\pi } \right )=2\left ( {a-{e}^{\pi }} \right )\geq 0\)​ ​， \(h\left( x \right) = g'\left( x \right)\) 在 \(\left [ {0，\pi } \right ]\) 上单调递增，故\(h\left ( {x} \right )={g}^{'}\left ( {x} \right )\ge {g}^{'}\left ( {0} \right )=1>0\) ，符合条件.

②当 \( \frac{{e}^{\pi }}{2\pi }\le a<{e}^{\pi }\) 时，同①，当 \(x\in \left [ {0，\frac {\pi } {4}} \right ]\) 时， \({h}^{'}\left ( {x} \right )\) 递增；当 \(x\in \left [ {\frac {\pi } {4}，\pi } \right ]\) 时， \({h}^{'}\left ( {x} \right )\) 递减；

∵ \({h}^{'}\left ( {\frac {\pi } {4}} \right )>{h}^{'}\left ( {0} \right )=2\left ( {a+1} \right )>0\)， \({h}^{'}\left ( {\pi } \right )=2\left ( {a-{e}^{\pi }} \right )<0\) ，

∴由连续函数零点存在性定理及单调性知， \(\exists {x}_{0}\in \left ( {\frac {\pi } {4}，\pi } \right )\)， \({h}^{'}\left ( {{x}_{0}} \right )=0\).

于是，当 \(x\in \left . {[0，{x}_{0}} \right )\) 时，\({h}^{'}\left ( {x} \right )>0\)，\(h\left( x \right) = g'\left( x \right)\) 单调递增；

当 \(x\in {(x}_{0}，\pi ]\) 时， \({h}^{'}\left ( {x} \right )<0\)， \(h\left( x \right) = g'\left( x \right)\) 单调递减.

∵\( h\left(0\right)=1>0\)，\( h\left(\pi \right)=-{e}^{\pi }+2a\pi \ge 0\)，\(\therefore g'\left ( {x} \right )=h\left ( {x} \right )\geq \text{min}\left \{ {h\left ( {0} \right )，h\left ( {\pi } \right )} \right \} \geq 0\)​ ，符合条件.

综上，实数 \( a\) 的取值范围是 \(\left . {[\frac {{e}^{\pi }} {2\pi }，+\infty } \right )\).

方法二：（参变分离）

由对称性，不妨设 \( 0\le {x}_{1}<{x}_{2}\le \pi \)，

则 \( \frac{f\left({x}_{1}\right)-f\left({x}_{2}\right)}{{x}_{1}^{2}-{x}_{2}^{2}}+a>0\) 即为 \( f\left({x}_{2}\right)+a{x}_{2}^{2}>f\left({x}_{1}\right)+a{x}_{1}^{2}\).

设 \( g\left(x\right)=f\left(x\right)+a{x}^{2}\)，则 \( g\left(x\right)\) 在 \(\left [ {0，\pi } \right ]\) 上单调递增，

故\(g'\left ( {x} \right )={e}^{x}\left ( {\sin {x+\cos {x}}} \right )+2ax\geq 0\)​ ​在 \(\left [ {0，\pi } \right ]\) 上恒成立.

\({\because g}^{'}\left ( {0} \right )=1>0\)，\(\therefore g'\left ( {x} \right )={e}^{x}\left ( {\sin {x+\cos {x}}} \right )+2ax\geq 0\)​​在 \(\left [ {0，\pi } \right ]\) 上恒成立

\( \iff -2a\le \frac{{e}^{x}\left(\mathit{sin}x+\mathit{cos}x\right)}{x}\)\(⇔-2a\leq \frac {{e}^{x}\left ( {\sin {x+\cos {x}}} \right )} {x}，\forall x\in (0，\pi ]\)​.

设 \(h\left ( {x} \right )=\frac {{e}^{x}\left ( {\sin {x+\cos {x}}} \right )} {x}，x\in (0，\pi ]\)​​，​则\(h'\left ( {x} \right )=\frac {{e}^{x}\left ( {2x\cos {x}-\sin {x-\cos {x}}} \right )} {{x}^{2}}，x\in (0，\pi ]\)​ ​​.

设\(φ\left ( {x} \right )=2x-\tan {x}-1，x\in (0，\frac {\pi } {2})\cup (\frac {\pi } {2}，\pi ]\)​ ​​，

则 \(φ'\left ( {x} \right )=2x-\frac {1} {\cos^{2} {x}}，x\in (0，\frac {\pi } {2})\cup (\frac {\pi } {2}，\pi ]\)​​.

由 \({φ}^{'}\left ( {x} \right )>0\)，\(x\in \left ( {0，\frac {\pi } {2}} \right )\cup (\frac {\pi } {2}，\pi ]\)，得 \(φ\left ( {x} \right )\) 在\(\left ( {0，\frac {\pi } {4}} \right )\)，\((\frac {3\pi } {4}，\pi ]\) 上单调递增；

由 \({φ}^{'}\left ( {x} \right )<0\)，\(x\in \left ( {0，\frac {\pi } {2}} \right )\cup (\frac {\pi } {2}，\pi ]\)，得 \(φ\left ( {x} \right )\) 在 \(\left ( {\frac {\pi } {4}，\frac {\pi } {2}} \right )\)，\((\frac {\pi } {2}，\frac {3\pi } {4}]\) 上单调递减.

故 \(x\in \left ( {0，\frac {\pi } {2}} \right )\) 时 \(φ\left ( {x} \right )\le φ\left ( {\frac {\pi } {4}} \right )=\frac {\pi } {2}-2<0\)； \(x\in (\frac {\pi } {2}，\pi ]\) 时 \(φ\left ( {x} \right )\ge φ\left ( {\frac {3\pi } {4}} \right )=\frac {3\pi } {2}>0\).

从而，\(φ\left ( {x} \right )\cos {x}=2x\cos {x}-\sin {x}-\cos {x<0，x\in \left ( {0，\frac {\pi } {2}} \right )}\cup (\frac {\pi } {2}，\pi ]\)​​，​

又 \( x=\frac{\pi }{2}\) 时， \(2x\cos {x}-\sin {x}-\cos {x=-1<0}\)，故 ​​​\(h'\left ( {x} \right )=\frac {{e}^{x}\left ( {2x\cos {x-\sin {x-\cos {x}}}} \right )} {{x}^{2}}<0，x\in (0，\pi ]\)​，​

​\(h\left ( {x} \right )=\frac {{e}^{x}\left ( {\sin {x+\cos {x}}} \right )} {x}，x\in (0，\pi ]\)​ 单调递减，\(h\left ( {x} \right )_{\text{min}}=h\left ( {\pi } \right )=-\frac {{e}^{\pi }} {\pi }，x\in (0，\pi ]\)​ ​​.

于是， \( -2a\le -\frac{{e}^{\pi }}{\pi }\iff a\ge \frac{{e}^{\pi }}{2\pi }\).

综上，实数\( a\)的取值范围是 \(\left . {[\frac {{e}^{\pi }} {2\pi }，+\infty } \right )\)