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：\({f}^{\, '}\left ( {x} \right )={e}^{x}-x-a\)，令 \(g\left ( {x} \right )={f}^{\, '}\left ( {x} \right )\) ，则\({g}^{\, '}\left ( {x} \right )={e}^{x}-1\)

所以\(g\left ( {x} \right )\)在\(\left ( {-\infty ，0} \right )\)上单调递减，在\(\left ( {0，\infty } \right )\)上单调递增

当\(a\leq 1\)时，\(g\left ( {x} \right )_{\text{min}}=g\left ( {0} \right )=1-a\geq 0\)，即 \({f}^{\, '}\left ( {x} \right )\geq 0\) ，所以\(f\left ( {x} \right )\)在\(\text{R}\)上递增．

不妨设\({x}_{1}<{x}_{2}\)，则 \({x}_{2}>0\)

要证： \({x}_{1}+{x}_{2}<0\)

只需证： \({x}_{1}<-{x}_{2}\)

只需证： \(f\left ( {{x}_{1}} \right )<f\left ( {{-x}_{2}} \right )\)

只需证： \(2-f\left ( {{x}_{2}} \right )<f\left ( {{-x}_{2}} \right )\)

令 \(F\left ( {x} \right )=f\left ( {x} \right )+f\left ( {-x} \right )-2\left ( {x>0} \right )\)，\({F}^{\, '}\left ( {x} \right )={f}^{\, '}\left ( {x} \right )-{f}^{\, '}\left ( {-x} \right )={e}^{x}-{e}^{-x}-2x\)

\({F}^{\, ''}\left ( {x} \right )={e}^{x}+{e}^{-x}-2>0\)，所以\({F}^{\, '}\left ( {x} \right )\)在\(\left ( {0，+\infty } \right )\)上递增，有\({F}^{\, '}\left ( {x} \right )>{F}^{\, '}\left ( {0} \right )=0\)

所以\(F\left ( {x} \right )\)在\(\left ( {0，+\infty } \right )\)上递增，即\(F\left ( {x} \right )>F\left ( {0} \right )=0\)，故\(F\left ( {{x}_{2}} \right )>0\)，即\(2-f\left ( {{x}_{2}} \right )<f\left ( {{-x}_{2}} \right )\)，所以\({x}_{1}+{x}_{2}<0\)．