{"ops":[{"attributes":{"class":"normal-img"},"insert":{"native-image":{"alt":"read-normal-img","url":"https://b2.sanwen.net/b_article/watermark/196566a729394d3c30c705b806417dab6e83b891.jpg","width":1078,"height":297,"size":87951,"status":"loaded"}}},{"insert":"第一问分类讨论较为繁琐，关键在于注意到x=lna是一个极值点，解答略去。\n第二问利用第一问结论可知t=lna＞1，代入后采用先猜后证的证明方法\n"},{"attributes":{"class":"normal-img"},"insert":{"native-image":{"alt":"read-normal-img","url":"https://b2.sanwen.net/b_article/watermark/f06219ce185ec89a2cb80713dee1dee165ad196b.jpg","width":1980,"height":784,"size":242846,"status":"loaded"}}},{"attributes":{"color":"#00aeec"},"insert":"蓝字部分为分析过程。注意到端点处（x=1）原函数与导函数的值都等于零，则端点处二阶导数必大于零，解出m的范围为m≥-e"},{"insert":"\n"},{"attributes":{"color":"#18191c"},"insert":"下对此加以证明"},{"insert":"\n"},{"attributes":{"class":"normal-img"},"insert":{"native-image":{"alt":"read-normal-img","url":"https://b2.sanwen.net/b_article/watermark/c8f7a7d7a9e0cbe17cbc8c34f9570b62fc865396.jpg","width":2060,"height":1548,"size":545294,"status":"loaded"}}},{"insert":"从正反两个方面进行证明，只需讨论二阶导数的正负性即可。\n本题关键在于讨论端点特殊情况后加以证明的先猜后证方法，如果直接对含有参数m的函数求导讨论会比较复杂。当然，蓝色部分的思考过程可以省去，那么卷面上就能呈现出更加简洁的过程；如果对于后续证明计算能力信心不足/考试时间不够了，建议将蓝色思考过程留着得过程分。\n（最后，做出第一小问是最关键的）\n"}]}
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