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共回答了23个问题采纳率:87% 举报
已知函数
(1)求曲线y=f(x)在(2,f(2))处的切线方程;
(2)若g(x)=f(x)一
![](https://img.yulucn.com/upload/e/1e/e1ef7e76551faf38d616c9fa42d0996c_thumb.jpg)
有两个不同的极值点.其极小值为M,试比较2M与一3的大小,并说明理由;
(3)设q>p>2,求证:当x∈(p,q)时,
![](https://img.yulucn.com/upload/8/9a/89ab9ee3ef3334ac5164eccbd5e05424_thumb.jpg)
.
(1)
![](https://img.yulucn.com/upload/6/dc/6dc04f6fe3c331cc89a28a06fde9492c_thumb.jpg)
;(2)
![](https://img.yulucn.com/upload/5/ab/5ab404f946aa8ddeb95b20d4ffde6c8b_thumb.jpg)
;(3)证明过程详见解析.
试题分析:本题主要考查导数的运算、利用导数研究函数的单调性、利用导数求函数的极值和最值、利用导数求曲线的切线方程等数学知识,考查学生分析问题解决问题的能力、转化能力和计算能力.第一问,先对
![](https://img.yulucn.com/upload/9/7c/97cd62a91975507fb8cb9444a0b5f778_thumb.jpg)
求导,将
![](https://img.yulucn.com/upload/9/bb/9bbffcf49902603cbc53cb8a6d633be1_thumb.jpg)
代入到
![](https://img.yulucn.com/upload/b/b5/bb56a2e0419aa4fe39ea96651db0fd33_thumb.jpg)
中得到切线的斜率,将
![](https://img.yulucn.com/upload/9/bb/9bbffcf49902603cbc53cb8a6d633be1_thumb.jpg)
代入到
![](https://img.yulucn.com/upload/9/7c/97cd62a91975507fb8cb9444a0b5f778_thumb.jpg)
中得到切点的纵坐标,最后利用点斜式,直接写出切线方程;第二问,对
![](https://img.yulucn.com/upload/e/f4/ef462a91e8baf896e3e2a4598c171143_thumb.jpg)
求导,由于
![](https://img.yulucn.com/upload/e/f4/ef462a91e8baf896e3e2a4598c171143_thumb.jpg)
有2个不同的极值点,所以
![](https://img.yulucn.com/upload/7/03/7033a5eb32f560be43d97ef6604d3fb9_thumb.jpg)
有2个不同的根,即
![](https://img.yulucn.com/upload/9/bd/9bd89d892bb36e312055f85a72170e32_thumb.jpg)
在
![](https://img.yulucn.com/upload/4/4a/44aeaae7ca97aa110b76b1c1cbf1da8a_thumb.jpg)
有两个不同的根,所以
![](https://img.yulucn.com/upload/0/c0/0c0f41ee34249bd57c270f9b7949deca_thumb.jpg)
且
![](https://img.yulucn.com/upload/4/09/409016d763ead8ac6b5426509a00b404_thumb.jpg)
,可以解出a的取值范围,所以根据
![](https://img.yulucn.com/upload/e/f4/ef462a91e8baf896e3e2a4598c171143_thumb.jpg)
的单调性判断出
![](https://img.yulucn.com/upload/0/b8/0b841c02d088e21133975db729a460b9_thumb.jpg)
为极小值,通过函数的单调性求最值,从而比较大小;第三问,用分析法证明分析出只须证
![](https://img.yulucn.com/upload/1/d1/1d143bb2409a2fa7cc7d9b6eb423575c_thumb.jpg)
,构造函数,利用函数的单调性证明,同理再证明
![](https://img.yulucn.com/upload/0/a2/0a263bd998b409a8778b00fff1b74464_thumb.jpg)
,最后利用不等式的传递性得到所证不等式.
试题解析:(1)易知
![](https://img.yulucn.com/upload/0/6e/06e497f7fe521d709a053ee9b07b9f4c_thumb.jpg)
,∴
∴所求的切线方程为
![](https://img.yulucn.com/upload/1/54/1548e82e79d693397c3fb5c80b08490e_thumb.jpg)
,即
![](https://img.yulucn.com/upload/6/dc/6dc04f6fe3c331cc89a28a06fde9492c_thumb.jpg)
4分
(2)易知
![](https://img.yulucn.com/upload/8/28/828abb2ce77b32dbc8152658a544d5bf_thumb.jpg)
,
∵
![](https://img.yulucn.com/upload/b/4d/b4d9f7014f8b4a941b3cc21bb9c5b5bd_thumb.jpg)
有两个不同的极值点
∴
![](https://img.yulucn.com/upload/9/bd/9bd89d892bb36e312055f85a72170e32_thumb.jpg)
在
![](https://img.yulucn.com/upload/4/4a/44aeaae7ca97aa110b76b1c1cbf1da8a_thumb.jpg)
有两个不同的根
则
![](https://img.yulucn.com/upload/0/c0/0c0f41ee34249bd57c270f9b7949deca_thumb.jpg)
且
![](https://img.yulucn.com/upload/4/09/409016d763ead8ac6b5426509a00b404_thumb.jpg)
解得
![](https://img.yulucn.com/upload/8/d3/8d37c76e38cdb5c93bc4f9c80605fa65_thumb.jpg)
6分
![](https://img.yulucn.com/upload/b/d4/bd48e8761025a9ef647dbff411c79b16_thumb.jpg)
在
![](https://img.yulucn.com/upload/e/21/e214e6187c6207e7c58f75ad2b787856_thumb.jpg)
递增,
![](https://img.yulucn.com/upload/e/0b/e0b205cc3149ba7dec86bb08bc033354_thumb.jpg)
递减,
![](https://img.yulucn.com/upload/4/4a/44aeaae7ca97aa110b76b1c1cbf1da8a_thumb.jpg)
递增
∴
![](https://img.yulucn.com/upload/b/d4/bd48e8761025a9ef647dbff411c79b16_thumb.jpg)
的极小值
又∵
∴
则
![](https://img.yulucn.com/upload/3/48/3483ae886719b3e73e10efffe68d2940_thumb.jpg)
,∴
![](https://img.yulucn.com/upload/0/ea/0eae36437fd4ca3eb37944e642f8611b_thumb.jpg)
在
![](https://img.yulucn.com/upload/3/ae/3aedfd8359809b7b51ab768a849203cb_thumb.jpg)
递减
∴
![](https://img.yulucn.com/upload/7/af/7af6e7f5b442e71de148104af9c5285e_thumb.jpg)
,故
![](https://img.yulucn.com/upload/7/c5/7c57835b4332052ac65f799abb8d3eac_thumb.jpg)
9分
(3)先证明:当
![](https://img.yulucn.com/upload/4/73/4735b7b7417650541d19819fa1255317_thumb.jpg)
时,
1年前
2