绝密★启用前

2018年普通高等学校招生全国统一考试

理科数学答案解析

一、选择题

1.【答案】C

【解析】da9fc945ca99ece65c3a4d44ef7b5097.png,则93e3833888a85413bdcd84e8c1966501.png,选C.

2.【答案】B

【解析】55aff0db633d281924b9224474b15660.png89f272ea96ce9e646ca83f3c72ddc5aa.png,故选B.

3.【答案】A

【解析】经过一年的新农村建设,农村的经济收入增加了一倍,所以建设前与建设后在比例相同的情况下,建设后的经济收入是原来的2倍,所以建设后种植收入为37%相当于建设前的74%,故选A.

4.【答案】B

【解析】令02731f49cef7ec135770140b199cf7cb.png的公差为8277e0910d750195b448797616e091ad.png,由821fd714273d12098a5abdc1f6fcfed2.png,4178829a8c36cd8fceb2b713b302671a.png得3a120d58a37e9ee99a35146bcc5f799b.png7b5251cfc42de7d531673524768d7f12.png,则fb01d0ee4ccab8202153419a519dfe7a.png,故选B.

5.【答案】D

【解析】b65d5152fd3b93a2f4bf4c94f5fc8cc3.png,cc4040ce7844196d71ad88e28a46e131.pngdac7a64a691a82377886976d23bf2f72.png4e0eea1bd7d6e46116f2b7fbbe8d4026.png,则3872c9ae3f427af0be0ead09d07ae2cf.png,则abd88c67b92ef4088621dc1af7795265.png,a7d70030a16ba97279215a9a1e71e57c.png,所以90df32c463e32184b18452d405b876b0.png,在点5c16f757233856dcf311176b7410d2d5.png处的切线方程为5dbad057040ec6eb5aa5841786e25d33.png,故选D.

6.【答案】A

【解析】0d3e5f6e44b22f1792b8d7c75a53b422.png,

则8d242e487dda54e28676c27a1f65df83.png,故选A.

7.【答案】B

【解析】将三视图还原成直观图,并沿点7fc56270e7a70fa81a5935b72eacbe29.png所在的母线把圆柱侧面展开成如图所示的矩形,从点69691c7bdcc3ce6d5d8a1361f22d04ac.png到点8d9c307cb7f3c4a32822a51922d1ceaa.png的运动轨迹在矩形中为直线段时路径最短,长度为be692c2967f164725701b759829cd1ca.png,故选B.

word/media/image28.gif

8.【答案】D

【解析】由方程组15ef1ad7dabc3bee92dca9e6ae483632.png,解得b2016a60f3d6352463984df49f45414a.png或3e558f074d34a031c5ba4777944fd63e.png,不妨记babdae3ca9b3d6b2acafaa79e9790198.png.

又800618943025315f869e4e1f09471012.png为ec6014ffc3d92709aa6a4fea11bb3788.png,所以5219db3dd1881dc0ee9ed36ee8e78ea8.png,故选D.

9.【答案】C

【解析】若e84fec1e074026d6fa8e3155482c35c3.png存在2个零点,即285078bef260094f6d15e3eff9cdad62.png有2个不同的实数根,即7c1c9491ba7c6e8d6d2cfa82e39b22ca.png与4f618ba1f7bbfd5fd1c0e41d44a2c73c.png的图像有两个交点,由图可知直线4f618ba1f7bbfd5fd1c0e41d44a2c73c.png不在直线45ce3926b05ebb26d70d35d66c6b1511.png的上方即可,即98145acba9121e3022278c1b40f7b2cc.png,则c368bd1ed46d32cc94f8eed036f959f9.png.故选C.

10.【答案】A

【解析】令fa1398f5ce14260b4f4fe75d7bdb5a8f.png角ce04be1226e56f48da55b6c130d45b94.png分别对应的边长为a44c56c8177e32d3613988f4dba7962e.png

Ⅰ,Ⅱ,Ⅲ对应的面积分别为4f61d84c8cdfc64d874576cf3d527da3.png.则6e171bb4f4c578f1acac3bb4d50615db.png；8799212ba4f07c894b8bad6a4c63268f.png；a83ccca7a3c689face9908e1e323566c.png,因为05e865eb1b3c33eb86dacd8325c0c1a7.png,所以821150b635fde918e9923869563d4a6a.png.所以171d9e6f06ae439f9258adac00ef1a00.png,故选A.

11.【答案】B

【解析】如图所示,不妨记f96fb6c789a71ebc5ddeffc531d5946d.png,800618943025315f869e4e1f09471012.png为d8888c9eb26f75e00c7f92f18903d91d.png,渐近线为27dc72501500973db52a73a3559cc0b6.png,所以06a5f36a616f89dd9af523b360d54512.png,则cac2626c3cb5cfec6d357fd20bc029af.png,故选B.

12.【答案】A

【解析】正方体中,连接顶点7efd1ee08558f56b05c301e5bc6ea944.png,三棱锥510cd5dc3d6a2c98379e369d8af7e73a.png为正三棱锥,侧棱与底面所成的角都相等,所以正方体的每条棱与平面b9d2a4133c595f02b0ca277c3497e4a7.png所成的角均相等,不妨令平面5bec3c8af21354246064b1aee139f5c2.png 070801cad1258f71c249b98c0af9ec0f.png.易知,当平面ab410a966ac148e9b78c65c6cdf301fd.png截得正方体的截面为如图所示的平行六边形8827a41122a5028b9808c7bf84b9fcf6.png时截面的面积可以取到最大值.不妨取d8f5729ceb054f0b00e7fcdc7c8da44e.png,则c82f5250779f60075645200ffe8ff9c8.png,bf027614b2651106475f341abef726ee.png,16ba6aeb77f402e1e68f387c354b56cc.png且16b9709cea3b15934ee53f667bcc32d3.png,

等腰梯形d51f782bc42c5e183ce16a74eb806397.png、a7ab494d16db35b2280164286592a6e4.png的高分别为847078ef645060fce6d09f801d5edeb9.png和b82a88186b057092381d6ac4227eb3b2.png

所以bc6ec5bcf2bf8de9d2d7c443fa285df2.png41c1eccc01a1281656815c7bc6ba22e3.png.

当8252d77bf5b096d504816a5094724abd.png时,截面面积的最大值为693c2f60b55e1499cd2dcbb9c2ce92d8.png.故选A.

二、填空题

13.【答案】1679091c5a880faf6fb5e6087eb1b2dc.png

【解析】可行域为75b781a7c7441078ffd5053329c34092.png及其内部,当直线e55bc262f2fdd4b3a7db779ecee5c7a6.png经过点38ab0777cd97cdac50771910593f235b.png时,f31d14dfa1f8344db0b3a4223613b15b.png.

14.【答案】e3303356345c610fe287ed26649f082b.png

【解析】由7b24ddbe7ae42813752d95287cc94d91.png得bc6baa69a2f1e56a04a43097d583a4e0.png,当a4f9f890663b2a0597c2fc0e75738ec8.png时,2dc50bad0e987339b0f4df49322b9e67.png,即b26c36f27c2f162071009d98d5688b77.png,所以02731f49cef7ec135770140b199cf7cb.png是等比数列,9cdb83a36c97357c54dc0ec59caf50df.png.

15.【答案】16

【解析】恰有1位女生的选法有e91df1c75eb03ec03710defa35c44ed1.png种,恰有2位女生的选法有6758453aba8608b11d5d999f3a435ab1.png种,所以不同的选法共有16种.

16.【答案】15e72fc96817d7d877848e8966f0b1cd.png

【解析】因为50bbd36e1fd2333108437a2ca378be62.png是奇函数,且0b9607851b414c5f73d649576ea755f5.png,即周期为a1c836d0e4b0212c512b016cba67a88b.png,

所以只需要研究50bbd36e1fd2333108437a2ca378be62.png在616fb95f67d851add0d7930d835802a0.png上的图像.

又b162eb5a94b5619c8149f88e0e8d4667.png,则50bbd36e1fd2333108437a2ca378be62.png在616fb95f67d851add0d7930d835802a0.png上的极值点为4d3f73154d358f1b8afe8465e0b4d992.png,因为73ea67268ba377d51947650d12e2a2c2.png,所以e6ee1304a0e3cdf76c1483d375622d07.png15e72fc96817d7d877848e8966f0b1cd.png.

三、解答题

（一）必考题：共60分。

17.【答案】(1)fd597b5df4a276310e535663bbd4e5de.png

(2)5

【解析】(1)如图所示,在c2196e00bc8bd87732d45cbe12a3522e.png中,由正弦定理ec704d3f85e6e2f2a08115b38dbba1d4.png,

word/media/image112.gif得8a734c13837ae0c1416c11301fe3cd98.png,

0b54171bff8a0e6ae5429de53e6eb9b3.png,37d33ca8972207ac433bda43ec3cea2c.png为锐角,

a3eb2f6197b0233e1a9516ff0435683f.png；

(2)0b54171bff8a0e6ae5429de53e6eb9b3.png,e610602474af3206d44647df037d7bbc.png,

若4ab075f124de6b2b16da745a65eda271.png,

则在508d5fbc02bff8b36bc1db81015e5b26.png中,由余弦定理b5aa1bca059004ffd8d9a2d5d2781d03.png,

得bd127bd9f408dadc2ed756846d9da992.png.

18.【答案】(1)见解析

(2)75276e5ef991e623587b42b4b08f7665.png

【解析】(1)证明：f2eabb677c74c2da92c37f96ef6bd9a2.png四边形cb08ca4a7bb5f9683c19133a84872ca7.png为正方形,e97844b7caa5578d2bc9c7637c0679cf.png分别为9c5c2555b85fd79114fc9fcd3628381f.png的中点,

word/media/image128.gifbf50d71a066aae72eda11b47088a7c2e.png且999f814fa0e7766978fa160e207ab452.png,

e108bece4f9a0470b204b25a8d834499.png,5a83d79f0464c7d03ee8e430cc543612.png平面5a424e50ffdca719329d9ab07625d5ef.png,

d8055244d77469dc0577c4cf4e28e17e.png平面9d19d7f014d8225539ffb111d82550c6.png,95e029696a8e77db6f75665e6464c095.png平面09061edd01d762301ad1ec00adcf47dd.png平面9d19d7f014d8225539ffb111d82550c6.png.

(2)方法1：由(1)知e0a206ad4e78ac287bba9f053500bb11.png平面5a424e50ffdca719329d9ab07625d5ef.png,5a83d79f0464c7d03ee8e430cc543612.png3acf83834396fa1c878707132ead62b8.png,

f2eabb677c74c2da92c37f96ef6bd9a2.pngf6aeb3cca45c78f15c7d0c8a54863a2c.png,b501ddb618123bdacdb58a9cf3d82d9c.png.

令正方形cb08ca4a7bb5f9683c19133a84872ca7.png的边长为2,

7f76971b75be1d1ca09abe248ebf80bb.png,1548e1257d278cf8b835f5e218403838.png.

作adde16bed287080091067dbaabb7b161.png交2c9b682412689d6723e3b31653b5774c.png于点f186217753c37b9b9f958d906208506e.png,连接75f75daed3373b39ee67e33c84afc37d.png,

由(1)知平面09061edd01d762301ad1ec00adcf47dd.png平面9d19d7f014d8225539ffb111d82550c6.png,89596dbb809b3f98189812c3c4610b9f.png平面5a424e50ffdca719329d9ab07625d5ef.png,平面789364a77f9ad6e778de4aca7ea06468.png平面a64453ed3244de7ce2005e04361eca30.png,

7c2fdc504d592bd25227b76b871b6523.png平面9d19d7f014d8225539ffb111d82550c6.png,斜线e2fca8135c2fadca093abd79a6b1c0d2.png在平面9d19d7f014d8225539ffb111d82550c6.png内的射影为75f75daed3373b39ee67e33c84afc37d.png,

aa4d95b9b1720d6906f718316b7974f3.png等于e2fca8135c2fadca093abd79a6b1c0d2.png与平面9d19d7f014d8225539ffb111d82550c6.png所成的角.

ba29cde923b07f8ef5efda332c67e18b.png,80b7471c8eb6c5c52557034dab5b51f9.png,即d074d33e514d6ae3b18037ad75362911.png且726d23120552eeb9fce3dd1cdf8a8250.png,

95e029696a8e77db6f75665e6464c095.png在921dc6d2301482e5b69551c3e1ce8089.png中,52dee34b5c3f5563140982ba6deea960.png.

95e029696a8e77db6f75665e6464c095.png在092fe4f412aa1735048fbced4af31f50.png中,d6f952d1227714bce88892eb3fb1cdeb.png,即e2fca8135c2fadca093abd79a6b1c0d2.png与平面9d19d7f014d8225539ffb111d82550c6.png所成角的正弦值为75276e5ef991e623587b42b4b08f7665.png.

方法2：作adde16bed287080091067dbaabb7b161.png交2c9b682412689d6723e3b31653b5774c.png于点f186217753c37b9b9f958d906208506e.png,连接75f75daed3373b39ee67e33c84afc37d.png,

由(1)知平面09061edd01d762301ad1ec00adcf47dd.png平面9d19d7f014d8225539ffb111d82550c6.png,89596dbb809b3f98189812c3c4610b9f.png平面5a424e50ffdca719329d9ab07625d5ef.png,平面789364a77f9ad6e778de4aca7ea06468.png平面a64453ed3244de7ce2005e04361eca30.png,

7c2fdc504d592bd25227b76b871b6523.png平面9d19d7f014d8225539ffb111d82550c6.png,斜线e2fca8135c2fadca093abd79a6b1c0d2.png在平面9d19d7f014d8225539ffb111d82550c6.png内的射影为75f75daed3373b39ee67e33c84afc37d.png,

aa4d95b9b1720d6906f718316b7974f3.png等于e2fca8135c2fadca093abd79a6b1c0d2.png与平面9d19d7f014d8225539ffb111d82550c6.png所成的角,

令正方形cb08ca4a7bb5f9683c19133a84872ca7.png的边长为2,92eab8d072e8930551fe3b4d4e76c70c.png,

则f1d444414f8a006428592904f23c4448.png,f4edc2102819c22045357f328459914f.png,6e5daa0d9d291187686d06ec82582c2c.png,

由7e50bfb231d3e764dfe9a422bd511c94.png得0248cf65f8be35fdbad89d22d92f09ee.png,解得a726d057f869a8a7dd4d67e2a02e606d.png.

95e029696a8e77db6f75665e6464c095.pngb283008cb7e31949d00466a073a3516d.png,9755be43eef8e9a5c0b1efed7fba68f3.png,则d6f952d1227714bce88892eb3fb1cdeb.png,即e2fca8135c2fadca093abd79a6b1c0d2.png与平面9d19d7f014d8225539ffb111d82550c6.png所成角的正弦值为75276e5ef991e623587b42b4b08f7665.png.

word/media/image212.gif方法3：作adde16bed287080091067dbaabb7b161.png交2c9b682412689d6723e3b31653b5774c.png于点f186217753c37b9b9f958d906208506e.png,

由(1)知平面09061edd01d762301ad1ec00adcf47dd.png平面9d19d7f014d8225539ffb111d82550c6.png,89596dbb809b3f98189812c3c4610b9f.png平面5a424e50ffdca719329d9ab07625d5ef.png,平面789364a77f9ad6e778de4aca7ea06468.png平面a64453ed3244de7ce2005e04361eca30.png,

7c2fdc504d592bd25227b76b871b6523.png平面9d19d7f014d8225539ffb111d82550c6.png,

以3a3ea00cfc35332cedf6e5e9a32e94da.png为坐标原点,建立如图所示的空间直角坐标系.

令正方形cb08ca4a7bb5f9683c19133a84872ca7.png的边长为2,92eab8d072e8930551fe3b4d4e76c70c.png,

则65efb10bf6d7c5a65dd94591ae4922db.png

bd9474d3dae02a558eb2518f3d4ecc0f.png,fbb5d58c0a3195b0121a7bf958cfe1d6.png,

即4b137aef093a67c926f9380e34d5205f.png,

即0cfcb00cb19e6ebb9c7d625a63586097.png,解得a726d057f869a8a7dd4d67e2a02e606d.png.

所以eec7e4f24dc0273f31d5ffc99910e4b8.png,

易知平面9d19d7f014d8225539ffb111d82550c6.png的一个法向量为ac19500f7574b00df8a16a3fef3a9654.png,故7e3bb58f769289361058f449fd04b392.png,

即e2fca8135c2fadca093abd79a6b1c0d2.png与平面9d19d7f014d8225539ffb111d82550c6.png所成角的正弦值为75276e5ef991e623587b42b4b08f7665.png.

19.【答案】(1)直线25ec916d56b8212e569dbf2e4e4b51d4.png的方程为：0aefb7867de9215c6458f8b0ff77fc06.png或e0f1441a2a61fb8ff43283083bb4c32a.png

(2)见解析

【解析】(1)右焦点为6794ce2527719431776852159e1bd001.png,当2db95e8e1a9267b7a1188556b2013b33.png与9dd4e461268c8034f5c8564e155c67a6.png轴垂直时有532fd20862862c72046f2e0c6a874f44.png,则7fc56270e7a70fa81a5935b72eacbe29.png为b235699db4d209069597fccb938b91f0.png或551fea8ca6774c245da983d098797231.png,

直线25ec916d56b8212e569dbf2e4e4b51d4.png的方程为：0aefb7867de9215c6458f8b0ff77fc06.png或e0f1441a2a61fb8ff43283083bb4c32a.png；

(2)方法1：令直线87a21f7c038d76487a4454312725cf06.png的斜率分别为3b18bd697b6d8ffe0b28fd24e1f9b861.png,

①当2db95e8e1a9267b7a1188556b2013b33.png与9dd4e461268c8034f5c8564e155c67a6.png轴重合时有09785c813f9fa3e75e6c6500c2bf8b47.png,所以df18b0a090065395e7c3ead365b19776.png；

②当2db95e8e1a9267b7a1188556b2013b33.png与9dd4e461268c8034f5c8564e155c67a6.png轴不重合时,令5a61c78d99e0681c5c568689877ebd67.png8abdc390cf62f2cb5c210564edb18b13.png,

由cc6ddac75ee8081647856238e707057b.png得2143dc1574a8081c0f4ebef4cb54bfe5.png,则09a254558c0715f22188cf42bcb39e6c.png,

因为d62c36215ac6edfc1fc66a77baf561fd.pngfe6e6ccafbcd9e23c8e68df5fc4461af.png,

所以d62c36215ac6edfc1fc66a77baf561fd.pngedc2d564d269a4d03bb6dcec53a0213d.png,即直线87a21f7c038d76487a4454312725cf06.png的倾斜角互补,得1814b2deb45b550ca358b4636a66ea9e.png.

综合①②所述,得1814b2deb45b550ca358b4636a66ea9e.png.

方法2：令直线87a21f7c038d76487a4454312725cf06.png的斜率分别为3b18bd697b6d8ffe0b28fd24e1f9b861.png,

①由(1)知,当2db95e8e1a9267b7a1188556b2013b33.png与9dd4e461268c8034f5c8564e155c67a6.png轴垂直时有21c1e6e71c8eb5f7ed14825401fc4371.png,即直线87a21f7c038d76487a4454312725cf06.png的倾斜角互补,得1814b2deb45b550ca358b4636a66ea9e.png；

②当2db95e8e1a9267b7a1188556b2013b33.png不与9dd4e461268c8034f5c8564e155c67a6.png轴垂直时,令85325fb8b72cd02dcccce13cc2d04db9.png8abdc390cf62f2cb5c210564edb18b13.png,

由249a48106686a67b894fc8c80874d1e9.png得85bbd037fe22ba1e1d6a0d5f843e7fee.png,则a53ddf94c55aaa5f8de77c7b27ac46fa.png,

因为d62c36215ac6edfc1fc66a77baf561fd.png36253ca9ceadf13109c1437de1c82ec1.png,

所以4d72206fff82876f55a7d4cbac25dd4a.png91ca7ada1ae68d6bd4b48a901136bd2a.png,

即直线87a21f7c038d76487a4454312725cf06.png的倾斜角互补,得1814b2deb45b550ca358b4636a66ea9e.png.

综合①②所述,得1814b2deb45b550ca358b4636a66ea9e.png.

20.【答案】(1)4410c2193913ae206bb44750d04c2368.png

(2)(ⅰ)df69ecf2f962526d951907b7d3d87961.png

(ⅱ)应该对这箱余下的所有产品都作检验.

【解析】(1)由n次独立重复事件的概率计算得f0c33e57a528b5439b68fbc76c27b19c.png,

470bd2674a39f476484c75b1df5a57b3.png且a2ce2d0e19ea7ec82c10e480f8475435.png,

1c4a9a271ff57e678801544c60ca1e95.png时,得022691647546622579253e6d0c6b66ee.png.

又当8800faefe1418c3a1e7eeb123aa77702.png时,261ad3a5686b969a3ebee8795ccd1d2d.png,e815091558c144801ee49e7a7cdc31ed.png单调递增；当1dccbc262f11872b70b7fcbcc6f6be2b.png时,56d9c899bfaade0de9c48c1314778c93.png,e815091558c144801ee49e7a7cdc31ed.png单调递减,

所以022691647546622579253e6d0c6b66ee.png是e815091558c144801ee49e7a7cdc31ed.png在b6dbc33006b907f2db1855810abfce98.png上唯一的极大值点,也是最大值点,即4410c2193913ae206bb44750d04c2368.png.

(2)(ⅰ)已检验的20件产品的检验费用为a190bd237d8b36c54de6c1e8b46a601b.png元.

该箱余下的产品的不合格品件数服从二项分布74a81f34e518ce9a246feca293ec2b6a.png,估计不合格品件数为a06f81d23a337afa5752f9aa0dae3a5f.png,

若不对该箱余下的产品作检验,余下的产品的赔偿费用估计为02d3f494fdafc5dd5cfad24dda0593f6.png元.

所以,若不对该箱余下的产品作检验,则33c5134c9623eb131658c847aba70a4a.png.

(ⅱ)若对该箱余下的产品都作检验,则只需支付检验费用,13fdbee9c8b1ed8282c1a62417c846ef.png.

因为aa7071189643d8bcb16e91bf3ef5f121.png,所以应该对这箱余下的所有产品都作检验.

21.【答案】(1)2dfc03408510a4cb71c6a9953a129532.png时,50bbd36e1fd2333108437a2ca378be62.png在定义域b921db311612fd3665c51872c7a83455.png上始终单调递减；

d834a943b0a92f67d91481b42bc89d84.png时,50bbd36e1fd2333108437a2ca378be62.png在372021e44d8f5b7031fd90c6549a38bc.png上递减,

在07137cf30494a01ba1179fde988bb39b.png上递增.

(2)见解析

【解析】(1)4b1994d0b08711cfa56ca81318376b36.png

令29f11b449e6a92ee08f3ef89b013d0cc.png,d6bf013cd8cd8181e880e881976f1f98.png.

①75de8ad3cf1b7a742b38f8a2f71e0017.png时,78ed716d96895ff1e60a8f90d8d92846.png,53b934a42252cd024c1b17395c8485d3.png恒成立,

所以50bbd36e1fd2333108437a2ca378be62.png在定义域b921db311612fd3665c51872c7a83455.png上始终单调递减.

②45e810433a9791ce184eacab3daabb09.png或d834a943b0a92f67d91481b42bc89d84.png时,55572ad9aeddcc415b04fa3e0e1cc5eb.png.

由c36a0c4a611d3144a5e9078a3aa481ac.png即06605d0b94674429f3ab62bec2350d50.png解得712ac7b0e7ab40030591e5cdff3f3f42.png,且645efd0472d3986609053a09a3abf22b.png.

45e810433a9791ce184eacab3daabb09.png时,7ccc1d11896e2314cbfd5c5e51f24cd1.png,99300cb144197ff8dc44e17568431c8f.png恒成立,所以50bbd36e1fd2333108437a2ca378be62.png在定义域b921db311612fd3665c51872c7a83455.png上始终单调递减.

d834a943b0a92f67d91481b42bc89d84.png时,e7496ac73b97c098ebfda0fbe320bc62.png,

在88bcb1bb3a9231499f499e0a7f0a92f5.png上99300cb144197ff8dc44e17568431c8f.png,50bbd36e1fd2333108437a2ca378be62.png单调递减；在682d2d6c6265678c975188fe26e60d8c.png上8baaf0ef5eac596ebf01a45909545fc1.png,50bbd36e1fd2333108437a2ca378be62.png单调递增.

综上所述,2dfc03408510a4cb71c6a9953a129532.png时,50bbd36e1fd2333108437a2ca378be62.png在定义域b921db311612fd3665c51872c7a83455.png上始终单调递减；

d834a943b0a92f67d91481b42bc89d84.png时,50bbd36e1fd2333108437a2ca378be62.png在372021e44d8f5b7031fd90c6549a38bc.png上递减,

在07137cf30494a01ba1179fde988bb39b.png上递增.

(2)证明：方法1：由(1)知d834a943b0a92f67d91481b42bc89d84.png时50bbd36e1fd2333108437a2ca378be62.png存在两个极值点,且e7496ac73b97c098ebfda0fbe320bc62.png.

欲证明d667c4bef60c332e77ad0ce23dcfcb0d.png等价于证明86be8d9f435a1f2b6ce09ce9509840b9.png.

即证明e379a2cac1c506e3bf3167076c839924.png,其中9865b118af4cfc107929ec116ab9eb80.png是方程f431e237b41116b911e3d009c1350015.png的两个根.

令4075a53b51c671ffe9a2f082b606dc8d.png,则满足2f271b9d193d938cc90fd8b26750569d.png,即4d80da4a4098365c02209d3511e2683e.png.

bade64772262bd0e1dcc762a255abbe3.png

dd1ae0a7f959633ccc1fbadb5e1a52a9.png,6041abeb8f59b64c9a70ddc13c81eaf5.png,4075a53b51c671ffe9a2f082b606dc8d.png在3e4aa66b985d086e99cc457f456c7814.png上为减函数.

因为e7496ac73b97c098ebfda0fbe320bc62.png,所以97b10fee5b0d384b55431a20fa245593.png,即e379a2cac1c506e3bf3167076c839924.png,得证.

方法2：由(1)知e7496ac73b97c098ebfda0fbe320bc62.png,84302175f3fde3e08923609e44a3ea6d.png,b5898a59b63d67ad659b03dcfe7abb50.png,从而有e9b0e6a89021719dc8b60a08817c53cf.png.

e53f9554405d20de7638b686fc99531b.png

4896fd55cc5b830198fa86d6601afff5.pngb7fabe241a4fc46fbc388e92854e1aee.png,

要证明d667c4bef60c332e77ad0ce23dcfcb0d.png等价于证明2402fef554fa89556f9ca2946cb96675.png,即证明10ff3b5bf51706ec30f6ae76fd810e46.png.

f28b859b1382084f4dae1caa901333b0.png,95e029696a8e77db6f75665e6464c095.png只需证明18ece505496ce036de04a7587833e47f.png,即证明6a3a92c10d4fd8e2a7d1279c53ab3df1.png成立即可.

令8a05b22e6fb94de9791d10e544bda927.png,

则98279df7194fb341c5c8194566289f24.png,e1ea486fdf4e0046705c807b9642c7bd.png在b6dbc33006b907f2db1855810abfce98.png上为减函数.

所以6eaaef88fbb0e958168937d2aa9bdad3.png,根据23e2b70795905afc0bcd0562c38db69e.png,证得6a3a92c10d4fd8e2a7d1279c53ab3df1.png成立,得证.

（二）选考题：共10分。

22.【答案】(1)932d0ec79260e01afd1dd960c7bc69bb.png的直角坐标方程为391de5e66405bfd4a34d57ea6e46dc6a.png

(2)9824b26a51714309aa4afd370035ce53.png的方程为：00f35f9ce32c396185ca65f4fd7525ed.png.

【解析】(1)9cdaecd00fbdfb09f1cbbb4434dfa684.png,

所以932d0ec79260e01afd1dd960c7bc69bb.png的直角坐标方程为391de5e66405bfd4a34d57ea6e46dc6a.png；

word/media/image394.gif (2)曲线9824b26a51714309aa4afd370035ce53.png：b49059cfda2cfdd3c4258c48b9ddca49.png,其图像是关于415290769594460e2e485922904f345d.png轴对称且以717fdb3c373d624e85a8a8c76ce80778.png为端点的两条射线.

932d0ec79260e01afd1dd960c7bc69bb.png：ce476df10da28c6a29a2ce19b5a92bc7.png,其图像是以929849a79d5999b8dad863fe1feabfda.png为圆心,半径为2的圆.

若9824b26a51714309aa4afd370035ce53.png与932d0ec79260e01afd1dd960c7bc69bb.png有且仅有三个公共点,

则7ac875d7d159339134b37ddd56563963.png且bd5212a78a5bd8e80e1824a8f596e3ea.png与932d0ec79260e01afd1dd960c7bc69bb.png相切(如图).

由7ad6d3239602961a4e2a3ed715f193b8.png且7ac875d7d159339134b37ddd56563963.png,解得8291d9f0e28185e57b9f1aafc1dd047f.png,则9824b26a51714309aa4afd370035ce53.png的方程为：00f35f9ce32c396185ca65f4fd7525ed.png.

23.【答案】(1)解集为95c6816f652321fd2e36bd6ffd250a61.png.

(2)daebe3322cb2c801dfc685f03565f904.png.

【解析】(1)当3872c9ae3f427af0be0ead09d07ae2cf.png时,86ded70119db4239ca8560296ca078fe.png,则

5c718cd8d0682b37d5444000b894095f.png时,ac78f66b11e11c72da8a883a3e51a345.png,则c5fa18e0f40aaa481bf8d30999c3cc80.png无解；

f440dc0236526042f52a3d40435a2d22.png时,080320743c76f725cd1f62a2c774c4e6.png,则c5fa18e0f40aaa481bf8d30999c3cc80.png的解集为e308e458bad0543579bd64f537f664d0.png；

e3f7f32051ade969e770ab04b5a4853b.png时,548ffbe8622c11798f7c2665c414ad67.png,则c5fa18e0f40aaa481bf8d30999c3cc80.png的解集为e2d902a9042923c98895938662785215.png.

综上所述,所求解集为95c6816f652321fd2e36bd6ffd250a61.png.

(2)54cdc91a376b36571d8cabd0d7ee441f.png时不等式81400a213f6566c44f41ed8e9250fcc0.png成立,即08a863471e19b41274c6901d2e6baae3.png,则c123da9c273e681068d781e4637a5bb2.png成立.

所以99a329da0b93da091f90d5d6bf9f5c33.png.

因为aa135e67321926f181d788c1a35afdf2.png时,有b50fce5537c3f757b44e18609a9bd02c.png,所以daebe3322cb2c801dfc685f03565f904.png.

2018年高考理科数学全国卷1-答案