高等数学I（下）学习通超星期末考试答案章节答案2024年

高等数学I（下）学习通超星期末考试章节答案2024年/star3/origin/21299483b4d3babc928c2f4efaa9dddb.png

答案:原式(2分)(2分)(1分)/star3/origin/5a92f80d4eade13f80693eddd2726e31.png

答案:原式(2分)(或将换成,将换成)(3分)/star3/origin/18ba5814873401bc09b9fbaeb0e690f3.png

答案:原式=(2分)=(3分)/star3/origin/bba4197ca271acfb84a81291ed0e26d8.png

答案:原式(2分)(3分)/star3/origin/faae229ac76d49aa320907b420327e6c.png

答案:原式(2分)(1分)(2分)/star3/origin/8e599fce5775c7985959bc0c78220dcd.png

答案:原式=(或)(2分)=(3分)/star3/origin/b28d7b7b07988ce44939ca9728d4952a.png

答案:原式=(2分)=(1分)=(2分)/star3/origin/98a07869fc9be3430564f4f416ab5a4c.png

答案:对/star3/origin/fff5df2f594148a00d13e0d3443dc69a.png

答案:对/star3/origin/5361661d700e10066de52befce356150.png

答案:错/star3/origin/7a03beba39bab8c56276fc6edbebe7a2.png

答案:错/star3/origin/b855df12e7f440e8596717c95c26e3dd.png

答案:原式=(3分)=(5分)/star3/origin/34063c5471592ef1f06cecba2bf21c59.png

答案:令则原式=(4分)=(4分)/star3/origin/500d5626718834deaf2d83cf41a4d613.png

答案:原式==(4分)(4分)/star3/origin/eee073e0c1e79a14c1d11a54ce3cc6ac.png

答案:原式==(4分)即,故(4分)/star3/origin/7afb29b9cb742870b66ec864c602ef7a.png

答案:对/star3/origin/8bff64ae9dfaa80f03289c1e61a59dfd.png

答案:对/star3/origin/b774433db495314d0a0fb1861f4e5cda.png

答案:错/star3/origin/dbcda81d2d6a8873ce34cd1548dcb118.png

答案:对/star3/origin/c81d89b831119de2486e55e84cdf39a2.png

答案:错/star3/origin/63c3b2348e6a8b59ef7ea8ca2bab614e.png

答案:错/star3/origin/81047f05bbd4ea887c8bf7d350b0d72b.png

答案:(5分)(3分)/star3/origin/6a3ea2ba2599aa70ac3aa60cbec43bef.png

答案:(5分)(3分)/star3/origin/2771fb5cb7a16b2102817dda83cd3373.png

答案:(5分)(3分)/star3/origin/783282fa91487ad13a6ede57367a2512.png

答案:原式=(4分)=(4分)/star3/origin/ac6e23c5002f69682dd92d9dd51d6518.png

答案:错/star3/origin/a95b7427e59d430dc2076e55d24a5f74.png

答案:对/star3/origin/c8f3bc735e16e862abc74fe21c1a4298.png

答案:对/star3/origin/ede10479ba41ba42a20e8c9cf40bab6c.png

答案:0/star3/origin/b8b82b13ba39c70ba6835bd8ebfdf153.png

答案:原式==/star3/origin/c128110b7be580dc825668879dd377e2.png

答案:原式====/star3/origin/8948413d7415069a9a89d635ed417de3.png

答案:原式=/star3/origin/34feec308c68b8b5ee98185218377e19.png

答案:对/star3/origin/10d4a2e8b908871b9086ab920c8219c1.png

答案:错/star3/origin/6c24b940e78ca59b8d534151713d1a70.png

答案:错/star3/origin/d1e9e44acb9ae642c94bb656202f587d.png

答案:对/star3/origin/79df823da77dd9b1cc77e97a4cb2a3ad.png

答案:错/star3/origin/a5d51db60be7347428ca611b4eb6faf7.png

答案:对/star3/origin/8c669ebe7142a2c1890c962e037df22f.png

答案:错/star3/origin/96b900524307e9466ac2cb9d6ab9e1c8.png

答案:对/star3/origin/f63301931164081fcd0e5b3d0893ea35.png

答案:对/star3/origin/ffa96851d454f7e322ea9fa9dd99a1f1.png

答案:(4分)(4分).(2分)/star3/origin/23e20475340bdecb4f151a62bd65f585.png

答案:由及导数的几何意义得,在点处的切线方程为,在点处的切线方程为.(4分)由解得交点坐标为.(2分)所以,所求面积为.(4分)/star3/origin/de99ac985fd159841ed6bb77f361f40f.png

答案:错/star3/origin/73ebd63f95543d0e4a0116291395c27f.png

答案:错/star3/origin/3b011b869c44469de816eb58675192e2.png

答案:错/star3/origin/571c65e2e4b0578413da8b14.png

答案:对/star3/origin/018779260577c4891356abae15685ea4.png

答案:对/star3/origin/da23528c32ef589aaa158df4503419fc.png

答案:发散/star3/750_1024/32e08a34d2c16e210380f70e13609af8.png

答案:收敛,且值为(ln2)/5/star3/origin/c010a3ed91c7ec0e5eb483ae8d758921.png

答案:收敛,且值为/star3/origin/61cfa98bed178d004a1660fed8cd746b.png

答案:收敛,且值为/star3/origin/344425933636bb57cffa2170a24fc3da.png

答案:由条件可设所求的点为,由,即(5分)因此所求点的坐标为.(3分)/star3/origin/b26946c97581ff2737137362667fa432.png

答案:向量(4分)模方向余弦:,,;(4分)/star3/origin/663b09e63cd4527602b43596f03d52ba.png

答案:(4分)(4分)/star3/origin/7a90dc1c3b3c439dea6467efe972d224.png

答案:向量垂直于、,则向量平行于.因此设向量,(5分)由题设得,得到,所以(3分)/star3/origin/7aa0a07be63003d27e2c2f860f559c24.png

答案:错/star3/origin/0900a509751050171def3c075783e80f.png

答案:-1.5/star3/origin/268f2396a487a99e3e3b5abbaad6be6b.png

答案:0/star3/origin/d867c2d1a0b997a93bcfac5fcec0ecb2.png

答案:10/star3/origin/932974df7c20fa178734a03a3d65240e.png

答案:平面上任一点满足平面方程为/star3/origin/05a0addffadfef9b3ff7f6969d38e2e3.png

答案:直线方向向量取取特殊点则对称式方程:,参数方程:/star3/origin/c17c27ac2b5d32719cb788700f6390ac.png

答案:直线方向向量取直线方程为/star3/origin/f58170cb5f045ff2d1056ecde42a786a.png

答案:平面法向量取平面方程为/star3/origin/fbb347346644087039605a31a2d6df25.png

答案:设方程为将点带入方程得故平面方程为过点空间上任意三个不同的点都可以确定唯一一个平面.

答案:错过已知点且与已知直线垂直的平面不止一个.

答案:错/star3/origin/73e354251d5c40f858e1d60021af6abe.png

答案:对/star3/origin/4564d6bbf2ce1211ed58f92355fd81bf.png

答案:对/star3/origin/722e47e2e1e992af483e7831777c3463.png

答案:错三元方程空间直角坐标系一定表示为一个平面.

答案:错/star3/origin/a999fca419a27fc0b693647a5db1916e.png

答案:-2/star3/origin/136a4e456129d25edcf3edd3b4ea883e.png

答案:平面上的圆/star3/origin/e71f79a0af914f791de2d4fb6861e43d.png

答案:不过原点但平行于轴/star3/origin/59336498d0f0d29954b1b1ebca2ea5bc.png

答案:过原点且垂直于轴/star3/origin/14dfd193649a5d6468a2f2a04d077343.png

答案:直线在平面上/star3/origin/26a532714a6f007d33c11df986ebf266.png

答案:/star3/origin/acaf9f2feedaac3ce187829f482bf384.png

答案:平行/star3/origin/19d1cb712bf56700cc1e5ae266e0fe6f.png

答案:圆柱面/star3/origin/506838992e545f37463463c78e59fa53.png

答案:错/star3/origin/862c72b96d1e7b46506c3b54817f0d4f.png

答案:对/star3/origin/65db26073a44a9093dab755e064e18f4.png

答案:对/star3/origin/9e4512cf3035c26cccc0f897069f8004.png

答案:对/star3/origin/b3eb9c9d9a04baa00a95194d2c25cff2.png

答案:错/star3/origin/27c554f77e53405694dc3a5543c38872.png

答案:对/star3/origin/11532bbbe695babdbe313e5149b79062.png

答案:必要非充分条件/star3/origin/b34fc998a0120b48e7d38a2907078034.png

答案:解方程组,得驻点在处,因=且故取得极大值/star3/origin/454366bb05743598b9dd3a7cc3fd2d81.png

答案:令/star3/origin/db9261977aaa68529f1da20cbef5d2f4.png

答案:对多元函数得极值一定是在它得驻点处取得.

答案:错/star3/origin/f6f430774c28cbd8c4078df1273e1035.png

答案:对/star3/origin/98a98d8e8e20e17d24c73ec48088c9bf.png

答案:极小值为/star3/origin/fb6b14a789fbe80fc8b0cff0475a9b9c.png

答案:;/star3/origin/df49a2cbfeca186b4043e4bda4110164.png

答案:;/star3/origin/d3d221b2d72e6489c09ebaaad87a8122.png

答案:充分条件/star3/origin/4415b3e0eb61b158abed2ec2737d174d.png

答案:令/star3/origin/6eb89fc4b64be8d7475c34bd6a36883c.png

答案:8/star3/origin/d9fc73ec12273b9361034f7abcb84f26.png

答案:因为,而发散,故原级数不绝对收敛;又因为数列单调递减,且,故由莱布尼茨判别法,级数收敛,综上,级数收敛,但不绝对收敛./star3/origin/73c4c54e8387a27a0252c013f026b3a2.png

答案:因为,而,由比值审敛法知,收敛;再由比较审敛法知,收敛/star3/origin/e3426131ff9ca22b251db1408fc6582e.png

答案:因为故原级数发散/star3/origin/c233bc8c5392ee21cca0642ecd605bf7.png

答案:因为,故原级数收敛/star3/origin/7a3364ba65b9ebf23a1710cc0663f08b.png

答案:取,由于又收敛,故由比较审敛法,级数收敛./star3/origin/3452c3197f314effb4a71186ac41d604.png

答案:错/star3/origin/22a3ccee22853efcebf1dba234e1440a.png

答案:错/star3/origin/8f43c4eb1a64707460d8e9b075ba3c2c.png

答案:对/star3/origin/d264db88c0ba3d8db5cf6b46c638292a.png

答案:错/star3/origin/21694b5adda2de2b2ea7396a5899f66d.png

答案:对/star3/origin/cf199312daecf5f6518a96e6bb57e09a.png

答案:当时发散,当时收敛/star3/origin/803458606de4d70fbb045e027a8d1205.png

答案:可能收敛,也可能发散/star3/origin/1873e6cf90ef504a8a2386934cd6463c.png

答案:幂级数的收敛域为,其和函数为.所以/star3/origin/fc5c4680e593ad9cdd3560e8947543af.png

答案:,,而级数发散,故级数发散.,且,由莱布尼茨定理得原级数收敛.故原级数条件收敛./star3/origin/dd7365d7b4f5c27820a27550b06c3992.png

答案:错/star3/origin/fe01ce73c51f608e2aa145f659a2d651.png

答案:对/star3/origin/bfb4907e2f20da8390f0b84e572415f6.png

答案:对/star3/origin/6799c56d537d45cc71d565cd32fe9ae9.png

答案:条件收敛/star3/origin/67e4d1a5dad4b1e1517df348e4c4041c.png

答案:绝对收敛/star3/origin/e54417eaf64ed1b835c7ae45f45c6b7d.png

答案:不能确定/star3/origin/757f937ad6ef5f7fc089c3a7ffa862c0.png

答案:左端点发散,右端点收敛/star3/origin/9a15737e0518427d88b0fdc238d93229.png

答案:将变量分离,得到,(2分)两边积分得为方程的通解,且是方程的两个解.(6分)/star3/origin/5f27f11f1190748401cf886d80dfddee.png

答案:令,(2分)于是,解出,即通解.(6分)/star3/origin/c496364d621b86a8646e2b2e1969f4ad.png

答案:即(2分)常数变易法,令则,代入,即,;故原方程通解为;由初始条件,得;于是,原方程的特解为(6分)/star3/origin/1250c4bd1a52041da82820a3a6e6e1dd.png

答案:(5分)(3分)/star3/origin/09e6bcd596246c7235da297f263e8f92.png

答案:错/star3/origin/1f1632c423bf301b5d307adae0bbddce.png

答案:错/star3/origin/50adbe33b892f189785082246b6aa749.png

答案:对/star3/origin/29780bb05273e2bc90b4426d702d9370.png

答案:对/star3/origin/cd409e177379c64c17226aaee01e9f8f.png

答案:错/star3/origin/47337837b12faf62b7ce99b33741bc0a.png

答案