高等數(shù)學(xué)I（下）學(xué)習(xí)通超星期末考試答案章節(jié)答案2024年

高等數(shù)學(xué)I（下）學(xué)習(xí)通超星期末考試章節(jié)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?qū)Q成,將換成)(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:對(duì)/star3/origin/fff5df2f594148a00d13e0d3443dc69a.png

答案:對(duì)/star3/origin/5361661d700e10066de52befce356150.png

答案:錯(cuò)/star3/origin/7a03beba39bab8c56276fc6edbebe7a2.png

答案:錯(cuò)/star3/origin/b855df12e7f440e8596717c95c26e3dd.png

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

答案:令則原式=(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:對(duì)/star3/origin/8bff64ae9dfaa80f03289c1e61a59dfd.png

答案:對(duì)/star3/origin/b774433db495314d0a0fb1861f4e5cda.png

答案:錯(cuò)/star3/origin/dbcda81d2d6a8873ce34cd1548dcb118.png

答案:對(duì)/star3/origin/c81d89b831119de2486e55e84cdf39a2.png

答案:錯(cuò)/star3/origin/63c3b2348e6a8b59ef7ea8ca2bab614e.png

答案:錯(cuò)/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:錯(cuò)/star3/origin/a95b7427e59d430dc2076e55d24a5f74.png

答案:對(duì)/star3/origin/c8f3bc735e16e862abc74fe21c1a4298.png

答案:對(duì)/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:對(duì)/star3/origin/10d4a2e8b908871b9086ab920c8219c1.png

答案:錯(cuò)/star3/origin/6c24b940e78ca59b8d534151713d1a70.png

答案:錯(cuò)/star3/origin/d1e9e44acb9ae642c94bb656202f587d.png

答案:對(duì)/star3/origin/79df823da77dd9b1cc77e97a4cb2a3ad.png

答案:錯(cuò)/star3/origin/a5d51db60be7347428ca611b4eb6faf7.png

答案:對(duì)/star3/origin/8c669ebe7142a2c1890c962e037df22f.png

答案:錯(cuò)/star3/origin/96b900524307e9466ac2cb9d6ab9e1c8.png

答案:對(duì)/star3/origin/f63301931164081fcd0e5b3d0893ea35.png

答案:對(duì)/star3/origin/ffa96851d454f7e322ea9fa9dd99a1f1.png

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

答案:由及導(dǎo)數(shù)的幾何意義得,在點(diǎn)處的切線方程為,在點(diǎn)處的切線方程為.(4分)由解得交點(diǎn)坐標(biāo)為.(2分)所以,所求面積為.(4分)/star3/origin/de99ac985fd159841ed6bb77f361f40f.png

答案:錯(cuò)/star3/origin/73ebd63f95543d0e4a0116291395c27f.png

答案:錯(cuò)/star3/origin/3b011b869c44469de816eb58675192e2.png

答案:錯(cuò)/star3/origin/571c65e2e4b0578413da8b14.png

答案:對(duì)/star3/origin/018779260577c4891356abae15685ea4.png

答案:對(duì)/star3/origin/da23528c32ef589aaa158df4503419fc.png

答案:發(fā)散/star3/750_1024/32e08a34d2c16e210380f70e13609af8.png

答案:收斂,且值為(ln2)/5/star3/origin/c010a3ed91c7ec0e5eb483ae8d758921.png

答案:收斂,且值為/star3/origin/61cfa98bed178d004a1660fed8cd746b.png

答案:收斂,且值為/star3/origin/344425933636bb57cffa2170a24fc3da.png

答案:由條件可設(shè)所求的點(diǎn)為,由,即(5分)因此所求點(diǎn)的坐標(biāo)為.(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、,則向量平行于.因此設(shè)向量,(5分)由題設(shè)得,得到,所以(3分)/star3/origin/7aa0a07be63003d27e2c2f860f559c24.png

答案:錯(cuò)/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:平面上任一點(diǎn)滿足平面方程為/star3/origin/05a0addffadfef9b3ff7f6969d38e2e3.png

答案:直線方向向量取取特殊點(diǎn)則對(duì)稱式方程:,參數(shù)方程:/star3/origin/c17c27ac2b5d32719cb788700f6390ac.png

答案:直線方向向量取直線方程為/star3/origin/f58170cb5f045ff2d1056ecde42a786a.png

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

答案:設(shè)方程為將點(diǎn)帶入方程得故平面方程為過點(diǎn)空間上任意三個(gè)不同的點(diǎn)都可以確定唯一一個(gè)平面.

答案:錯(cuò)過已知點(diǎn)且與已知直線垂直的平面不止一個(gè).

答案:錯(cuò)/star3/origin/73e354251d5c40f858e1d60021af6abe.png

答案:對(duì)/star3/origin/4564d6bbf2ce1211ed58f92355fd81bf.png

答案:對(duì)/star3/origin/722e47e2e1e992af483e7831777c3463.png

答案:錯(cuò)三元方程空間直角坐標(biāo)系一定表示為一個(gè)平面.

答案:錯(cuò)/star3/origin/a999fca419a27fc0b693647a5db1916e.png

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

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

答案:不過原點(diǎn)但平行于軸/star3/origin/59336498d0f0d29954b1b1ebca2ea5bc.png

答案:過原點(diǎn)且垂直于軸/star3/origin/14dfd193649a5d6468a2f2a04d077343.png

答案:直線在平面上/star3/origin/26a532714a6f007d33c11df986ebf266.png

答案:/star3/origin/acaf9f2feedaac3ce187829f482bf384.png

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

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

答案:錯(cuò)/star3/origin/862c72b96d1e7b46506c3b54817f0d4f.png

答案:對(duì)/star3/origin/65db26073a44a9093dab755e064e18f4.png

答案:對(duì)/star3/origin/9e4512cf3035c26cccc0f897069f8004.png

答案:對(duì)/star3/origin/b3eb9c9d9a04baa00a95194d2c25cff2.png

答案:錯(cuò)/star3/origin/27c554f77e53405694dc3a5543c38872.png

答案:對(duì)/star3/origin/11532bbbe695babdbe313e5149b79062.png

答案:必要非充分條件/star3/origin/b34fc998a0120b48e7d38a2907078034.png

答案:解方程組,得駐點(diǎn)在處,因=且故取得極大值/star3/origin/454366bb05743598b9dd3a7cc3fd2d81.png

答案:令/star3/origin/db9261977aaa68529f1da20cbef5d2f4.png

答案:對(duì)多元函數(shù)得極值一定是在它得駐點(diǎn)處取得.

答案:錯(cuò)/star3/origin/f6f430774c28cbd8c4078df1273e1035.png

答案:對(duì)/star3/origin/98a98d8e8e20e17d24c73ec48088c9bf.png

答案:極小值為/star3/origin/fb6b14a789fbe80fc8b0cff0475a9b9c.png

答案:;/star3/origin/df49a2cbfeca186b4043e4bda4110164.png

答案:;/star3/origin/d3d221b2d72e6489c09ebaaad87a8122.png

答案:充分條件/star3/origin/4415b3e0eb61b158abed2ec2737d174d.png

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

答案:8/star3/origin/d9fc73ec12273b9361034f7abcb84f26.png

答案:因?yàn)?而發(fā)散,故原級(jí)數(shù)不絕對(duì)收斂;又因?yàn)閿?shù)列單調(diào)遞減,且,故由萊布尼茨判別法,級(jí)數(shù)收斂,綜上,級(jí)數(shù)收斂,但不絕對(duì)收斂./star3/origin/73c4c54e8387a27a0252c013f026b3a2.png

答案:因?yàn)?而,由比值審斂法知,收斂;再由比較審斂法知,收斂/star3/origin/e3426131ff9ca22b251db1408fc6582e.png

答案:因?yàn)楣试?jí)數(shù)發(fā)散/star3/origin/c233bc8c5392ee21cca0642ecd605bf7.png

答案:因?yàn)?故原級(jí)數(shù)收斂/star3/origin/7a3364ba65b9ebf23a1710cc0663f08b.png

答案:取,由于又收斂,故由比較審斂法,級(jí)數(shù)收斂./star3/origin/3452c3197f314effb4a71186ac41d604.png

答案:錯(cuò)/star3/origin/22a3ccee22853efcebf1dba234e1440a.png

答案:錯(cuò)/star3/origin/8f43c4eb1a64707460d8e9b075ba3c2c.png

答案:對(duì)/star3/origin/d264db88c0ba3d8db5cf6b46c638292a.png

答案:錯(cuò)/star3/origin/21694b5adda2de2b2ea7396a5899f66d.png

答案:對(duì)/star3/origin/cf199312daecf5f6518a96e6bb57e09a.png

答案:當(dāng)時(shí)發(fā)散,當(dāng)時(shí)收斂/star3/origin/803458606de4d70fbb045e027a8d1205.png

答案:可能收斂,也可能發(fā)散/star3/origin/1873e6cf90ef504a8a2386934cd6463c.png

答案:冪級(jí)數(shù)的收斂域?yàn)?其和函數(shù)為.所以/star3/origin/fc5c4680e593ad9cdd3560e8947543af.png

答案:,,而級(jí)數(shù)發(fā)散,故級(jí)數(shù)發(fā)散.,且,由萊布尼茨定理得原級(jí)數(shù)收斂.故原級(jí)數(shù)條件收斂./star3/origin/dd7365d7b4f5c27820a27550b06c3992.png

答案:錯(cuò)/star3/origin/fe01ce73c51f608e2aa145f659a2d651.png

答案:對(duì)/star3/origin/bfb4907e2f20da8390f0b84e572415f6.png

答案:對(duì)/star3/origin/6799c56d537d45cc71d565cd32fe9ae9.png

答案:條件收斂/star3/origin/67e4d1a5dad4b1e1517df348e4c4041c.png

答案:絕對(duì)收斂/star3/origin/e54417eaf64ed1b835c7ae45f45c6b7d.png

答案:不能確定/star3/origin/757f937ad6ef5f7fc089c3a7ffa862c0.png

答案:左端點(diǎn)發(fā)散,右端點(diǎn)收斂/star3/origin/9a15737e0518427d88b0fdc238d93229.png

答案:將變量分離,得到,(2分)兩邊積分得為方程的通解,且是方程的兩個(gè)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)常數(shù)變易法,令則,代入,即,;故原方程通解為;由初始條件,得;于是,原方程的特解為(6分)/star3/origin/1250c4bd1a52041da82820a3a6e6e1dd.png

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

答案:錯(cuò)/star3/origin/1f1632c423bf301b5d307adae0bbddce.png

答案:錯(cuò)/star3/origin/50adbe33b892f189785082246b6aa749.png

答案:對(duì)/star3/origin/29780bb05273e2bc90b4426d702d9370.png

答案:對(duì)/star3/origin/cd409e177379c64c17226aaee01e9f8f.png

答案:錯(cuò)/star3/origin/47337837b12faf62b7ce99b33741bc0a.png

答案