高等數(shù)學(xué)A(2)學(xué)習通超星期末考試答案章節(jié)答案2024年

高等數(shù)學(xué)A(2)學(xué)習通超星期末考試章節(jié)答案2024年/star3/origin/0b1a4810d01ff302e95e33b51f409aba.png

答案:錯/star3/origin/70fd6604d08bf9f0f10a91c974cdef06.png

答案:對/star3/origin/36835544535ad9f0b86f21e9246e22d6.png

答案:對/star3/origin/bb52d9f68ddfda546a53e7791e1d500b.png

答案:對/star3/origin/cf34ddea32dc3a3c111bc169b2c6d456.png

答案:對/star3/origin/dbdd4d61f5bb68f12a0c3601300235ae.png

答案:x+y=1/star3/origin/3faa62ba19211bdc86f18790f77cf63a.png

答案:函數(shù)及區(qū)域有關(guān)/star3/origin/7388fbb59ddcaaccb8aa3d547efb2199.png

答案:可確定倆個具有連續(xù)偏導(dǎo)數(shù)的隱函數(shù)和/star3/origin/0466ed3ddaab4dd948dd3c92ba8277e6.png

答案:既非充分條件也非必要條件/star3/origin/8214b0f695aa62f6e206c5c57f6fd3be.png

答案:取得極小值/star3/origin/1452f7d5dc8afc4eda148dbafa4deb0a.png

答案:無極值,而有極小值/star3/origin/ccf9a3953c96f5779cbab69b70e4d56a.png

答案:駐點但非極值點/star3/origin/a178444a6316cce7b3472dc6d4204c9c.png

答案:點是極大值點/star3/origin/00a0040467b32686774e93bc2fd74fbb.png

答案:11/15/star3/origin/ee8a2605af56cf34ff84ceb295a95adc.png

答案:設(shè)過該直線的平面束方程為即投影平面法向量為而已知平面法向量為根據(jù)題意可知,從而解得從而所求投影平面方程為./star3/origin/7f1ed180a40800d54a04799233d9fe07.png

答案:設(shè)過該兩平面交線的平面束方程為即因為平面過原點,所以解得從而所求平面方程為即./star3/origin/2f430302e0152f08cfffd2a97277ce36.png

答案:所求直線與已知直線平行,故取所求直線方程為./star3/origin/9bc2d4b94d52e7221e5fbf7faff99b75.png

答案:設(shè)過直線的平面束方程為,即所求平面的法向量,直線的方向向量由平面與直線平行知,得方程,解得,所求平面方程為/star3/origin/d9c4534e07958d02c261922223c0b202.png

答案:直線方向向量且過點,直線方向向量根據(jù)題意取從而所求平面方程為即./star3/origin/bcb57b851d6a090f7d36011f28528d9b.png

答案:所求直線平行于已知的兩平面,從而取所求直線方程為./star3/origin/cf34b6024e5fe3e9ba1dfb9b3c586c3d.png

答案:令,則,所以直線過點所求直線平行于已知的兩平面,從而取所求直線方程為,參數(shù)方程為./star3/origin/445db5e37d306798bd9a81691855b9f5.png

答案:所求平面垂直于已知的兩平面,從而取所求平面方程為即/star3/origin/b2da94164ffe7b5d868382a230c10568.png

答案:AB中點坐標即,取所求平面為即./star3/origin/1eb47bf2937caa76bedaaf91a0b94db7.png

答案:取所求平面為即/star3/origin/a1c312fe04b719190a218bf4e2723aa5.png

答案:重合/star3/origin/1c16ebefab2a2eb3862f8d0b04c545c8.png

答案:(1,-6,2)/star3/origin/80f0568ba868a7e34334c89740577f43.png

答案:L的方程可為/star3/origin/7bbf290eef77f5335a325005ddf22d3e.png

答案:(a,-b,-c)在平面xOy上求一點,使它到x=0,y=0及x+2y-16=0三直線距離平方之和為最小.

答案:設(shè)所求點為(x,y),則此點到x=0的距離為|y|,到y(tǒng)=0的距離為|x|,到x+2y-16=0的距離為,而距離平方之和為.解方程組,即.得唯一的駐點,根據(jù)問題的性質(zhì)可知,到三直線的距離平方之和最小的點一定存在,故即為所求./star3/origin/e3dcf099c33256672ba0c9e5905d18a9.png

答案:設(shè),,代入點,所以切平面方程為:,即法線方程為:。/star3/origin/336556b9ba1ce38ccb0a044cf7e68807.png

答案:設(shè)/star3/origin/038882700710a7a1fac5d94290f74b79.png

答案:,,由得由得/star3/origin/8c44e7bbb3c528266d009898f8a6ba9c.png

答案:;68/13/star3/origin/336f83c87636715c7b54ca2a1ebe3ff8.png

答案:極大值點且是最大值點/star3/origin/b0b25f55c3e080ace2a5cd1e61638595.png

答案:,,/star3/origin/88fe81d77eed4dbb32a14a0fee0ca904.png

答案:./star3/origin/525f5b1e9742a640edcc0972d6fe9bdc.png

答案:./star3/origin/cbb2c785565443cc4c7fb9e202d91eab.png

答案:./star3/origin/3543e0de921b8c2ebe4c40459282e7c9.png

答案:./star3/origin/29c0c25acdd9b696564b82dc058917db.png

答案:由高斯公式原式./star3/origin/ee138b3341b09090ae439b14ca0a8145.png

答案:由高斯公式/star3/origin/a8bd1a3f4a9582b9161c9e761c1dc3c3.png

答案:由高斯公式原式(這里用了對稱性)./star3/origin/f36bb093d74799ec97dd314c62f5c56b.png

答案:取下側(cè),且于是/star3/origin/0addd1cfbd4ed08b7d8c02657b6cd9ef.png

答案:,取上側(cè)所以/star3/origin/331112bfb62c7250e2e905e4e145a509.png

答案:由輪換對稱性可得/star3/origin/c319cd755c8f1a3c513d9a10730e1769.png

答案:.因此./star3/origin/3caa5481bcd91cb4306ce9cfd104133b.png

答案:在面內(nèi)成立,故為某函數(shù)的全微分../star3/origin/5d3be71634a9f0c23e46a459dbb7614e.png

答案:設(shè),則./star3/origin/be9b682b3439a79f436c90c94b5e8434.png

答案:記線段OA的方程為,圓弧AB的方程為.線段OB的方程為。./star3/origin/42ce6b942891523bc72fcd36ad3661e8.png

答案:；1/2；0.5/star3/origin/6c87252f7e4f9f4411530f05cc83af68.png

答案:0；0/star3/origin/bdecd4f8e7da51aa8c7dfd337df3bcd9.png

答案:向量穿過曲面的流量/star3/origin/02e8d4c240aa951149b41fc9cccae48f.png

答案:先將奇延拓成周期為的周期函數(shù),且。故/star3/origin/f30e29e5df93fdb3e332845b3f285056.png

答案:先將奇延拓為周期的周期函數(shù),且。故當時,/star3/origin/2746eab243a3b7b0993b1d66b442d06d.png

答案:先將偶延拓為周期的周期函數(shù),且。故/star3/origin/b962972322cfed9ebff5343fe90b6fb7.png

答案:當即時級數(shù)收斂,所以收斂區(qū)間為/star3/origin/bd452f87cf0a0ed89722a14bc1f6c670.png

答案:,收斂域為./star3/origin/90f35e0212caf5f587958d491d6c78ed.png

答案:.根據(jù)比值審斂法:當即時,級數(shù)收斂;當即時,級數(shù)發(fā)散.當或時,級數(shù)都是發(fā)散的.該級數(shù)的收斂半徑為,收斂域為./star3/origin/919742836eea1c7099eb0a801bbbc416.png

答案:因為.根據(jù)比值審斂法:當即時,級數(shù)收斂;當即時,級數(shù)發(fā)散.當或時,級數(shù)也是收斂的.該級數(shù)的收斂半徑為,收斂域為./star3/origin/1a0a278f5ef2e874b9eaf0fd435fd9e4.png

答案:;-1/2;-0.5/star3/origin/05243ffa7602f074a69d2bd812e18270.png

答案:;0/star3/origin/5adedb0d5a55350f59a87197ed493339.png

答案:;0/star3/origin/d756cb88a34c021ee2ab5741f82ef76b.png

答案:;-1/star3/