專升本高等數(shù)學(xué)測試題答案

1、專升本高等數(shù)學(xué)測試題專升本高等數(shù)學(xué)測試題1.函數(shù)是（D）xysin1??（A）奇函數(shù)；（B）偶函數(shù)；（C）單調(diào)增加函數(shù)；（D）有界函數(shù)解析解析因?yàn)椋此院瘮?shù)為有界函數(shù)1sin1???x2sin10???xxysin1??2.若可導(dǎo)，且，則有（B）；)(uf)e(xfy?（A）；（B）；xfyxd)e(d?xfyxxde)e(d?（C）；（D）xfyxxde)e(d?xfyxxde)]e([d?解析解析可以看作由和復(fù)合而成的復(fù)合函數(shù))e

2、(xfy?)(ufy?xue?由復(fù)合函數(shù)求導(dǎo)法，??xxufufye)(e)(???????所以xfxyyxxde)e(dd????3.=(B)????0dexx(A)不收斂(B)1(C)－１(D)0.解析解析????0dexx?????0ex110???4.的特解形式可設(shè)為（A）；2(1)exyyyx???????(A)；(B)；2()exxaxb?()exxaxb?(C)；(D)()exaxb?2)(xbax?解析解析特征方程為，特

3、征根為==1=1是特征方程的特征重根，于是有0122???rr1r2r?2()expyxaxb??5.(C)，其中：≤≤；????yxyxDdd22D122yx?4(A)；(B)；2π4201ddrr???2π401ddrr???(C)；(D)2π2201ddrr???2π201ddrr???解析解析此題考察直角坐標(biāo)系下的二重積分轉(zhuǎn)化為極坐標(biāo)形式當(dāng)時，，由于≤≤，表示為，，故???????sincosryrxddddxyrr??122y

4、x?4D21??r02π???????yxyxDdd22ddDrrr?????2π2201ddrr???11.交錯級數(shù)的斂散性為)1(1)1(11??????nnnn（4）=??????11)1(1)1(nnnn?????1)1(1nnn而級數(shù)收斂，故原級數(shù)絕對收斂.????1)1(1nnn12..（第二個重要極限第二個重要極限）xxx)11(lim2???解一解一原式==，10])11[(lim)11(lim)11()11(lim??

5、???????????xxxxxxxxxxx1ee1??解二解二原式==)1()(2])11[(lim2xxxx?????1e0?13.)]1ln(11[lim20xxxx???解所求極限為型，不能直接用洛必達(dá)法則，通分后可變成或型.???00??)]1ln(11[lim20xxxx???xxxxxxx2111lim)1ln(lim020????????.21)1(21lim)1(211lim00?????????xxxxxx14.設(shè)，

6、求.xxxfe)(?)(xf解：解：令兩邊取對數(shù)得：xxye?xyxlneln?兩邊關(guān)于求導(dǎo)數(shù)得：xxxyyxxelne1????)elne(xxyyxx??即.)elne(exxxyxxx??15.求在閉區(qū)間上的極大值與極小值，最大值與最小值.3)(xxf?23x??55?解：解：令得xxxf63)(2???0)(??xf2021???xx66)(????xxf06)0(????f06)2(??????f∴的極大值為4，極小值為.)(