发布时间 : 星期二 文章2014-2015年考研数学二真题及答案解析更新完毕开始阅读
?z?z?2yze?2y??1??0??x?x? ??z?z?e2yz(2z?2y)?2y??0?y?y??当x?11,y?时,z?0 2211(,)22故
?z?x1?z??,2?y11(,)221??
2故dz11(,)22111??dx?(?)dy??(dx?dy)
222(12) 曲线limnSn的极坐标方程是r??，则L在点(r,?)?(n????,)处的切线的直角坐标方程是
22__________. 【答案】y??2?x??2
?x?rcos???cos?【解析】由直角坐标和极坐标的关系 ?，
y?rsin???sin??于是?r,???????????,?,对应于?x,y???0,?, ?22??2?2dydydyd??cos??sin?切线斜率 ???dxdxdxcos???sin?d??2所以切线方程为y????x?0?
2?2?即y=?x?
?2????0,??2????
(13) 一根长为1的细棒位于x轴的区间[0,1]上,若其线密度??x???x?2x?1,则该细棒的质
2心坐标x?__________. 【答案】
11 2010x??x?dx?【解析】质心横坐标x? ???x?dx10 9
?x3?152?xdx=?x?2x?1dx???x?x????3?0??0?0?3??
4211?x23x?1112x?xdx=x?x?2x?1dx????4?3x?2?0?12?0???0???1111?x?12=
52031122(13) 设二次型f?x1,x2,x3??x12?x2?2ax1x3?4x2x3的负惯性指数是1，则a的取值范围
_________. 【答案】??2,2?
222【解析】配方法：f?x1,x2,x3???x1?ax3??ax3??x2?2x3??4x3
22由于二次型负惯性指数为1，所以4?a?0，故?2?a?2.
三、解答题：15～23小题,共94分.请将解答写在答题纸指定位置上.解答应写出文字说明、证．．．明过程或演算步骤. (15)(本题满分10分)
2?求极限limx???x1?2?1??t?t?e?1??t?dt??????.
?1?x2ln?1???x?x1?21???dt2dttt??t(e?1)?tt(e?1)?t???1?1????lim?【解析】lim
x???x???11x2ln(1?)x2?xxx?lim[x(e?1)?x]
x???21x1?txet?1?tet?1t1?lim?lim?lim?. 2??t?0?t?0t?0t2t2t222(16)(本题满分10分)
已知函数y?y?x?满足微分方程x?yy??1?y?，且y?2??0，求y?x?的极大值与极小 值.
【解析】 由x?yy??1?y?，得
(y?1)y??1?x?????????????????????①
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2222 此时上面方程为变量可分离方程，解的通解为
131y?y?x?x3?c 332 由y(2)?0得c?
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1?x2 又由①可得 y?(x)?2
y?1 当y?(x)?0时，x??1，且有：
x??1,y?(x)?0?1?x?1,y?(x)?0 x?1,y?(x)?0所以y(x)在x??1处取得极小值，在x?1处取得极大值
y(?1)?0,y(1)?1
即：y(x)的极大值为1，极小值为0. (17)(本题满分10分)
设平面区域D???x,y?1?x2?y?4,x?0,y?0,计算??D2?xsin?x2?y2x?y??dxdy.
【解析】D关于y?x对称，满足轮换对称性，则：
xsin(?x2?y2)ysin(?x2?y2)dxdy???dxdy ??x?yx?yDDxsin(?x2?y2)1?xsin(?x2?y2)ysin(?x2?y2)??I???dxdy??????dxdy
x?y2D?x?yx?y?D???122sin(?x?y)dxdy ??2D21?2??d??sin?r?rdr120
2?1?(?)?rdcos?r4?121?2??cos?r?r|1??cos?rdr?
?1??4?1?12? ???2?1?sin?r|1?4???
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3??
4(18)(本题满分10分)
?2z?2z设函数f(u)具有二阶连续导数，z?f(ecosy)满足2?2?(4z?excosy)e2x，若
?x?yx，求f(u)的表达式. f(0)?0,f'(0)?0【解析】由z?fecosy,?x??z?z?f?(excosy)?excosy,?f?(excosy)???exsiny? ?x?y?2zxxxxx????f(ecosy)?ecosy?ecosy?f(ecosy)?ecosy， 2?x?2zxxxxx????f(ecosy)??esiny??esiny?f(ecosy)??ecosy? ?????2?y?2z?2zx2x由 ，代入得， +?4z?ecosye??22?x?yf???excosy??e2x?[4f?excosy??excosy]e2x
即
f???excosy??4f?excosy??excosy，
令excosy=t,得f???t??4f?t??t 特征方程
*?2?4?0,???2 得齐次方程通解y?c1e2t?c2e?2t
11,b?0，特解y*??t 4412t?2t则原方程通解为y=f?t??c1e?c2e?t
411',c2??, 则 由f?0??0,f?0??0，得c1?1616111y=f?u??e2u?e?2u?u.
16164设特解y?at?b，代入方程得a??(19)(本题满分10分)
设函数f(x),g(x)在区间[a,b]上连续，且f(x)单调增加，0?g(x)?1，证明：（I）
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