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# 98数学试题(上〕

?x x?011. f(x)?(x?|x|),g(x)??2 则f[g(x)]?_______________ 2?x x?0

?0?x x]?2 f[g(x)? x x?0?

2. 设y?xe则y2x(n)?_____________

y(n)?2nxe2x?2n?1ne2x

3. 设x?0时,cosx?cos2x与kx是等价无穷小, 则k?________ 2

3k? 2

4. 曲线x?y?xy?7?0在点(1,2)处的切线斜率是_______________ ?

22331 1125. 设函数f(u)可微,且y?f(sin(x))则dy?____. dy?2xf'(sin(x))cos(x)dx

6.

3

?_________

C tsinu?du?x??17. 设?则y'?______,u??y?sint?tcosty"?_______

y'?t,

ex?b1. 试确定常数a,b的值,使 f(x)?有无穷间断点 x?0, 有可去间断点(x?a)(x?1)

x?1

ex?b解 欲使 x?0为间断点,则 a?0, 若x?1是可去间断点, 则 lim存在, x?1(x?a)(x?1)

ex?b(x?1)?0, 于是 b?e. 因此 a?0,b?e 于是 lime?b?limx?1x?1x(x?1)x

2. 求极限lim(x?011?)cotx sinxx

x?sinx1?lim? 3x?0x6

2??ex x?13. 确定a,b 的值使f(x)在x?1处可导 f(x)?? ??ax?b x?1

f??(1)?limf(x)?f(1)ax?b?e?limx?1?0x?1?0x?1x?1

ax?b?(a?b)a(x?1)?lim?lim?a, x?1?0x?1?0x?1x?1

2

2f(x)?f(1)ex?ef??(1)?lim?lim?lim2xex?2e x?1?0x?1?0x?1x?1?0x?1

4.

2??cos2t?6??sintdt??6(sint?sin2t)dt?1 01?sint0125.

?dx ln(x?dx?xln(x?

?

?xln(xC

6.

.

?e1??e1??1

?10?lim?1????0?0?limarcsin(1??)???0??2

?2x解 y'?,(1?x2)2

x1?2(3x2?1)y"?, (1?x2)3 x2?当

x?是下凸函数. , y"?0,于是 f(x)在区间

(??,当

?是下凹函数. ?x?时, y"?0,于是 f(x)在区间

(?3333

??)是下凸函数. 时, y"?0,于是 f(x)在区间

33

3) 34

2当

x?拐点为

(?四. (12分)抛物线y?3ax?2bx?c通过原点(0,0) 并且当0?x?1时y?0. 若它与直线

x?1,y?0围成的曲边三角形的面积等于1 ,试确定a,b,c 使此曲边三角形绕X轴旋转所得旋转体体积最小.

?1

03ax2?2bxdx?a?b?1

194Vx??(3ax2?2bx)2dx?a2?3ab?b2, 053

94Vx?a2?3a(1?a)?(1?a)2 53

188841Vx'?a?3?6a??a?a??0 533153

59于是 a??,b? 44

1sinx2dx?五. (7分) 证明 ??2 2x24

?tanx?sinx?xcosx?sinxcosx??(1?)?0,故 解 在区间[,)上，???2xxxx42?????

f(x)??22sinx?2在区间上递减，即最大值M?f()?最小值m?f()?, 4?x2?

??

22??sinx22??12sinx2(?)???(?), 即 ??? ?24?x?242?x2

44?

. F'(a)?0?F'(b), 因此只要证明 (a,b) 内至少存在一点? 使得 F'(?)?0

F(x)?F(a)?0,x?a(?a,?1 ,于是) F(x)?F(a),x?(a,a??1) x?a

F(x)?F(b)F'(b)?lim?0,于是存在b点的右邻域,使得x?a?x?b

F(x?)F(b)F(x)?F(b),x?(b??1,b) ?0,x??b?(2 ,于是b, )x?b