(3)y?(1?10(4)y?1?x52)是由y?u,u?1?v,v?10w,w??x5复合而成.
121是由y?u?1,u?1?v,v?arcsinw,w?2x复合而成.
1?arcsin2x16. 设f(x)定义在(-∞,+∞)上,证明:
(1) f(x)?f(?x)为偶函数; (2)f(x)?f(?x)为奇函数. 证: (1)设F(x)?f(x)?f(?x),则?x?(??,??), 有F(?x)?f(?x)?f(x)?F(x) 故f(x)?f(?x)为偶函数.
(2)设G(x)?f(x)?f(?x),则?x?(??,??),
有G(?x)?f(?x)?f(?x)??[f(x)?f(?x)]??G(x) 故f(x)?f(?x)为奇函数.
17. 某厂生产某种产品,年销售量为106件,每批生产需要准备费103元,而每件的年库存费为0.05元,如果销售是均匀的,求准备费与库存费之和的总费用与年销售批数之间的函数(销售均匀是指商品库存数为批量的一半). 解: 设年销售批数为x, 则准备费为103x;
106106106?0.05元. 又每批有产品件,库存数为件,库存费为2xx2x106?0.05设总费用为,则y?10x?.
2x318. 邮局规定国内的平信,每20g付邮资0.80元,不足20 g按20 g计算,信件重量不得超过2kg,试确定邮资y与重量x的关系. 解: 当x能被20整除,即[xxxx]??0.80?时,邮资y?; 20202025xx?x?]?时,由题意知邮资y??0.80. ?1??2020?20?当x不能被20整除时,即[?x?x?,0?x?2000且?????25?20?综上所述有y????x?1??0.80,0?x?2000且?x?????20???20????其中
x;20 x.20?x?,?x?分别表示不超过x,x?1的最大整数.
?1?2020?20????20??6
19. 证明:
(1)arcsinhx?ln(x?1?x2); (2)arctanhx?11?xln,?1?x?1 21?xex?e?x证: (1)由y?sinhx?得e2x?2yex?1?0
2x2解方程e2x?2yex?1?0得e?y?1?y, x22x因为e?0,所以e?y?1?y,x?ln(y?1?y)
所以y?sinhx的反函数是y?arcsinhx?ln(x?1?x2) (???x???).
ex?e?x1?y1?y11?y2xe?2x?ln,x?ln(2)由y?tanhx?x得,得;
e?e?x1?y21?y1?y又由
1?y?0得?1?y?1, 1?y所以函数y?tanhx的反函数为
y?arctanhx?11?xln (?1?x?1). 21?x20. 已知水渠的横断面为等腰梯形,斜角?=40°,如图所示.当过水断面ABCD的面积为定值S0时,求湿周L(L=AB+BC+CD)与水深h之间的函数关系式,并指明其定义域.
图1-1
解:
S0?从而 BC?11h(AD?BC)?h(2hcot??BC?BC)?h(BC?hcot?) 22S0?hcot?. hL?AB?BC?CD (AB?CD)?2Shh?BC?2?0?hcot? sin?sin?hS02?cos?S02?cos40???h??h?hsin?hsin40由h?0,BC?S0?hcot??0得定义域为(0,S0tan40?). h21. 写出下列数列的通项公式,并观察其变化趋势:
1234579(1) 0,,,,,?; (2) 1,0,?3,0,5,0,?7,0,?; (3) ?3,,?,,?.
3456357
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解: (1)xn?n?1,当n??时,xn?1. n?1n?1(2)xn?ncosπ,
2(3)xn?(?1)n2n?1,当n无限增大时,变化趁势有两种,分别趋于1,-1. 2n?1n??当n无限增大时,有三种变化趋势:趋向于??,趋向于0,趋向于??.
22. 对下列数列求a?limxn,并对给定的?确定正整数N(?),使对所有n?N(?),有
xn?a??:
(1)xn?1nπsin,??0.001; (2)xn?n?2?n,??0.0001. n211nπ1?1????,只须n?.取N???,则sin?nn2???解: (1)a?limxn?0,???0,要使xn?0?n??当n?N时,必有xn?0??. 当??0.001时,N??1??1000或大于1000的整数.
??0.001??n?2?n?221????
n?2?n2nn(2)a?limxn?0,???0,要使xn?0?n??只要n?取N?1?即n?1?2即可.
?1?,则当n?N时,有
xn?0??. 2??????1?88
或大于10的整数. ?102???0.0001?当??0.0001时, N?23. 根据数列极限的定义证明:
13n?13?0; (2)lim?;n??n2n??2n?12 n个22???n?a(3)lim?1; (4)lim0.99?9?1.n??n??n(1)lim证: (1)???0,要使
?1?111,只要.取n????N??0??,则当n>N时,恒有22n?n???11lim?0. .故???022n??nn 8
(2) ???0,要使
55553n?13?5?????,只要n?,取N???,则当???2n?122(2n?1)4nn???n>N时,恒有
3n?133n?13?. ???.故limn??2n?122n?12(3) ???0,要使
a2a2a2n2?a2?2??,只要n?,取?1?22n?n(n?a?n)n?a2?n???,则当n>N时,恒有???n2?a2n2?a2?1. ?1??,从而limn??nnn个??????1,故???0,不防设??1,要使(4)因为对于所有的正整数n,有
0.999?9?1n个1?ln??ln??????n??,只要n?,取N??,则当n?N时,恒有
??0.99?9?110ln10?ln10?n个????????,故lim0.99?9?1. 0.9?9?91n??n个24. 若limxn?a,证明limxn?a,并举反例说明反之不一定成立.
n??n??证: ?limxn?0,由极限的定义知,???0,?N?0,当n?N时,恒有xn?a??.
n??而 xn?a?xn?a??
????0,?N?0,当n?N时,恒有xn?a??,
由极限的定义知limxn?a.
n??但这个结论的逆不成立.如xn?(?1),limxn?1,但limxn不存在.
n??n??n25. 利用夹逼定理求下列数列的极限:
(1)lim[(n?1)k?nk],0?k?1;
n??nn(2)limna1n?a2???am,其中a1,a1,?,am为给定的正常数;
n??1nn(3)lim(1?2?3);n??n1(4)lim1?.n??nkk
解: (1)?0?(n?1)?n?n
k11k?1??k? ?n?(1?)?1(1?)?11?k????nn????n9
而lim0?0,当k?1时,limn??1?0
n??n1?k?lim[(n?1)k?nk]?0.
n??(2)记a?max{a1,a2,?,am} 则有
nan?nannan1?a2???m?nm?an 1即 a?an?ann12???am?mn?a
1而 limnn??a?a, limn??m?a?a,
故 limnannnn??1?a2???am?a
即 limnannnn??1?a2???am?max{a1,a2,?,am}.
111(3)?(3n)n?(1?2n?3n)n?(3?3n)n 1n?1即 3?(1?2n?3n)n?3n
n?1而 lim3nn???3, lim3n???3
1故 lim(1n???2n?3n)n?3.
(4)?1?1?1n?1?1n 而 nlim1???0, nlim(1???1n)?1
故 lim11n???n?1. 26. 利用单调有界准则证明下列数列有极限,并求其极限值:
(1)xxn1?2,xn?1?2xn,n?1,2,?; (2)x1?1,xn?1?1?1?x,n?1,2,?.n证: (1)?x1?2?2,不妨设xk?2,则
xk?1?2xk?2?2?2.
故对所有正整数n有xn?2,即数列?xn?有上界. 又xn?1?xn?2xn?xn?xn(2?xn)
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