题目列表(包括答案和解析)
8.解:(1)∵反比例函数的图像经过点A(1,3),
∴,即m=-3.
∴反比例函数得表达式为. ……3分
∵一次函数y=kx+b的图像经过A(1,-3)、C(0,-4),
∴ 解得
∴一次函数的表达式为y=x-4 ……3分
(2)由消去y,得x2-4x+3=0.
即(x-1)(x-3)=0.
∴x=1或x=3.
可得y=-3或y=-1.
于是或
而点A的坐标是(1,-3),
∴点B的坐标为(3,-1)。 ……2分
7. 解:(1)设药物燃烧阶段函数解析式为,由题意得:
························································································································ 2分
.此阶段函数解析式为······································································· 3分
(2)设药物燃烧结束后的函数解析式为,由题意得:
·························································································································· 5分
.此阶段函数解析式为······································································ 6分
(3)当时,得···················································································· 7分
························································································································· 8分
·························································································································· 9分
从消毒开始经过50分钟后学生才可回教室.···························································· 10分
6. 解 (Ⅰ)∵点P(2,2)在反比例函数的图象上,
∴.即. ···································································································· 2分
∴反比例函数的解析式为.
∴当时,. ···························································································· 4分
(Ⅱ)∵当时,;当时,, ····················································· 6分
又反比例函数在时值随值的增大而减小, ············································· 7分
∴当时,的取值范围为.································································ 8分
5. (1)证明:分别过点C、D作
垂足为G、H,则
(2)①证明:连结MF,NE
设点M的坐标为,点N的坐标为,
∵点M,N在反比例函数的图象上,
∴,
由(1)中的结论可知:MN∥EF。
②MN∥EF。
4. 解:(1)由题意可知,.
解,得 m=3. ………………………………3分
∴ A(3,4),B(6,2);
∴ k=4×3=12. ……………………………4分
(2)存在两种情况,如图:
①当M点在x轴的正半轴上,N点在y轴的正半轴
上时,设M1点坐标为(x1,0),N1点坐标为(0,y1).
∵ 四边形AN1M1B为平行四边形,
∴ 线段N1M1可看作由线段AB向左平移3个单位,
再向下平移2个单位得到的(也可看作向下平移2个单位,再向左平移3个单位得到的).
由(1)知A点坐标为(3,4),B点坐标为(6,2),
∴ N1点坐标为(0,4-2),即N1(0,2); ………………………………5分
M1点坐标为(6-3,0),即M1(3,0). ………………………………6分
设直线M1N1的函数表达式为,把x=3,y=0代入,解得.
∴ 直线M1N1的函数表达式为. ……………………………………8分
②当M点在x轴的负半轴上,N点在y轴的负半轴上时,设M2点坐标为(x2,0),N2点坐标为(0,y2).
∵ AB∥N1M1,AB∥M2N2,AB=N1M1,AB=M2N2,
∴ N1M1∥M2N2,N1M1=M2N2.
∴ 线段M2N2与线段N1M1关于原点O成中心对称.
∴ M2点坐标为(-3,0),N2点坐标为(0,-2). ………………………9分
设直线M2N2的函数表达式为,把x=-3,y=0代入,解得,
∴ 直线M2N2的函数表达式为.
所以,直线MN的函数表达式为或. ………………11分
(3)选做题:(9,2),(4,5). ………………………………………………2分
3. 解:(1) ………(每个点坐标写对各得2分)………………………4分
(2) ∵ ∴…1分
∴ …………………1分
∴ …………………2分
(3) ① ∵
∴相应B点的坐标是 …………………………………………1分
∴. ………………………………………………………………1分
② 能 ……………………………………………………………………1分
当时,相应,点的坐标分别是,
经经验:它们都在的图象上
…………………………………………………………………1分
2. 解:(1)(-4,-2);(-m,-)
(2) ①由于双曲线是关于原点成中心对称的,所以OP=OQ,OA=OB,所以四边形APBQ一定是平行四边形
②可能是矩形,mn=k即可
不可能是正方形,因为Op不能与OA垂直。
解:(1)作BE⊥OA,
∴ΔAOB是等边三角形
∴BE=OB·sin60o=,
∴B(,2)
∵A(0,4),设AB的解析式为,所以,解得,的以直线AB的解析式为
(2)由旋转知,AP=AD, ∠PAD=60o,
∴ΔAPD是等边三角形,PD=PA=
1. 证明:(1)分别过点C,D,作CG⊥AB,DH⊥AB,
垂足为G,H,则∠CGA=∠DHB=90°.……1分
∴ CG∥DH.
∵ △ABC与△ABD的面积相等,
∴ CG=DH. …………………………2分
∴ 四边形CGHD为平行四边形.
∴ AB∥CD. ……………………………3分
(2)①证明:连结MF,NE. …………………4分
设点M的坐标为(x1,y1),点N的坐标为(x2,y2).
∵ 点M,N在反比例函数(k>0)的图象上,
∴ ,.
∵ ME⊥y轴,NF⊥x轴,
∴ OE=y1,OF=x2.
∴ S△EFM=, ………………5分
S△EFN=. ………………6分
∴S△EFM =S△EFN. ……
由(1)中的结论可知:MN∥EF. ………8分
② MN∥EF. …………………10分
(若学生使用其他方法,只要解法正确,皆给分.
24.(1,1);
17. 18. 19. 20. 21. 22. 51.2 23.
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