英语辅导报材料答案

答 案

一、选择 C D B D 、 BD C C 二、填空 9）（-1，-4） 10）

π

π2

- 11）065 12）4π-，31

22n n π

---

.

三、解答题 13）6

13

1±=

x 14）5 15）（1）证明：∵∠A =∠A ,∠ABD =∠ACB , ……… 1分

∴△ABD ∽△ACB . ………………… 2分

（2）解: ∵△ABD ∽△ACB ，

∴AB AD AC

AB

=. ……………………………3分

∴757

AC

=. ………………………………4分

∴495

AC =. ……………………………5分

16）图略

π2

5 17）4.5米

18）①对称轴：直线1=x 顶点坐标（1，2）

A （-1，0）、

B （3，0） 图略

② 令23212++-x x m x +-= 一个交点0=??，解得m =2

7

A B C ?的面积为3

四、解答题（本题共20分，每小题5分）

19）4.101)1%)(201(502=++x )(3.23.0舍或-=x

20）解：①证明：∵∠BED =∠BAD ，∠C =∠BED ∴∠BAD =∠C

∵OC ⊥AD 于点F ∴∠BAD +∠AOC =90o ∴∠C +∠AOC =90o ∴∠OAC =90o ∴OA ⊥AC ∴AC 是⊙O 的切线.

②∵OC ⊥AD 于点F ，∴AF =2

1

AD =4 Rt △OAF 中，OF=22AF OA -=2

Rt △OA F ∽Rt △OCA

OA AF

AC OF

?=

=

A

D

B

21）解：①30；20． ②

1

2

．

共有16种可能的结果，且每种的可能性相同，其中小张获得车票的结果有6种： （2，1），（3，1），（3，2），（4，1），（4，2），（4，3），

∴小张获得车票的概率为63168P ==；则小李获得车票的概率为35

188

-=．

∴这个规则对小张、小李双方不公平．

22）解：①由题意：22125338124448b c b c ?=?++????=?++??

解得718

1

292b c ?=-????=??

②12y y y =-231

151362988

82x x x ??=-+--+ ???21316822x x =-++

③21316822y x x =-++2111(1236)46822x x =--+++21

(6)118

x =--+

∵1

08

a =-<， ∴抛物线开口向下．

在对称轴6x =左侧y 随x 的增大而增大．

由题意5x <，所以在4月份出售这种水产品每千克的利润最大．

最大利润211

(46)111082=--+=（元）．

1 2 3 4 1 1 2 3 4 2 1 2 3 4 3 1 2 3 4 4 开始 小张 小李

五、解答题（本题共22分，第23题7分，第24题8分，第25题7分） 23）解：（1）∵一元二次方程2220kx x k ++-=有实数根，

()2k 0,2420.k k ≠??∴?-??-≥??()2

k 0,410.

k ≠??∴?-≥?? ∴当0k ≠时，一元二次方程2220kx x k ++-=有实数根． （2）①由求根公式，得1(1)

k x k

-±-=

． 122

1k x k k

-∴=

=-，21x =-． 要使1x ，2x 均为整数，2

k

必为整数，

所以，当k 取12±±、时，1x ，2x 均为整数． ②将12

1x k

=-

，21x =-代入方程 1210x x k ++-=中，得2

1k k

=-．

设12y k =，21y k =-，并在同一平面直角坐标系中分别画出12

y k

=与21y k =-的

图象．

由图象可得，关于k 的方程1210x x k ++-=的解为11k =-，22k =．

24）解：解：（1）∵抛物线的顶点为Q （2，-1） ∴设()122

--=x a y

将C （0，3）代入上式，得 ()12032

--=a 解得1=a

∴()122

--=x y ， 即342+-=x x y

（2）分两种情况：

①当点P 1为直角顶点时,点P 1与点B 重合(如图) 令y =0, 得0342=+-x x 解之得11=x , 32=x

∵点A 在点B 的右边, ∴B(1,0), A(3,0)∴P 1(1,0) ②解:当点A 为△APD 2的直角顶点是(如图) ∵OA=OC, ∠AOC= 90, ∴∠OAD 2= 45

当∠D 2AP 2= 90时, ∠OAP 2= 45, ∴AO 平分∠D 2AP 2

又∵P 2D 2∥y 轴, ∴P 2D 2⊥AO, ∴P 2、D 2关于x 轴对称. 设直线AC 的函数关系式为b kx y += 将A(3,0), C(0,3)代入上式得

??

?=+=b

b

k 330, ∴???=-=31b k ∴3+-=x y

∵D 2在3+-=x y 上, P 2在342+-=x x y 上, ∴设D 2(x ,3+-x ), P 2(x ,342+-x x ) ∴(3+-x )+(342+-x x )=0

0652=+-x x , ∴21=x , 32=x (舍) ∴当x =2时, 342+-=x x y =32422+?-=-1 ∴P 2的坐标为P 2(2,-1)(即为抛物线顶点) ∴P 点坐标为P 1(1,0), P 2(2,-1)

(3)解: 由题(2)知,当点P 的坐标为P 1(1,0)时,不能构成平行四边形

当点P 的坐标为P 2(2,-1)(即顶点Q)时,

平移直线AP(如图)交x 轴于点E,交抛物线于点

F.

当AP=FE 时,四边形PAFE 是平行四边形 ∵P(2,-1), ∴可令F(x ,1) ∴1342=+-x x

解之得: 221-=x , 222+=x

∴F 点有两点,即F 1(22-,1), F 2(22+,1)

25）解：（1）当''A B 边经过点B 时，α= 60 °； （2）猜想：①如图1，点D 在AB 边上时，m =2； ②如图2，点D 在AB 的延长线上时，m =4.

（阅卷说明：为与后边证明不重复给分，猜想结论不设给分点）

证明：① 当090α?<

CD CE

CA CB =

''

. 由旋转性质可知，CA ='CA ，CB ='CB ，∠ACD=∠BCE . ∴

CD CE

CA CB

=

. ∴ △CAD ∽△CBE . ……………2分 ∴ ∠A =∠CBE=30°. ∵ 点D 在AB 边上，∠CBD=60°， ∴ 2CBD CBE ∠=∠，即 m =2. ② 当90120α?<

∵ 点D 在AB 的延长线上，180120CBD CBA ∠=?-∠=?，

∴ 4CBD CBE ∠=∠，即 m =4. ……………………………………4分 （阅卷说明：第（2）问用四点共圆方法证明的扣1分.）

（

3）解：在Rt △ABC 中，∠ACB=90°，∠A ∴ AB = 2 ，AC =ABC S

. 由 △CAD ∽△CBE 得

AD BE

AC BC

=

. ∵ AD =x ， ∴

1BE =，

3BE x =. ①当点D 在AB 边上时，

AD =x ，2BD AB AD x =-=-，∠DBE=90°. 此时，11(2)22BDE

S

S BD BE x

==?=-= .

当S =1

3

ABC S ?=. 整理，得 2210x x -+=.

解得 121x x ==，即AD =1.…………………5分

此时D 为AB 中点，∠DCB=60°，∠BCE=30°=∠CBE .∴ EC = EB .

∵ ''90A CB ∠=?，点E 在'CB 边上， ∴ 圆心E 到'A C 的距离EC 等于⊙E 的半径EB . ∴ 直线'A C 与⊙E 相切

. …………………………………………………6分 ②当点D 在AB

的延长线上时，AD =x ，2BD x =-，∠DBE=90°.（如图2）.

11(2)22BDE

S S BD BE x

==?=-= .

当S =1

3

ABC S ?

=. 整理，得 2210x x --=

. 解得 11x =，21x =. 即AD =.……………………………………………………………… 7分 此时∠BCE=α，而90120α?<

∴ ∠CBE ＜∠BCE .

∴ EC ＜EB ，即圆心E 到'A C 的距离EC 小于⊙E 的半径EB . ∴ 直线'A C 与⊙E 相交.

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i on? o n A 1 (1a-2d) 一、教学目标： 1. 语言知识目标： 1) 能掌握以下单词： e lf 能掌握以下句型： — i o — a ins. — — . — e? —Y ./N ’ . 2) 能了解以下语法： —复合不定代词some i ng等的用法。 — f等反身代词的用法。 3）一般过去时态的特殊疑问句，一般疑问句及其肯定、否定回答。 2. 情感态度价值观目标： 学会用一般过去时进行信息交流，培养学生的环保意识，热爱大自然。 二、教学重难点

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n ing . n. r e . a ins. r e? . time. . d time. . . r e. . r s. IV．Pair work . . . . N . r e. . .r oom and give any help they need. . o ns. n ing . ’ . .Yes, or No.

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