題目列表(包括答案和解析)
A. | B. | ||
C. | D. | ||
E. | F. |
請閱讀以下的網(wǎng)站簡介,然后匹配簡介與網(wǎng)站圖片信息。
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從A,B,C,D四個(gè)選項(xiàng)中,找出其劃線部分與所給單詞的劃線部分讀音相同的選項(xiàng)。
polite A.possible B.climate C.province D.terrible
houses A.potatoes B.teaches C.knives D.cakes
piece A.believe B.friendly C.tie D.society
social A.cinema B.percent C.bicycle D.a(chǎn)ncient
repair A.culture B.inspire C.compare D.heart
從A,B,C,D四個(gè)選項(xiàng)中,選出其畫線部分與所給單詞的畫線部分讀音相同的選項(xiàng),并將答案寫在答題卡上。
bother A. faith B. method C northern D. threat
bow A. sow B. shower C. flow D. grown
prove A. nod B. plot C. remove D. roll
surface A .discovery B. condition C. special D.certain
speech A.church B.stomach C.technology D.machine
從A B C D中找出其劃線部分與所給單詞劃線部分讀音相同的項(xiàng)
satisfy A. branch B. engage C. graduate D. around
expedition A. exist B. Exercise C. expense D. explore
whole A. where B. which C. why D. who
chemical A. church B. child C. chips D. Christmas
south A. courage B. soup C. southern D. trousers
從A,B,C,D四個(gè)選項(xiàng)中,找出其劃線部分與所給單詞的劃線部分讀音相同的選項(xiàng),并在答題卡上將該項(xiàng)涂黑。
【小題1】idea
A.mean | B.pleasant | C.healthy | D.a(chǎn)rea |
A.shock | B.provide | C.remove | D.moment |
A.exercise | B.explore | C.examine | D.expression |
A.persuade | B.a(chǎn)dvise | C.nervous | D.increase |
A.a(chǎn)chieve | B.friend | C.science | D.believe |
1.解析:,故選A。
2.解析:∵
,
故選B。
3.解析:由,得,此時(shí),所以,,故選C。
4.解析:顯然,若與共線,則與共線;若與共線,則,即,得,∴與共線,∴與共線是與共線的充要條件,故選C。
5.解析:設(shè)公差為,由題意得,;,解得或,故選C。
6.解析:∵雙曲線的右焦點(diǎn)到一條漸近線的距離等于焦距的,∴,又∵,∴,∴,∴雙曲線的離心率是。故選B.
7.解析:∵、為正實(shí)數(shù),∴,∴;由均值不等式得恒成立,,故②不恒成立,又因?yàn)楹瘮?shù)在是增函數(shù),∴,故恒成立的不等式是①③④。故選C.
8.解析:∵,∴在區(qū)間上恒成立,即在區(qū)間上恒成立,∴,故選D。
9.解析:∵
,此函數(shù)的最小值為,故選C。
10.解析:如圖,∵正三角形的邊長為,∴,∴,又∵,∴,故選D。
11.解析:∵在區(qū)間上是增函數(shù)且,∴其反函數(shù)在區(qū)間上是增函數(shù),∴,故選A
12.解析:如圖,①當(dāng)或時(shí),圓面被分成2塊,涂色方法有20種;②當(dāng)或時(shí),圓面被分成3塊,涂色方法有60種;
③當(dāng)時(shí),圓面被分成4塊,涂色方法有120種,所以m的取值范圍是,故選A。
13.解析:做出表示的平面區(qū)域如圖,當(dāng)直線經(jīng)過點(diǎn)時(shí),取得最大值5。
14.解析:∵,∴時(shí),,又時(shí),滿足上式,因此,,
∴。
15.解析:設(shè)正四面體的棱長為,連,取的中點(diǎn),連,∵為的中點(diǎn),∴∥,∴或其補(bǔ)角為與所成角,∵,,∴,∴,又∵,∴,∴與所成角的余弦值為。
16.解析:∵,∴,∵點(diǎn)為的準(zhǔn)線與軸的交點(diǎn),由向量的加法法則及拋物線的對稱性可知,點(diǎn)為拋物線上關(guān)于軸對稱的兩點(diǎn)且做出圖形如右圖,其中為點(diǎn)到準(zhǔn)線的距離,四邊形為菱形,∴,∴,∴,∴,∴,∴向量與的夾角為。
17.(10分)解析:(Ⅰ)由正弦定理得,,,…2分
∴,,………4分
(Ⅱ)∵,,∴,∴,………………………6分
又∵,∴,∴,………………………8分
∴!10分
18.解析:(Ⅰ)∵,∴;……………………理3文4分
(Ⅱ)∵三科會考不合格的概率均為,∴學(xué)生甲不能拿到高中畢業(yè)證的概率;……………………理6文8分
(Ⅲ)∵每科得A,B的概率分別為,∴學(xué)生甲被評為三好學(xué)生的概率為!12分
(理)∵,,,!9分
∴的分布列如下表:
0
1
2
3
∴的數(shù)學(xué)期望!12分
19.(12分)解析:(Ⅰ)時(shí),
,,
由得, 或 ………3分
+
0
-
0
+
遞增
極大值
遞減
極小值
遞增
, ………………………6分
(Ⅱ)在定義域上是增函數(shù),
對恒成立,即
………………………9分
又(當(dāng)且僅當(dāng)時(shí),)
………………………4分
20.解析:(Ⅰ)∵∥,,∴,∵底面,∴,∴平面,∴,又∵平面,∴,∴平面,∴!4分
(Ⅱ)∵平面,∴,,∴為二面角的平面角,………………………6分
,,∴,又∵平面,,∴,∴二面角的正切值的大小為。………………………8分
(Ⅲ)過點(diǎn)做∥,交于點(diǎn),∵平面,∴為在平面內(nèi)的射影,∴為與平面所成的角,………………………10分
∵,∴,又∵∥,∴和與平面所成的角相等,∴與平面所成角的正切值為。………………………12分
解法2:如圖建立空間直角坐標(biāo)系,(Ⅰ)∵,,∴點(diǎn)的坐標(biāo)分別是,,,∴,,設(shè),∵平面,∴,∴,取,∴,∴!4分
(Ⅱ)設(shè)二面角的大小為,∵平面的法向量是,平面的法向量是,∴,∴,∴二面角的正切值的大小為!8分
(Ⅲ)設(shè)與平面所成角的大小為,∵平面的法向量是,,∴,∴,∴與平面所成角的正切值為。………………………12分
21.(Ⅰ) 解析:如圖,設(shè)右準(zhǔn)線與軸的交點(diǎn)為,過點(diǎn)
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