題目列表(包括答案和解析)
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My mother really knows how to get others to do what she suggests. Besides my father, her neighbors and even a stranger in a grocery store, has fallen under her 36 .
One day we were shopping in a supermarket. When we were passing by the baby aisle(過(guò)道), she 37 a young mother browsing through different brands of powdered milk. Her baby was playing with a toy. My mother 38 babies so she went up to the woman and chatted with her.
My mother is an easy person to like. She has the kindest eyes in the world, so it isn’t 39
to trust her. She 40 with the baby a little bit and told the woman about how I used to be cute at that age, too. They laughed together.
The woman was about to put Brand X Milk Powder in her cart when my mother suggested Brand Y 41 because it obviously had better vitamins. Would you believe the woman actually
42 brands just like that?
She really does know how to influence people. But what about her getting other people to
43 whatever she tells them?
My father and I actually had a long conversation about her 44 on how to influence people. According to my father, my mother is just a really likeable person, and she has 45 been
like that, even when he first met her. Chatting people up is natural for her.
During our conversation, I realized that what my mother has is the 46 to build rapport(親善) with other people. With rapport comes 47 . People are more likely to be influenced by those who they think understand their viewpoints or share their opinions.
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1.解析:,故選A。
2.解析:抽取回族學(xué)生人數(shù)是,故選B。
3.解析:由,得,此時(shí),所以,,故選C。
4.解析:∵∥,∴,∴,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(xiàn)的右焦點(diǎn)到一條漸近線(xiàn)的距離等于焦距的,∴,又∵,∴,∴雙曲線(xiàn)的漸近線(xiàn)方程是,故選D.
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)的最小正周期是,故選C。
10.解析:如圖,∵正三角形的邊長(zhǎng)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:將代入結(jié)果為,∴時(shí),表示直線(xiàn)右側(cè)區(qū)域,反之,若表示直線(xiàn)右側(cè)區(qū)域,則,∴是充分不必要條件。
14.解析:∵,∴時(shí),,又時(shí),滿(mǎn)足上式,因此,。
15.解析:設(shè)正四面體的棱長(zhǎng)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(xiàn)與軸的交點(diǎn),由向量的加法法則及拋物線(xiàn)的對(duì)稱(chēng)性可知,點(diǎn)為拋物線(xiàn)上關(guān)于軸對(duì)稱(chēng)的兩點(diǎn)且做出圖形如右圖,其中為點(diǎn)到準(zhǔn)線(xiàn)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(huì)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(píng)為三好學(xué)生的概率為!12分
19.(12分)解析:(Ⅰ)∵,∴,
,,……………3分
(Ⅱ)∵,∴,
∴,
又,∴數(shù)列自第2項(xiàng)起是公比為的等比數(shù)列,………………………6分
∴,………………………8分
(Ⅲ)∵,∴,………………10分
∴!12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(guò)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.解析:(Ⅰ)設(shè)拋物線(xiàn)方程為,將代入方程得
所以?huà)佄锞(xiàn)方程為。………………………2分
由題意知橢圓的焦點(diǎn)為、。
設(shè)橢圓的方程為,
∵過(guò)點(diǎn),∴,解得,,,
∴橢圓的方程為。………………………5分
(Ⅱ)設(shè)的中點(diǎn)為,的方程為:,
以為直徑的圓交于兩點(diǎn),中點(diǎn)為。
設(shè),則
∵
………………………8分
∴
………………………10分
當(dāng)時(shí),,,
此時(shí),直線(xiàn)的方程為!12分
22.(12分)解析:(Ⅰ)∵是偶函數(shù),∴,
又∵∴,,………………………2分
由得,,
∵時(shí),;時(shí),;時(shí),;∴時(shí),函數(shù)
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