題目列表(包括答案和解析)
.To his great excitement, the day he looked forward to _______ at last!
A. had come B. coming C. came D. to come
.— Do you like a house with no garden?
—________ , But anyhow, it's better to have one than none.
A. Not really B. Not especially
C. Not a bit D. Not a little
.—When can we come to visit you?
—Anytime you feel like________.
A. one B. it C .so D. that
.______was known to them all that William had broken his promise______ he would give each of them a gift.
A. As, which B. What, that C. It , that D. It , which
.—Do you feel like ________ there or shall we take a bus?
—I’d like to walk. But since there isn't much time left, I'd rather we________ a taxi.
A. walking; hire B. to walk; hire C. to walk; hired D. walking; hired
1.C 2.A 3.B 4.D 5.C 6.B 7.D 8.C 9.B 10.A
11.120° 12.3x+y-1=0 13. 14.10 15.100 16.(1),(4)
17.解:(1)設(shè)拋物線,將(2,2)代入,得p=1. …………4分
∴y2=2x為所求的拋物線的方程.………………………………………………………5分
(2)聯(lián)立 消去y,得到. ………………………………7分
設(shè)AB的中點(diǎn)為,則.
∴ 點(diǎn)到準(zhǔn)線l的距離.…………………………………9分
而,…………………………11分
,故以AB為直徑的圓與準(zhǔn)線l相切.…………………… 12分
(注:本題第(2)也可用拋物線的定義法證明)
18.解:(1)在△ACF中,,即.………………………………5分
∴.又,∴.…………………… 7分
(2)
. ……………………………14分
(注:用坐標(biāo)法證明,同樣給分)
19.
解法一:(1)連OM,作OH⊥SM于H.
∵SM為斜高,∴M為BC的中點(diǎn),∴BC⊥OM.
∵BC⊥SM,∴BC⊥平面SMO.
又OH⊥SM,∴OH⊥平面SBC.……… 2分
由題意,得.
設(shè)SM=x,
則,解之,即.………………… 5分
(2)設(shè)面EBC∩SD=F,取AD中點(diǎn)N,連SN,設(shè)SN∩EF=Q.
∵AD∥BC,∴AD∥面BEFC.而面SAD∩面BEFC=EF,∴AD∥EF.
又AD⊥SN,AD⊥NM,AD⊥面SMN.
從而EF⊥面SMN,∴EF⊥QS,且EF⊥QM.
∴∠SQM為所求二面角的平面角,記為α.……… 7分
由平幾知識(shí),得.
∴,∴.
∴,即所求二面角為. ……………… 10分
(3)存在一點(diǎn)P,使得OP⊥平面EBC.取SD的中點(diǎn)F,連FC,可得梯形EFCB,
取AD的中點(diǎn)G,連SG,GM,得等腰三角形SGM,O為GM的中點(diǎn),
設(shè)SG∩EF=H,則H是EF的中點(diǎn).
連HM,則HM為平面EFCB與平面SGM的交線.
又∵BC⊥SO,BC⊥GM,∴平面EFCB⊥平面SGM. …………… 12分
在平面SGM中,過O作OQ⊥HM,由兩平面垂直的性質(zhì),可知OQ⊥平面EFCB.
而OQ平面SOM,在平面SOM中,延長OQ必與SM相交于一點(diǎn),
故存在一點(diǎn)P,使得OP⊥平面EBC. ……………………… 14分
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