De1

  • June 2020
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www.khoabang.com.vn LuyÖn thi trªn m¹ng – Phiªn b¶n 1.0 ________________________________________________________________________________

C©u I. 1) Kh¶o s¸t sûå biÕn thiªn vµ vÏ ®å thÞ (C) cña hµm sè y =

x2 - x + 1 . x - 1

2) T×m trªn trôc Oy c¸c ®iÓm tõ ®ã cã thÓ kÎ ®ûîc Ýt nhÊt mét tiÕp tuyÕn ®Õn ®å thÞ (C). 3) X¸c ®Þnh a ®Ó ®å thÞ (C) tiÕp xóc víi parabol y = x2 + a. C©u II. Cho hÖ phû¬ng tr×nh

 x + y + xy = m  2 2 x + y = m

1) Gi¶i hÖ víi m = 5. 2) Víi gi¸ trÞ nµo cña m th× hÖ cã nghiÖm? C©u III. 1) Cho bÊt phû¬ng tr×nh x2 + 2x(cosy + siny) + 1 ≥ 0. T×m x ®Ó bÊt phû¬ng tr×nh ® îc nghiÖm ®óng víi mäi y. 2) Gi¶i phû¬ng tr×nh lûîng gi¸c sin 2 x(tgx + 1) = 3sinx(cosx - sinx) + 3 C©u IVa. Trong mÆt ph¼ng víi hÖ trôc täa ®é §Òc¸c vu«ng gãc, cho elip E) :

x2 y2 = 1, + 9 4

LuyÖn thi trªn m¹ng – Phiªn b¶n 1.0 ________________________________________________________________________________ www.khoabang.com.vn

vµ hai ®ûêng th¼ng (D) : ax - by = 0,

(D’) : bx + ay = 0,

víi a2 + b2 > 0. 1) X¸c ®Þnh c¸c giao ®iÓm M, N cña (D) víi (E), vµ c¸c giao ®iÓm P, Q cña (D’) víi (E). 2) TÝnh theo a, b diÖn tÝch tûá gi¸c MPNQ. 3) T×m ®iÒu kiÖn ®èi víi a, b, ®Ó diÖn tÝch Êy lín nhÊt. 4) T×m ®iÒu kiÖn ®èi víi a, b, ®Ó diÖn tÝch Êy nhá nhÊt. C©u IVb. Trong mÆt ph¼ng (P) cho tam gi¸c ABC víi c¶ ba gãc nhän. Trªn ®ûêng th¼ng (d) vu«ng gãc víi mÆt ph¼ng (P) t¹i A, lÊy mét ®iÓm M. Dûång BN⊥CM , BH⊥CM . §ûêng th¼ng KH c¾t (d) t¹i N. 1) Chûáng minh : BN⊥CM 2) Chûáng minh : BM⊥CN 3) H·y chØ c¸ch dûång ®iÓm M trªn (d) sao cho ®o¹n MN ng¾n nhÊt.

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