tuyển tập Đề dự bị môn toán thi Đh 2002 - 2008 và hd giảichi tiet
TRANSCRIPT
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
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D B THI I HC 2002 - 2008 RA V HNG DN GII
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
2
PHN TH NHT
D B THI I HC 2002 - 2008
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 1 Cu I: Cho hm s y= x4 - mx2 + m - 1 (1)(m l tham s) 1. Kho st s bin thin v v th hm s (1) khi m = 8. 2. Xc nh m sao cho th hm s (1) ct trc honh ti 4 im phn bit. Cu II: 1. Gii bt phng trnh x 2x + 1 x1 1
2 2
log (4 + 4) log (2 - 3.2 )
2. Xc nh m phng trnh 2(sin4x + cos4x) + cos4xx + 2sin2x + m = 0 c t
nht mt nghim thuc on 0; 2
.
Cu III: 1. Cho hnh chp S.ABC c y ABC l tam gic u cnh a v cnh bn SA vung gc vi mt phng y (ABC). Tnh khong cch t im A n mt phng
(SBC) theo a, bit rng a 6SA = 2
.
2. Tnh tch phn 1 3
20
xI = dxx + 1 .
Cu IV: 1. Trong mt phng Oxy cho hai ng trn: (C1): x2 + y2 - 10x = 0 (C2): x2 + y2+ 4x - 2y - 20 = 0 Vit phng trnh ng trn i qua cc giao im ca (C1) , (C2) v c tm nm trn ng thng x + 6y - 6 = 0. 3. Vit phng trrnh ng tip tuyn chung ca hai ng trn (C1) v (C2). Cu V: 1. Gii phng trnh 24 4 2 12 2 16x x x x . 2. i tuyn hc sinh gii ca mt trng gm 18 em, trong c 7 hc sinh khi 12, 6 hc sinh khi 11 v 5 hc sinh khi 10. Hi c bao nhiu cch c 8 hc sinh i d tri h sao cho mi khi c t nht mt em c chn. Cu VI: Gi x, y, z l khong cch t im M thuc min trong ca tam gic ABC c ba gc nhn n cc cnh BC, CA, AB. Chng minh rng:
2 2 2a + b + cx + y + z
2R ; a, b, c l cnh tam gic, R l bn knh
ng trn ngoi tip. Du ng thc xy ra khi no ?
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 2 Cu I: 1. Tm s nguyn dng tho mn bt phng trnh: 3 n-2n nA + 2C 9n, trong
k kn nA , C ln lt l s chnh hp v s t hp chp k ca n.
2. Gii phng trnh 84 221 1log (4x + 3) + log (x - 1) log (4 )2 4
x
Cu II:
Cho hm s 2x - 2x + my =
x - 2 (1)(m l tham s).
1. Xc nh m hm s (1) nghch bin trn on [- 1; 0]. 2. Kho st s bin thin v v th hm s (1) khi m = 1. 3. Tm a phng trnh sau c nghim: 2 21 + 1 - t 1 + 1 - t9 - (a + 2).3 + 2a + 1 = 0 Cu III:
1. Gii phng trnh 4 4sin x + cos x 1 1 = cotx - 5sin2x 2 8sin2x
2. Xt tam gic ABC c di cc cnh AB = c, BC = a, CA = b. Tnh din tch tam gic ABC, bit rng: bsinC(b.cosC + c.cosB) = 20. Cu IV: 1. Cho t din OABC c cc cnh OA, OB v OC i mt vung gc. Gi , , ln lt lcc gc gia mt phng (ABC) vi cc mt phng (OBC), (OCA) v (OAB), chng minh rng: cos + cos + cos 3 . 2.Trong khng gian Oxyz cho mf(P): x - y + z + 3 = 0 v hai im A(- 1; - 3; - 2), B( - 5; 7; 12). a) Tm to im A' i xng im A qua mf(P). b) Gi s M l mt im chy trn mf(P), tm gi tr nh nht ca MA + MB. Cu V:
Tnh ln3 x
x 30
I = .(e 1)e dx
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 3
Cu I: Cho hm s y = 13
x3 + mx2 -2x - 2m - 13
(1)(m l tham s)
1. Cho m = 12
: a) Kho st s bin thin v v th (C) hm s (1) .
b) Vit phng trnh tip tuyn ca th (C), bit rng tip tuyn song song vi ng thng y = 4x + 2.
2. Tm m thuc khong 50; 6
sao cho hnh phng gii hn bi th hm s (1)
v cc ng x = 0, x = 2, y = 0 c din tch bng 4. Cu II:
1. Gii h phng trnh 4 2
4 3 0
log log 0
x y
x y
2. Gii phng trnh 2
44
(2 - sin 2x)sin3xtan x + 1 = cos x
.
Cu III: 1. Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA vung gc vi mt phng (ABCD) v SA = a. Gi E l trung im ca cnh CD. Tnh theo a khong cch t im S n ng thng BE. 2. Trong khng gian Oxyz cho ng thng v mt phng (P).
2x + y + z + 1 = 0
: (P): 4x - 2y + z - 1 = 0x + y + z + 2 = 0
Vit phng trnh hnh chiu vung gc ca ng thng v mf(P). Cu IV:
1. Tm gii hn 3
x 0
x + 1 + x - 1L = limx
2. Trong mt phng Oxy cho hai ng trn: (C1): x2 + y2 - 4y - 5 = 0 (C2): x2 + y2 - 6x + 8y + 16 = 0 Vit phng trrnh ng tip tuyn chung ca hai ng trn (C1) v (C2). Cu V:
Cho x, y l hai s dng thay i tho mn iu kin x + y = 54
.
Tm gi tr nh nht ca biu thc 4 14
Sx y
.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 4 Cu I: 1. Gii bt phng trnh: x + 12 x - 3 + 2x + 1
2. Gii phng trnh tanx + cosx - cos2x = sinx(1 + tanx.tan x2
).
Cu II: Cho hm s y = (x - m)3 - 3x (m l tham s). 1. Xc nh m hm s cho t cc tiu tai im c honh x = 0. 2. Kho st s bin thin v v th hm s cho khi m = 1. 3. Tm k h bt phng trnh sau c nghim:
3
2 32 2
x - 1 - 3x - k < 01 1log x + log (x - 1) 12 3
Cu III: 1. Cho tam gic ABC vung cn c cnh huyn BC = a. Trn ng thng vung gc vi mt phng(ABC) ti A ly im S sao cho gc gia hai mt phng (ABC) v (SBC) bng 600. Tnh di SA theo a. 2. Trong khng gian Oxyz cho hai ng thng:
d1: 2x - az - a = 0 ax + 3y - 3 = 0
d :y - z + 1 = 0 x - 3z - 6 = 0
a) Tm a hai ng thng d1 v d2 ct nhau. b) Vi a = 2, vit phng trnh mt phng(P) cha d2 v song song d1 v tnh khong cch gia d1 v d2. Cu IV: 1. Gi s n l s nguyn dng v (1 + x)n = a0 + a1x + a2x2 + ...+akx2k +...+anxn.
Bit rng tn ti s nguyn k( 0 k n - 1 sao cho 1 12 9 24k k ka a a . Hy tnh n ?
2. Tnh tch phn 0
2x 3
- 1
I = x(e + x + 1)dx
Cu V: Gi A, B, C l ba gc ca tam gicABC. Chng minh rng tam gic ABC u th iu kn cn v l:
2 2 2A B C 1 A - B B - C C - Acos + cos + cos - 2 = cos cos cos 2 2 2 4 2 2 2
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 5
Cu I: Cho hm s y = 2x + mx1 - x
(1)(m l tham s)
1. Cho m = 12
. a) Kho st s bin thin v v th hm s (1) khi m = 0.
b) Vit phng trnh tip tuyn ca th (C), bit rng tip tuyn song song vi ng thng y = 4x + 2. 2. Tm m hm s (1) cc tr. Vi gi tr no ca m th khong cch gia hai im cc tr ca th hm s (1) bng 10. Cu II: 1. Gii phng trnh 3 232716 log 3log 0xx x x .
2. Cho phng trnh 2sinx + cosx+1sinx-2cosx+3
a (2)(a l tham s)
a) Gii phng trnh (2) khi a = 13
. b) Tm a phng trnh (2) c nghim.
b) Tm a phng trnh (2) c nghim. Cu III: 1. Trong mt phng Oxy cho ng thng d: x - y + 1 = 0 v ng trn (C): x2 + y2 + 2x - 4y = 0. Tm ta im M thuc ng thng d m qua k c hai ng thng tip xc vi ng trn (C) ti A v B sao cho gc AMB bng 600.
2. Trong khng gian Oxyz cho ng thng 2x - 2y - z + 1 = 0
d: x + 2y - 2z - 4 = 0
v mt
cu (S): x2 + y2 + z2 + 4x - 6y + m = 0. Tm m ng thng d ct mt cu ti hai im M, N sao cho MN = 9. 3. Tnh th tch ca khi t din ABCD, bit AB = a, AC = b, AD = c v cc gc BAC, CAD, DAB u bng 600. Cu IV:
1. Tnh tch phn
2
6 3 5
0
I = 1 - cos x .sinxcos xdx .
2. Tm gii hn 3 2 2
x 0
3x - 1 2 1L = lim1 - cosx
x
Cu V: Gi s a, b, c l bn s nguyn thay i tho mn 1 a < b < c < d 50 .
Chng minh bt ng thc 2a c b + b + 50 +
b d 50b v tm gi tr nh nht ca biu
thc a c + b d
.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 6 Cu I:
1. Kho st v v thi hm s y = 3 21 2 33
x x x (1)
2. Tnh din tch hnh phng gii hn bi th hm s (1) v trc honh. Cu II:
1. Gii phng trnh 21 s inx
8 osc x .
2. Gii h phng trnh 3 2
3 2
log ( 2 3 5 ) 3
log ( 2 3 5 ) 3x
y
x x x yy y y x
Cu III: 1. Cho hnh t din u ABCD, cnh a = 6 2 cm. Hy xc nh v tnh di on vung gc chung ca ng thng AD v ng thng BC.
2. Trong mt phng Oxy cho elip (E) : 2 2x + = 1
9 4y
v ng thng dm: mx - y - 1 = 0 a) Chng minh rng vi mi gi tr ca m, ng thng dm lun ct elip (E) ti hai im phn bit. b) Vit phng trnh tip tuyn ca (E) , bit rng tip tuyn i qua im N(1; - 3). Cu IV: Gi a1, a2, ..., a11 l cc h s trong khai trin (x + 1)10 (x + 2) = x11 + a1x10+ ...+ a11. Hy tnh h s a5. Cu V:
1. Tm gii hn 6
2x 1
x - 6x + 5L = lim(x - 1)
.
2. Cho tam gic ABC c din tch bng 32
. Gi a, b, c ln lt l di cc cnh
BC, CA, AB v ha, hb, hc tng ng l di cc ng cao k t cc nh A, B, C ca tam gic. Chng minh rng:
1 1 1 1 1 1 3a b ca b c h h h
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 7 Cu I:
1. Kho st v v thi hm s 22x - 4x - 3y =
2(x - 1).
2. Tm m phng trnh 2x2 - 4x - 3 + 2m 1x = 0 c hai nghim phn bit. Cu II: 1. Gii phng trnh 3 - tanx(tanx + 2sinx) + 6cosx = 0 .
2. Gii h phng trnh y xx y
log xy = log y
2 + 2 = 3
Cu III: 1. Trong mt phng Oxy cho parabol (P): 2y x v im I(0; 2). Tm to hai im M, N thuc (P) sao cho IM = 4IN
.
2. Trong khng gian Oxyz cho t din ABCD vi A(2; 3; 2), B(6; - 1; - 2), C( - 1; - 4; 3), D(1; 6; -5). Tnh gc gia hai ng thng AB v CD. Tm to im M thuc ng thng CD sao cho tam gic ABM c chu vi nh nht. 3. Cho lng tr ng ABCA'B'C' c y ABC l tam gic cn vi AB = AC = a v gc 0BAC = 120 , cnh bn BB' = a. Gi I l trung im CC' . Chng minh rng tam gic AB'I vung A. Tnh cosin ca gc gia hai mt phng (ABC) v (AB'I). Cu IV: 1. C bao nhiu s t nhin chia ht cho 5 m mi s c 4 ch s khc nhau.
2. Tnh tch phn:
4
0
xdxI = 1 + cos2x
Cu V: Tm gi tr ln nht v gi tr nh nht ca hm s: y = sin5x + 3 cosx.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 8 Cu I:
Cho hm s 2 2x + (2m + 1)x + m 4y =
2(x + m)m (1)(m l tham s).
1. Tm m hm s (1) c cc tr v tm khong cch gia hai im cc tr ca th hm s (1). 2. Kho st s bin thin v v th hm s (1) khi m = 0. Cu II: 1. Gii phng trnh cos2x + cosx(2tan2x - 1) = 2 . 2. Gii bt phng trnh x + 1 x x + 115.2 + 1 2 - 1 + 2 . Cu III: 1. Cho t din ABCD vi AB = AC = a, BC = b. Hai mt phng (BCD) v (ABC) vung gc nhau v gc 090BDC . Xc nh tm v bn knh mt cu ngoi tip t din ABCD theo a v b. 2. Trong khng gian Oxyz cho hai ng thng :
11:
1 2 1x y zd 2
3 1 0:
2 1 0x z
dx y
a) Chng minh rng, d1 v d2 cho nhau v vung gc nhau. b) Vit phng trnh tng qut ca ng thng d ct c hai ng v song
song vi ng thng : 4 7 31 4 2
x y z
.
Cu IV: 1. T cc ch s 0, 1, 2, 3, 4, 5 c th lp c bao nhiu s t nhin m mi s c 6 ch s khc nhau v ch s 2 ng cnh ch s ba.
2. Tnh tch phn: 1
3 2
0
I = x 1 - x dx
Cu V:
Tnh cc gc ca tam gic ABC bit rng 4 ( )
2 3 3sin sin sin2 2 2 8
p p a bc
A B C
trong BC = a, CA = b, AB = c v a + b +cp = 2
.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 9 Cu I: Cho hm s 2y = (x - 1)(x + mx + m) (1)(m l tham s). 1. Tm m hm s (1) ct trc honh ti ba im phn bit. 2. Kho st s bin thin v v th hm s (1) khi m = 4. Cu II: 1. Gii phng trnh 3cos4x - 8cos6x + 2cos2x + 3 = 0. 2. Tm m phng trnh 22 1
2
4 log x - log x + m = 0 c nghim thuc (0; 1).
Cu III: 1. Trong mt phng Oxy cho ng thng d : x - 7y + 10 = 0. Vit phng trnh ng trn c tm thuc ng thng : 2x + y = 0 v tip xc vi ng thng d ti im A(4; 2) 2. Cho hnh lp phng ABCD.A'B'C'D'. Tm im M thuc cnh AA' sao cho mt phng (BD'M) ct hnh lp phng theo mt thit din c din tch nh nht. 3. Trong khng gian Oxyz cho t din OABC vi A(0; 0; a 3 ), B(a; 0; 0), C(0; a 3 ; 0) (a > 0). Gi M l trung im BC. Tnh khong cch gia hai ng thng AB v OM. Cu IV: 1. Tm gi tr ln nht v gi tr nh nht ca hm s y = x6 + 4(1 - x2)3 trn on [- 1; 1].
2. Tnh tch phn: ln5 2x
xln2
eI = dxe 1
Cu V: T cc ch s 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s t nhin, mi s c 6 ch s v tho mn iu kin: Su ch s ca mi s l khc nhau v trong mi s tng ca ba ch s u nh hn tng ca ba ch s cui mt n v?
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 10 Cu I:
Cho hm s 2x - 1y = x - 1
(1)
1. Kho st s bin thin v v th (C) ca hm s (1). 2. Gi I l giao im hai ng tim cn ca (C). Tm im M thuc (C) sao cho tip tuyn ca (C) ti M vung gc vi ng thng IM. Cu II:
1. Gii phng trnh 2 x 2 - 3 cosx - 2sin - 2 4 = 1
2cosx - 1
.
2. Gii bt phng trnh 1 1 22 4
log x + 2log (x - 1) + log 6 0 .
Cu III:
1. Trong mt phng Oxy cho elip (E): 2 2x y + = 1
4 1, M( - 2; 3), N(5; n). Vit
phng trnh cc ng thng d1, d2 i qua M v tip xc vi (E). Tm n trong s cc tip tuyn ca (E) qua N c mt tip tuyn song song vi d1 hoc d2. 2. Cho hnh chp u S.ABC, y ABC c cnh bng a, mt bn to vi y mt gc bng 0 0(0 90 ) . Tnh th tch khi chp S.ABC v khong cch t nh A n mt phng (SBC). 3. Trong khng gian Oxyz cho hai im I(0; 0; 1), K(3; 0; 0). Vit phng trnh mt phng i qua hai im I, K v to vi mt phng Oxy mt gc 300. Cu IV: 1. T mt t gm 7 hc sinh n v 5 hc sinh nam cn chn ra 6 em trong s hc sinh n phi nh hn 4. Hi c bao nhiu cch chn nh vy.
2. Cho hm s x3af(x) = + bxe
(x + 1). Tm a v b bit rng:
f '(0) = - 22 v 1
0
(x)dx = 5f
Cu V:
Chng minh rng 2
x xe + cosx 2 + x - 2
, x .
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 11 Cu I:
Cho hm s 2 2x + 5x + m 6y =
x + 3 (1)( m l tham s)
1. Kho st s bin thin v v th ca hm s (1) khi m = 1. 2. Tm m hm s (1) ng bin trn khong (1; + ) . Cu II:
1. Gii phng trnh 2cos x(cosx - 1) = 2(1 + sinx)
sinx + cosx.
2. Cho hm s xf(x) = xlog 2, (x > 0, x 1) . Tnh f '(x) v gii bt phng trnh f '(x) 0. Cu III: 1. Trong mt phng Oxy cho tam gic ABC c nh A(1; 0) v hai ng thng ln lt cha cc ng cao v t B v C c phng trnh tng ng l x - 2y + 1 = 0 v 3x + y - 1 = 0. Tnh din tch tam gic ABC. 2. Trong khng gian Oxyz cho mt phng (P): 2x + 2y + z - m2 - 3m = 0(m l tham s) v mt cu (S): 2 2 2x - 1 + y + 1 + z - 1 = 9 . Tm m mt phng (P) tip xc mt cu (S). Vi m va tm c, hy xc nh to tip im ca mt phng (P) v mt cu (S). 3. Cho hnh chp S.ABC c y ABC l tam gic vung ti B, AB = a, BC = 2a, cnh SA vung gc vi y v SA = 2a. Gi M l trung im ca SC. Chng minh rng tam gic AMB cn ti M v tnh din tch tam gic AMB theo a. Cu IV: 1. T 9 ch s 0, 1, 2, 3, 4, 5, 6, 7, 8 c th lp c bao nhiu s t nhin chn m mi s gm 7 ch s khc nhau?
2. Tnh tch phn I = 21
3 x
0
x e dx .
Cu V: Tnh cc gc A, B, C ca tam gic ABC biu thc: 2 2 2Q = sin A + sin B - sin C t gi tr nh nht.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 12 Cu I: 1. Kho st s bin thin v v th (C) ca hm s: y = 2x3 - 3x2 - 1. 2. Gi dk l ng thng i qua M(0; - 1) v c h s gc bng k. Tm k ng thng dk ct (C) ti ba im phn bit. Cu II:
1. Gii phng trnh 2cos4x cotx = tanx + sin2x
2. Gii phng trnh x5log 5 4 = 1 - x Cu III: 2. Trong khng gian Oxyz cho hai im A( 2; 1; 1), B(0; - 1; 3) v ng thng
d: 3x - 2y - 11 = 0y + 3z - 8 = 0
a) Vit phng trnh mt phng (P) i qua trung im I ca AB v vung gc vi AB. Gi K l giao im ca ng thng d v mt phng (P), chng minh rng d vung gc vi IK. b) Vit phng trnh tng qut ca hnh chiu vung gc ca d trn mt phng c phng trnh x + y - z + 1 = 0. 2. Cho t din ABCD c AD vung gc vi mt phng (ABC) v tam gic ABC vung ti A, AD = a, AC = b, AB = c. Tnh din tch ca tam gic BCD theo a, b, c v chng minh 2S abc(a + b + c). Cu IV: 1. Tm s t nhin n tho mn: 2 n - 2 2 3 3 n - 3n n n n n nC C + 2C C + C C = 100 , trong knC l s t hp cp k ca n.
2. Tnh tch phn I = 2
1
x + 1lnxdx. x
e
Cu V: Xc nh tam gic ABC bit rng : 2 2(p - a)sin A + (p - b)sin B = csinAsinB .
trong BC = a, CA = b, AB = c, a + b + cp =2
.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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S 13 Cu I: Cho hm s y = x4 - 2m2 x2 + 1 (1)(m l tham s ) 1. Kho st hm s (1) khi m = 1. 2. Tm m th hm s (1) c ba im cc tr l ba nh ca mt tam gic vung cn. Cu II: 1. Gii phng trnh 4(sin3 x + cos3 x) = cosx + 3sinx.
2. Gii bt phng trnh log 4
[ log 2(x + 22 - xx )] < 0.
Cu III: 1. Trong mt phng Oxy cho ng thng d: x - y + 1 - 2 = 0 v im A(-1; 1). Vit phng trnh ng trn i qua A, qua gc to O v tip xc vi ng thng d. 2. Trong khng gian Oxyz cho hnh hp ch nht ABCD.A 1B 1C 1D 1 c A trng vi gc to O, B(1; 0; 0), D(0; 1; 0), A 1(0; 0; 2 ). a) Vit phng trnh mt phng (P) i qua ba im A 1, B, C v vit phng trnh hnh chiu vung gc ca ng thng B 1D 1 trn mt phng (P). b) Gi (Q) l mt phng qua A v vung gc vi A 1C. Tnh din tch thit din ca hnh chp A 1ABCD vi mt phng (Q). Cu IV: 1. Tnh th tch ca vt th trn xoay sinh ra bi php quay xung quanh trc Ox ca hnh phng gii hn bi trc Ox v ng y = x sinx (0 x ) 2. Cho tp hp A gm n phn t, n 7. Tm n, bit rng s tp con gm 7 phn t ca tp A bng hai ln s tp con gm ba phn t ca tp A. Cu V:
Gi (x; y) l nghim ca h phng trnh x - my = 2 - 4mmx + y = 3m + 1
(m l tham s). Tm
gi tr ln nht ca biu thc A = x2 + y2 - 2x, khi m thay i.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
16
S 14 Cu I: Cho hm s y = 2x3 - 2mx2 + m2 x - 2 (1)(m l tham s). 1. Kho st hm s (1) khi m = 1. 2. Tm m hm s (1) t cc tiu ti x = 1. Cu II:
1. Gii phng trnh 2 2 cos(x + 4 ) + 1
sinx = 1
cosx .
2. Gii bt phng trnh x - 12 + 6x - 11 > 4
x - 2
Cu III: 1. Trong mt phng Oxy cho im I(- 2; 0) v hai ng thng d 1 : 2x - y + 5 = 0 v d 2: x + y - 3 = 0. Vit phng trnh ng thng d i qua im I v ct hai ng thng d 1, d 2 ln
lt ti A, B sao cho
IA = 2.
IB .
2. Trong khng gian Oxyz cho A(4 ; 2; 2), B( 0 ; 0; 7) v ng thng
d: x - 3 y - 6 z - 1 = - 2 2 1
Chng minh rng hai ng thng d v AB cng thuc mt mt phng. Tm im C trn ng thng d sao cho tam gic ABC cn ti nh A. 3. Cho hnh chp S.ABC c SA = 3a v vung gc vi y ABC, tam gic ABC c AB = BC = 2a, gc B bng 1200. Tnh khong cch t nh A n mt phng (SBC). Cu IV:
1. Tnh tch phn I = 3
31
dxx + x .
2. Bit rng (2 + x)100 = a 0 + a 1x + a 2x2 + ... + a 100 x100 . Chng minh a 2 < a 3. Vi gi tr no ca k th a k < a k+1 (0 k 99)? Cu V:
Cho hm s f(x) = ex - sinx + x2 2 . Tm gi tr nh nht ca f(x) v chng minh
rng phng trnh f(x) = 3 c ng hai nghim.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
17
S 15 Cu I:
Cho hm s 2x - 2mx + 2y =
x - 1 (1)(m l tham s).
1. Kho st hm s (1) khi m = 1. 2. Tm m hm s (1) c hai im cc tr A, B. Chng minh rng khi ng thng AB song song vi ng thng d: 2x - y - 10 = 0. Cu II: 1. Gii phng trnh sin4xsin7x = cos3xcos6x. 2. Gii bt phng trnh log 3x > log x3. Cu III:
1. Trong mt phng Oxy cho elip (E): x2
8 + y2 4 = 1. Vit phng trnh cc tip
tuyn ca (E) song song vi ng thng d: x + 2 y - 1 = 0 2. Trong khng gian Oxyz cho A(2 ; 0; 0) v M( 1 ; 1; 1). a) Tm to O' i xng O qua ng thng AM. b) Gi (P) l mt phng thay i i qua ng thng AM, ct cc trc Oy, Oz ln lt ti cc im B, C. Gi s B(0; b; 0), C(0; 0; c), b > 0, c > 0. Chng minh
rng b + c = bc2 . Xc nh b, c sao cho din tch tam gic ABC nh nht.
Cu IV:
1. Tnh tch phn I =
3
cosx
0
e sin2xdx .
2. Bit rng (1 + 2x)n = a 0 + a 1x + a 2x2 + ... + a n xn . Chng minh a 2 < a 3. Bit rng a 0 + a 1 + a 2 + ... + a n = 729. Tm n v s ln nht trong cc s a 0, a 1, a 2, ..., a n Cu V:
Cho tam gic ABC tho mn A 900 v sinA = 2sinBsinCtanA2 . Tm gi tr nh
nht ca biu thc A1 - sin2S =
sinB.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
18
S 16 Cu I:
Cho hm s 2x + x + 4y = x + 1
(1) c th (C).
1. Kho st hm s (1) . 2. Vit phng trnh tip tuyn ca (C), bit tip tuyn vung gc vi ng thng d: x - 3y + 3 = 0. Cu II: 1. Gii phng trnh 2sinxcos2x + sin2xcosx = sin4xcosx.
2. Gii h phng trnh 2 2
x + y x - 1
x + y = y + x2 - 2 = x - y.
Cu III: 1. Trong mt phng Oxy cho tam gic ABC vung A. Bit A( - 1; 4), B( 1; -
4), ng thng BC i qua im K( 73 ; 2). Tm to C.
2. Trong khng gian Oxyz cho A(2 ; 0; 0) , B(2; 2; 0), C(0; 0; 2). a) Tm to O' i xng O qua mf(ABC). b) Cho im S di chuyn trn trc Oz, gi H l hnh chiu vung gc ca O trn ng thng SA. Chng minh rng din tch tam gic OBH nh hn 4. Cu IV:
1. Tnh tch phn I = 2
0
xsin xdx .
2. Bit rng trong khai trin nh thc Niutn ca (x + 1x )n tng cc h s ca hai
s hng u tin bng 24, tnh tng cc h s ca cc s hng cha xk vi k > 0 v chng minh rng tng ny l mt s chnh phng. Cu V:
Cho phng trnh x2 + ( m2 - 53 )
2x + 4 + 2 - m2 = 0.
Tm tt c cc gi tr m phng trnh c nghim.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
19
S 17 Cu I:
Cho hm s xy = x + 1
(1) c th (C).
1. Kho st hm s (1) . 2. Tm trn (C) nhng im M sao cho khong cch t M n ng thng d: 3x + 4y = 0 bng 1. Cu II: 1. Gii phng trnh sinx + sin2x = 3(cosx + cos2x) 2. Tm gi tr ln nht v gi tr nh nht ca hm s y = (x + 1) 21 - x . Cu III: 1. Trong mt phng Oxy cho im A(2; 3) v hai ng thng d 1: x + y + 5 = 0 v d2: x + 2y - 7 = 0. Tm to cc im B trn d 1 v C trn d 2 sao cho tam gic ABC c trng tm l G(2; 0). 2. Cho hnh vung ABCD c cnh AB = a. trn cc na ng thng Ax, By vung gc vi mf(ABCD) v nm v cng mt pha i vi mf(ABCD), ln lt ly cc im M, N sao cho tam gic MNC vung ti M. t AM = m, BN = n. Chng minh rng, m(n - m) = a2 v tm gi tr nh nht ca din tch hnh thang ABNM.
3. Trong khng gian Oxyz cho A(0 ; 1; 1) v ng thng d: x + y = 02x - z - 2 = 0
Vit phng trnh mt phng (P) i qua A v vung gc vi ng thng d. Tm to hnh chiu vung gc B' ca im B(1; 1; 2) trn mt phng (P). Cu IV:
1. Tnh tch phn I = ln8
2x x
ln3
e e 1dx .
2. C bao nhiu s t nhin tho mn ng thi ba im kin sau: gm ng 4 ch s i mt khc nhau; l s chn; nh hn 2158 ? Cu V:
Tm tt c cc gi tr m h sau c nghim: 2
2
x - 5x + 4 0
3x - mx x + 16 = 0
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
20
S 18
Cau I:
Goi (Cm) la o th cua ham so : y = 2 22 1 3x mx m
x m
(*) (m la tham so)
1. Khao sat s bien thien va ve o th cua ham so (*) ng vi m = 1. 2. Tm m e ham so (*) co hai iem cc tr nam ve hai pha truc tung. Cau II:
1. Giai he phng trnh : 2 2 4( 1) ( 1) 2
x y x yx x y y y
2. Tm nghiem tren khong (0; ) cua phng trnh :
2 2 34sin 3 cos 2 1 2cos ( )2 4x x x
Cau III: 1. Trong mat phang vi he toa o Oxy cho tam giac ABC can tai nh A co
trong tam G 4 1( ; )3 3
, phng trnh ng thang BC la 2 4 0x y va phng
trnh ng thang BG la 7 4 8 0x y .Tm toa o cac nh A, B, C. 2. Trong khong gian vi he toa o Oxyz cho 3 iem A(1;1; 0),B(0; 2; 0), C(0; 0; 2) . a) Viet phng trnh mat phang (P) qua goc toa o O va vuong goc vi BC. Tm toa o giao iem cua AC vi mat phang (P). b) Chng minh tam giac ABC la tam giac vuong. Viet phng trnh mat cau ngoai tiep t dien OABC. Cau IV:
1. Tnh tch phan 3
2
0
sin .I x tgxdx
.
2. T cac ch so 1, 2, 3, 4, 5, 6, 7, 8, 9 co the lap c bao nhieu so t nhien, moi so gom 6 ch so khac nhau va tong cac ch so hang chuc, hang tram hang ngan bang 8.
Cau V: Cho x, y, z la ba so thoa x + y + z = 0. Chng minh rang : 3 4 3 4 3 4 6x y z
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
21
S 19 Cau I:
1. Khao sat s bien thien va ve o th ( C ) cua ham so 2 1
1x xy
x
.
2. Viet phng trnh ng thang i qua iem M (- 1; 0) va tiep xuc vi o th ( C ) .
Cau II:
1. Giai he phng trnh : 2 1 13 2 4
x y x yx y
2. Giai phng trnh : 32 2 cos ( ) 3cos sin 04
x x x
Cau III: 1. Trong mat phang vi he toa o Oxy cho ng tron (C): x2 + y2 12 4 36 0x y . Viet phng trnh ng tron (C1) tiep xuc vi hai truc toa o Ox, Oy ong thi tiep xuc ngoai vi ng tron (C). 2. Trong khong gian vi he toa o ecac vuong goc Oxyz cho 3 iem A(2;0;0), C(0; 4; 0), S(0; 0; 4). a) Tm toa o iem B thuoc mat phang Oxy sao cho t giac OABC la hnh ch nhat. Viet phng trnh mat cau qua 4 iem O, B, C, S. b) Tm toa o iem A1 oi xng vi iem A qua ng thang SC.
Cau IV: 1. Tnh tch phan 7
30
21
xI dxx
.
2. Tm he so cua x7 trong khai trien a thc 2(2 3 ) nx , trong o n la so nguyen dng thoa man: 1 3 5 2 12 1 2 1 2 1 2 1... nn n n nC C C C = 1024. ( knC la so to hp chap k cua n phan t)
Cau V: Cm rang vi moi x, y > 0 ta co : 29(1 )(1 )(1 ) 256yx
x y . ang thc xay ra khi nao?
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
22
S 20 Cau I: 1. Khao sat s bien thien va ve o th ( C ) cua ham so 4 26 5y x x 2. Tm m e phng trnh sau co 4 nghiem phan biet : 4 2 26 log 0x x m . Cau II:
1. Giai he phng trnh : 2 1 13 2 4
x y x yx y
2. Giai phng trnh : 32 2 cos ( ) 3cos sin 04
x x x
Cau III:
1. Trong mat phang vi he toa o Oxy cho elip (E) : 2 2
64 9x y
= 1. Viet phng
trnh tiep tuyen d cua (E) biet d cat hai hai truc toa o Ox, Oy lan lt tai A, B sao cho AO = 2BO.
2. Trong khong gian vi he toa o Oxyz cho hai ng thang 1x y z: 1 1 2
d va
2
1 2:
1
x td y t
z t
( t la tham so )
a) Xet v tr tng oi cua d1 va d2 . b) Tm toa o cac iem M thuoc d1 va N thuoc d2 sao cho ng thang MN song song vi mat phang (P) : 0x y z va o dai oan MN = 2 . Cau IV:
1. Tnh tch phan 20
lne
x xdx .
2. Mot i van nghe co 15 ngi gom 10 nam va 5 n. Hoi co bao nhieu cach lap mot nhom ong ca gom 8 ngi biet rang trong nhom o phai co t nhat 3 n.
Cau V: Cho a, b, c la ba so dng thoa man : a + b + c = 34
. Chng minh rang :
3 3 33 3 3 3a b b c c a . Khi nao ang thc xay ra ?
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
23
S 21
Cau I: Cho ham so : y = 2 2 2
1x x
x
(*)
1. Khao sat s bien thien va ve o th ( C ) cua ham so (*) . 2. Goi I la giao iem cua hai tiem can cua ( C ). Chng minh rang khong co tiep tuyen nao cua (C ) i qua iem I . Cau II: 1. Giai bat phng trnh : 28 6 1 4 1 0x x x
2. Giai phng trnh : 2 2cos 2 1( ) 3
2 cosxtg x tg x
x
Cau III: 1. Trong mat phang vi he toa o Oxy cho 2 ng tron : (C1 ): x2 + y2 9 va (C2 ): x2 + y2 2 2 23 0x y . Viet phng trnh truc ang phng d cua 2 ng tron (C1) va (C2). Chng minh rang neu K thuoc d th khoang cach t K en tam cua (C1) nho hn khong cach t K en tam cua (C2 ). 2. Trong khong gian vi he toa o Oxyz cho iem M(5;2; - 3) va mat phang (P): 2 2 1 0x y z . a) Goi M1 la hnh chieu cua M len mat phang ( P ). Xac nh toa o iem M1
va tnh o dai oan MM1. b) Viet phng trnh mat phang ( Q ) i qua M va cha ng thang :
x - 1 y - 1 z - 52 1 - 6
Cau IV:
1.Tnh tch phan 4
sin
0
(tan cos )xx e x dx
.
2. T cac ch so 1, 2, 3, 4, 5, 6, 7 co the lap c bao nhieu so t nhien, moi so gom 5 ch so khac nhau va nhat thiet phai co 2 ch 1, 5 ?
Cau V: Chng minh rang neu 0 1y x th
14
x y y x . ang thc xay ra khi nao?
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
24
S 22 Cau I: Goi (Cm) la o th cua ham so y= x3+ ( 2m + 1) x2 m 1 (1) (m la tham so). 1) Khao sat s bien thien va ve o th cua ham so (1) khi m 1 . 2) Tm m e o th (Cm) tiep xuc vi ng thang y= 2mx m 1. Cau II: 1. Giai bat phng trnh : 2 7 5 3 2x x x
2. Giai phng trnh : 3 sin( ) 22 1 cos
xtg xx
Cau III: 1. Trong mat phang vi he toa o Oxy cho ng tron (C): x2 + y2 4 6 12 0x y . Tm toa o iem M thuoc ng thang d : 2 3 0x y sao cho MI = 2R, trong o I la tam va R la ban knh cua ng tron (C). 2. Trong khong gian vi he toa o Oxyz cho lang tru ng OAB.O1A1B1 vi A(2;0;0), B(0; 4; 0), O1(0; 0; 4) a) Tm toa o cac iem A1, B1. Viet phng trnh mat cau qua 4 iem O, A, B, O1. b) Goi M la trung iem cua AB.Mat phang ( P ) qua M vuong goc vi O1A va cat OA, OA1 lan lt tai N, K. Tnh o dai oan KN. Cau IV:
1. Tnh tch phan 3 2
1
lnln 1
e xI dxx x
.
2. Tm k 0;1;2;...;2005 sao cho 2005kC at gia tr ln nhat. ( knC la so to hp chap k cua n phan t)
Cau V: Tm m e he phng trnh sau co nghiem:
2 1 2 1
2
7 7 2005 2005( 2) 2 3 0
x x x xx m x m
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
25
S 23 Cau I:
1. Khao sat s bien thien va ve o th cua ham so 2 3 3
1x xy
x
.
2. Tm m e phng trnh 2 3 3
1x x m
x
co 4 nghiem phan biet
Cau II:
1. Giai bat phng trnh : 2
22
2 19 2 33
x xx x
.
2. Giai phng trnh :sin 2 cos 2 3sin cos 2 0x x x x Cau III: 1. Trong mat phang vi he toa o Oxy cho 2 iem A(0;5), B(2; 3) . Viet phng trnh ng tron i qua hai iem A, B va co ban knh R = 10 . 2. Trong khong gian vi he toa o Oxyz cho 3 hnh lap phng ABCD.A1B1C1D1 vi A(0;0;0), B(2; 0; 0), D1(0; 2; 2) a) Xac nh toa o cac iem con lai cua hnh lap phng ABCD.A1B1C1D1. Goi M la trung iem cua BC. Chng minh rang hai mat phang ( AB1D1) va ( AMB1) vuong goc nhau. b) Chng minh rang t so khong cach t iem N thuoc ng thang AC1 ( N A ) ti 2 mat phang ( AB1D1) va ( AMB1) khong phu thuoc vao v tr cua iem N. Cau IV:
1. Tnh tch phan
2
2
0
I = ( 2x - 1)cos xdx .
2. Tm so nguyen n ln hn 1 thoa man ang thc : 2 22 6 12n n n nP A P A .
( Pn la so hoan v cua n phan t va knA la so chnh hp chap k cua n phan t)
Cau V: Cho x, y, z la ba so dng va xyz = 1. Chng minh rang :
2 2 2 3
1 1 1 2x y z
y z x
.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
26
S 24 Cu I:
1. Kho st v v th hm s 2x 2x 5y
x 1
(C)
2. Da vo th (C), tm m phng trnh sau c hai nghim dng phn bit 2 22 5 ( 2 5)( 1)x x m m x Cu II:
1. Gii phng trnh: 3 3 2 3 2cos3xcos x sin 3x sin x8
2. Gii h phng trnh: 2
2
( 1) ( ) 4 ( , )
( 1)( 2)x y y x y
x y Rx y x y
Cu III: Trong khng gian Oxyz cho hnh lng tr ng ABCA'B'C' c A(0; 0; 0), B(2; 0; 0), C(0; 2; 0), A'(0; 0; 2). 1. Chng minh A'C vung gc vi BC'. Vit phng trnh mt phng (ABC'). 2. Vit phng trnh hnh chiu vung gc ca ng thng B'C' trn mf(ABC') Cu IV:
1. Tnh 6
2
dxI2x 1 4x 1
2. Cho x, y l cc s thc tho mn iu kin: x2 + xy + y2 3. Chng minh rng: 2 24 3 3 3 4 3 3x xy y . Cu Va:
1. Trong mt phng Oxy cho elip (E) 2 2
112 2x y
. Vit phng trnh ca hypebol
(H) c hai dng tim cn l 2y x v c hai tiu im l hai tiu im ca (E).
2. p dng khai trin ca nh thc Newton ca 1002x x , chng minh rng:
99 100 198 199
0 1 99 100100 100 100 100
1 1 1 1100 101 ... 199 200 02 2 2 2
C C C C
Cu Vb: 1. Gii bt phng trnh: x 1log ( 2x) 2 2. Cho hnh hp ng ABCD.A'B'C'D' c y l hnh thoi, cc cnh AB = AD =
a, BAD = 600 , cnh bn bng 32
a . Gi M, N ln lt l trung im ca cc
cnh A'D' v A'B'. Chng minh AC'mf(BDMN). Tnh th tch khi chp A.BDMN.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
27
S 25 Cu I:
1. Kho st v v th (C) ca hm s 4
2xy 2(x 1)2
(C)
2. Vit phng trnh cc ng thng i qua im A(0; 2) v tip xc vi (C). Cu II:
1. Gii phng trnh: 2sin 2x 4s inx +1=06
2. Gii h phng trnh: 3 3
2 2
8 2 ( , )
3 3( 1)x x y y
x y Rx y
Cu III: Trong khng gian Oxyz cho mf( ): 3x + 2y - z + 4 = 0 v hai im A(4; 0; 0), B(0; 4; 0). Gi I l trung im ca on thng AB. 1. Tm giao im ca ng thng AB vi mf( ). 2. Xc nh to im K sao cho KI vung gc vi mf( ) ng thi K cch u gc to O v mf( ) . Cu IV: 1. Tnh din tch hnh phng gii hn bi parabol y = x2 - x + 3 v ng thng d: y = 2x + 1 2. Cho x, y, z tho mn cc iu kn 3 3 3 1x y z . Chng minh rng:
9 9 9 3 3 33 3 3 3 3 3 4
x y z x y z
x y z y z x z x y
Cu Va: 1. Trong mt phng Oxy cho tam gic ABC c nh A thuc ng thng d: x - 4y - 2 = 0 , cnh BC song song vi d. Phng trnh ng cao BH: x + y + 3 = 0 v trung im ca cnh AC l M(1; 1). Tm to A, B, C. 2. T cc ch s 0, 1, 2, 3, 4 c th lp c bao nhiu s t nhin c 5 ch s khc nhau ? Tnh tng ca tt c cc s t nhin . Cu Vb: 1. Gii bt phng trnh: x 2x 2xlog 2 2log 4 log 8
2. Cho hnh chp S.ABCD c y ABCD l hnh ch nht vi AB = a, AD = 2a, cnh SA vung gc vi y, cnh SB to vi mt phng y mt gc 600. Trn
cnh SA ly im M sao cho AM = 33
a . Mt phng (BCM) ct SD ti N.
Tnh th tch khi chp S.BCNM.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
28
S 26 Cu I:
1. Kho st v v th (C) ca hm s 2x x 1yx 1
(C)
2. Vit phng trnh tip tuyn ca (C) i qua A(0; - 5). Cu II: 1. Gii phng trnh: 2 2 2(2sin x 1) tan 2x 3(2cos x -1) = 0 2. Gii phng trnh: 23 2 1 4 9 2 3 5 2 , )x x x x x x R Cu III: Trong khng gian Oxyz cho hai ng thng:
11
: 12
x ty tz
v 23 1:
1 2 1x y z
1. Vit phng trnh mt phng cha 1 v song song 2 . 2. Xc nh to im A trn 1 v im B trn 2 sao cho on thng AB c di nh nht. Cu IV:
1. Tnh 10
5
dxIx 2 x 1
2. Tm gi tr nh nht ca hm s: 211 74 1 , 02
y x xx x
Cu Va: 1. Trong mt phng Oxy cho tam gic ABC cn ti B, vi A(1; -1), C(3; 5). im B thuc ng thng d: 2x - y = 0. Vit phng trnh cc ng thng AB, BC. 2. T cc ch s 0, 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s chn, mi s c 5 ch s khc nhau trong c ng hai ch s l v hai ch s l ng cnh nhau. Cu Vb: 1. Gii phng trnh: 31 82
2
log x 1 log (3 x) log (x 1) 0
2. Cho hnh chp S.ABCD c y ABCD l hnh thoi cnh a. 060BAD . SA vung gc vi mf(ABCD), SA = a. Gi C' trung im ca SC. Mt phng (P) di qua AC' v song song BD, ct cc cnh SB, SD ca hnh chp ln lt ti B', D'. Tnh th tch khi chp S.AB'C'D'.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
29
S 27 Cu I: Cho hm s y = x3 + (1 - 2m)x2 + (2 - m)x + 2 (1) 1. Kho st v v th ca hm s (1) khi m = 2. 2. Tm tt c cc gi tr m th hm s (1) c cc i, cc tiu, ng thi honh ca im cc tiu nh hn 1. Cu II: 1. Gii phng trnh: cos2x + (1 + 2cosx)(sinx - cosx) = 0
2. Gii h phng trnh: 2 2
2 2
( )( ) 13 ( , )
( )( ) 25x y x y
x y Rx y x y
Cu III: Trong khng gian Oxyz cho mt phng (P): 2x + y - z + 5 = 0 v cc im A(0; 0; 4), B(2; 0; 0) 1. Vit phng trnh hnh chiu vung gc ca ng thng AB trn mf(P). 2. Vit phng trnh mt cu i qua O, A, B v tip xc vi mt phng (P). Cu IV:
1. Tnh e
1
3 2 ln xI dxx 1 2 ln x
2. Cho hai s dng x, y thay i th mn iu kin x + y 4. Tm gi tr nh
nht ca biu thc A = 2 3
2
3 4 24
x yx y
Cu Va: 1. Trong mt phng Oxy cho tam gic ABC c A(2; 1), ng cao qua nh B c phng trnh x - 3y - 7 = 0 v ng trung tuyn qua nh C c phng trnh x + y + 1 = 0. Xc nh to cc nh B, C ca tam gic.. 2. Cho hai ng thng song song d1 v d2. Trn ng thng d1 c 10 im phn bit, trn ng thng d2 c n im phn bit(n 2). Bit rng c 2800 tam gic c nh l cc im cho. Tm n. Cu Vb:
1. Gii phng trnh: 2 2x x 1 x x 29 10.3 1 0
2. Cho hnh lng tr ABC.A'B'C' c A'ABC l hnh chp tam gic u, cnh y AB = a, cnh bn AA' = b. Gi l gc gia hai mt phng (ABC) v (A'BC). Tnh tan v th tch khi chp A'BB'C'C.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
30
S 28
Cu I: Cho hm s 3
2 1133 3xy x x
1. Kho st v v th (C) ca hm s cho. 2. Tm trn th (C) hai im phn bit M, N i xng nhau qua trc tung. Cu II: 1. Gii phng trnh: 3 3 2cos x sin x 2sin x 1
2. Gii h phng trnh: 2 2
2 2 2
3( ) ( , )
7( )x xy y x y
x y Rx xy y x y
Cu III: Trong khng gian Oxyz cho mt phng (P): 4x - 3y + 11z - 26 = 0 v hai
ng thng 1 23 1 4 3: , :
1 2 3 1 1 2x y z x y zd d
1. Chng minh rng d1 v d2 cho nhau. 2. Vit phng trnh ng thng nm trn (P), ng thi ct c d1 v d2 . Cu IV:
1. Tnh 2
0
I (x 1) sin 2xdx
2. Gii phng trnh 14 2 2(2 1)sin(2 1) 2 0x x x x y Cu Va: 1. Trong mt phng Oxy cho ng thng d: x - y + 1 - 2 = 0 v im A(- 1; 1). Vit phng trnh ng trn (C) i qua A, O v tip xc d. 2. Mt lp hc c 33 hc sinh, trong c 7 n. Cn chia lp thnh 3 t, t 1 c 10 hc sinh, t 2 c 11 hc sinh v t 3 c 12 hc sinh sao cho trong mi t c t nht 2 hc sinh n. Hi c bao nhiu cch chia nh vy ? Cu Vb: 1. Gii phng trnh: x x 13 3log (3 1) log (3 3) 6
2. Cho hnh chp t gic u S.ABCD c cnh y bng a. Gi SH l ng cao ca hnh chp. Khong cch t trung im I ca SH n mt bn (SBC) bng b. Tnh th tch khi chp S.ABCD.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
31
S 29
Cu I: Cho hm s 31
xyx
(C)
1. Kho st v v th (C) ca hm s cho. 2. Cho im 0 0 0( ; )M x y thuc (C). Tip tuyn ca (C) ti 0 0 0( ; )M x y ct cc tim cn ca (C) ti cc im A, B. Chng minh 0 0 0( ; )M x y l trung im AB. Cu II: 1. Gii phng trnh: 3 24sin x 4sin x 3sin 2x 6cosx 0 2. Gii phng trnh: 2x + 2 7 - x = 2 x - 1 + - x 8 7 +1 ( )x x R Cu III: Trong khng gian Oxyz cho A(1; 2; 0), B(0; 4; 0), C(0; 0; 3) 1. Vit phng trnh ng thng i qua O v vung gc vi mf(ABC). 2. Vit phng trnh mt phng (P) cha OA, sao cho khong cch t B n (P) bng khong cch t C n (P). Cu IV:
1. Tnh 2
1
I (x 2) ln xdx
2. Gii hphng trnh: 2 2ln(1 ) ln(1 )
( , ) 12 20 0x y x y
x y Rx xy y
Cu Va: 1. Trong mt phng Oxy lp phng trnh chnh tc ca elip (E) c di trc ln bng 4 2 , cc nh trn trc nh v cc tiu im ca (E) cng thuc mt ng trn. 2. T cc ch s 0, 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s t nhin chn c 5 ch s khc nhau v mi s lp nn u nh hn 2500 ? Cu Vb:
1. Gii phng trnh: 2 4 212(log x 1)log x log 04
2. Cho hnh lp phng ABCD.A'B'C'D' c cnh bng a v im K thuc cnh
CC' sao cho 23
CK a . Mt phng ( ) i qua A, K v song song vi BD, chia
khi lp phng thnh hai khi a din. Tnh th tch ca hai khi a din .
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
32
S 30
Cu I: Cho hm s 2x 4x 3yx 2
1. Kho st v v th hm s. 2. Chng minh rng tch cc khong cch t mt im bt k trn th hm s n cc ng tim cn ca n l hng s. Cu II:
1. Gii phng trnh: 1 1sin 2x sin x 2cot g2x2sin x sin 2x
2. Tm m phng trnh: 2m x 2x 2 1 x(2 x) 0 (2) c nghim x 0;1 3
Cu III: Trong khng gian Oxyz cho hai im A (-1;3;-2), B (-3;7;-18) v mt phng (P): 2x - y + z + 1 = 0 1. Vit phng trnh mt phng cha AB v vung gc vi mp (P). 2. Tm ta im M (P) sao cho MA + MB nh nht. Cu IV:
1. Tnh 4
0
2x 1I dx1 2x 1
2. Gii h phng trnh: )Ry,x( 132y2yy
132x2xx1x2
1y2
Cu Va: 1. Trong mt phng Oxy cho ng trn (C) : x2 + y2 = 1. ng trn (C') tm I (2; 2) ct (C) ti cc im A, B sao cho AB 2 . Vit phng trnh ng thng AB. 2. C bao nhiu s t nhin chn ln hn 2007 m mi s gm 4 ch s khc nhau ? Cu Vb: 1. Gii bt phng trnh: 2x 4 2(log 8 log x ) log 2x 0 2. Cho lng tr ng ABCA1B1C1 c AB = a, AC = 2a, AA1 2a 5 v
o120BAC
. Gi M l trung im ca cnh CC1. Chng minh MBMA1 v tnh khong cch d t im A ti mt phng (A1BM).
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
33
S 31
Cu I: Cho hm s my x m (Cm)x 2
1. Kho st v v th hm s vi m = 1. 2. Tm m th (Cm) c cc tr ti cc im A, B sao cho ng thng AB i qua gc ta O. Cu II: 1. Gii phng trnh: 22 cos x 2 3 sin x cosx 1 3(sin x 3 cosx)
2. Gii h phng trnh 4 3 2 2
3 2
x x y x y 1
x y x xy 1
Cu III: Trong khng gian Oxyz cho cc im A(2,0,0); B(0,4,0); C(2,4,6) v
ng thng (d) 6x 3y 2z 06x 3y 2z 24 0
1. Chng minh cc ng thng AB v OC cho nhau. 2. Vit phng trnh ng thng // (d) v ct cc ng AB, OC.
Cu IV: 1. Trong mt phng Oxy cho hnh phng (H) gii hn bi cc ng 2xy4 v y = x. Tnh th tch vt th trn trong khi quay (H) quanh trc Ox trn mt vng. 2. Cho x, y, z l cc bin s dng. Tm gi tr nh nht ca biu thc:
33 3 3 3 3 33 32 2 2
x y zP 4(x y ) 4(y z ) 4(z x ) 2y z x
Cu Va: 1. Trong mt phng Oxy cho tam gic ABC c trng tm G(2, 0) bit phng trnh cc cnh AB, AC theo th t l 4x + y + 14 = 0; 02y5x2 . Tm ta cc nh A, B, C. 2. Trn cc cnh AB, BC, CD, DA ca hnh vung ABCD ln lt cho 1, 2, 3 v n im phn bit khc A, B, C, D. Tm n bit s tam gic c ba nh ly t n + 6 im cho l 439. Cu Vb:
1. Gii phng trnh 4 22x 1
1 1log (x 1) log x 2log 4 2
2. Cho hnh chp SABC c gc o60ABC,SBC
, ABC v SBC l cc tam gic u cnh a. Tnh theo a khong cch t nh B n mp(SAC).
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
34
S 32 Cu I: Cho hm s y = 2x3 + 6x2 5 1. Kho st v v th hm s (C). 2. Vit phng trnh tip tuyn ca (C), bit tip tuyn i qua A(1, 13). Cu II:
1. Gii phng trnh: 2x3cos2
42xcos
42x5sin
2. Tm m phng trnh: mx1x4 2 c nghim. Cu III: Trong khng gian Oxyz cho cc im A(3; 5; 5); B(5; 3; 7) v mt phng (P): x + y + z = 0 1. Tm giao im I ca ng thng AB vi mt phng (P). 2. Tm im M (P) sao cho MA2 + MB2 nh nht. Cu IV:
1. Tnh din tch hnh phng gii hn bi cc ng thng y = 0 v 1xx1xy 2
.
2. Chng minh rng h
1xx2007e
1yy2007e
2
y
2
x
c ng 2 nghim tha mn iu kin
x > 0, y > 0 Cu Va:
1. Tm x, y N tha mn h
66CA
22CA2x
3y
3y
2x
2. Cho ng trn (C): x2 + y2 8x + 6y + 21 = 0 v ng thng d: 01yx . Xc nh ta cc nh hnh vung ABCD ngoi tip (C) bit A d Cu Vb: 1. Gii phng trnh 21x2log1xlog 3
23
2. Cho hnh chp SABCD c y ABCD l hnh vung tm O, SA vung gc vi y hnh chp. Cho AB = a, SA = a 2 . Gi H v K ln lt l hnh chiu ca A ln SB, SD. Chng minh SC (AHK) v tnh th tch hnh chp OAHK.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
35
S 33
Cu I: Cho hm s x2
m1xy
(Cm)
1. Kho st v v th hm s vi m = 1 2. Tm m th (Cm) c cc i ti im A sao cho tip tuyn vi (Cm) ti A ct trc Oy ti B m OBA vung cn. Cu II:
1. Gii phng trnh: gxcottgxxsinx2cos
xcosx2sin
2. Tm m phng trnh : 01xmx13x4 4 c ng 1 nghim Cu III: Trong khng gian Oxyz cho cc im A(2;0;0); M(0;3;6) 1. Chng minh rng mt phng (P): x + 2y 9 = 0 tip xc vi mt cu tm M, bn knh MO. Tm ta tip im. 2. Vit phng trnh mt phng (Q) cha A, M v ct cc trc Oy, Oz ti cc im tng ng B, C sao cho VOABC = 3. Cu IV: 1. Trong mt phng ta Oxy, tnh din tch hnh phng gii hn bi cc ng y = x2 v 2x2y .
2. Gii h phng trnh:
xy9y2y
xy2y
yx9x2x
xy2x
2
3 2
23 2
Cu Va: 1. Tm h s ca x8 trong khai trin (x2 + 2)n, bit: 49CC8A 1n2n3n . 2. Cho ng trn (C): x2 + y2 2x + 4y + 2 = 0. Vit phng trnh ng trn (C') tm M(5, 1) bit (C') ct (C) ti cc im A, B sao cho 3AB . Cu Vb:
1. Gii phng trnh: 1xlog1
43logxlog23
x93
2. Trong mt phng (P) cho na ng trn ng knh AB = 2R v im C thuc na ng trn sao cho AC = R. Trn ng thng vung gc vi (P) ti A ly
im S sao cho o60SBC,SAB
. Gi H, K ln lt l hnh chiu ca A trn SB, SC. Chng minh AHK vung v tnh VSABC ?
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
36
S 34
Cu I: Cho hm s 1x21xy
(C)
1. Kho st v v th hm s. 2. Vit phng trnh tip tuyn vi (C), bit rng tip tuyn i qua giao im ca ng tim cn v trc Ox.
Cu II: 1. Gii phng trnh: 1xcos12
xsin22
2. Tm m phng trnh: m54x6x4x23x c ng 2 nghim
Cu III: Cho ng thng d: 11z
12y
23x
v mt phng (P):
02zyx 1. Tm giao im M ca d v (P). 2. Vit phng trnh ng thng nm trong (P) sao cho d v khong cch t M n bng 42 . Cu IV:
1. Tnh
1
02 dx4x
1xxI
2. Cho a, b l cc s dng tha mn ab + a + b = 3.
Chng minh: 23ba
baab
1ab3
1ba3 22
.
Cu Va: 1. Chng minh vi mi n nguyn dng lun c 2 10 1 2 11 ... 2 1 1 0n nn nn n n nnC n C C C
. 2. Trong mt phng Oxy cho im A(2, 1) ly im B thuc trc Ox c honh x 0 v im C thuc trc Oy c trung y 0 sao cho ABC vung ti A. Tm B, C sao cho din tch ABC ln nht. Cu Vb:
1. Gii bt phng trnh: 221 22
1 1log 2x 3x 1 log x 12 2
.
2. Cho lng tr ng ABCA1B1C1 c y ABC l tam gic vung aACAB , AA1 = a 2 . Gi M, N ln lt l trung im ca on AA1 v BC1. Chng minh MN l ng vung gc chung ca cc ng thng AA1 v BC1. Tnh 11BCMAV .
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
37
S 35
Cu I: Cho hm s 1x
xy
(C)
1. Kho st v v th hm s. 2. Vit phng trnh tip tuyn d ca (C) sao cho d v hai tim cn ca (C) ct nhau to thnh mt tam gic cn. Cu II: 1. Gii phng trnh: (1 tgx)(1 + sin2x) = 1 + tgx
2. Tm m h phng trnh :
1xyx
0myx2 c nghim duy nht
Cu III: Cho mt phng (P): x 2y + 2z 1 = 0 v cc ng thng:
2z
33y
21x:d1
v 55z
4y
65x:d2
1. Vit phng trnh mt phng (Q) cha d1 v (Q) (P). 2. Tm cc im M d1, N d2 sao cho MN//(P) v cch (P) mt khong bng 2.
Cu IV:
1. Tnh
2
0
2 xdxcosxI
2. Gii phng trnh: xx
2 2x1x12log .
Cu Va: 1. T cc ch s 0, 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s t nhin chn m mi s gm 4 ch s khc nhau. 2. Trong mt phng Oxy cho cc im A(0, 1) B(2, 1) v cc ng thng: d1: (m 1)x + (m 2)y + 2 m = 0
d2: (2 m)x + (m 1)y + 3m 5 = 0 Chng minh d1 v d2 lun ct nhau. Gi P = d1 d2. Tm m sao cho PBPA ln nht Cu Vb: 1. Gii phng trnh: 022.72.72 xx21x3 . 2. Cho lng tr ng ABCA1B1C1 c tt c cc cnh u bng a. M l trung im ca on AA1. Chng minh BM B1C v tnh d(BM, B1C).
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
38
S 36 PHN CHUNG CHO TT C CC TH SINH Cu I (2im) Cho hm s y = x3 + 3mx2 + (m + 1)x + 1 (1), m l tham s thc 1. Kho st s bin thin v v th ca hm s (1) khi m = - 1. 2. Tm cc gi tr ca m tip tuyn ca th hm s (1) ti im c honh x = -1 i qua A(1; 2). Cu II (2im). 1. Gii phng trnh: tanx cotx = 4cos22x
2. Gii phng trnh : 2(2x - 1)2x + 1 + 3 - 2x =
2
Cu III (2im). Trong khng gian Oxyz, cho hai ng thng:
1 25x - 6y - 6z + 13 = 0x - 3 y - 3 z - 3d : = = , d : x - 6y + 6z - 7 = 02 2 1
1. Chng minh d 1 v d 2 ct nhau. 2. Gi I l giao im ca d 1 v d 2 . Tm cc im A, B ln lt thuc d 1 , d 2 sao
cho tam gic IAB cn ti I v c din tch bng 4142 .
Cu IV (2im)
1. Tnh tch phn 3
312
2 2xdxIx
.
2. Gii phng trnh: sin x - 4e = tanx
PHN RING - TH SINH CH C LM 1 TRONG 2 CU: V.a HOC V.b Cu V.a. Theo chng trnh KHNG phn ban (2 im) 1. Cho tp hp E = {0, 1, 2, 3, 4, 5, 7}. Hi c bao nhiu s t nhin chn gm 4 ch s khc nhau c thnh lp t cc ch s ca E. 2. Trong mt phng Oxy, cho tam gic ABC vi ng cao k t nh B v ng phn gic trong gc A ln lt c phng trnh 3x + 4y + 10 = 0 v x - y + 1 = 0, im M(0; 2) thuc ng thng AB ng thi cch C mt khong bng
2 . Tm to cc nh ca tam gic ABC. Cu V.b. Theo chng trnh phn ban (2 im)
1. Gii bt phng trnh logarit 1 22
2x + 3log log 0x + 1
.
2. Cho hnh chp S.ABC c y ABC vung cn ti nh B, BA = BC = 2a, hnh chiu vung gc ca S trn mt phng y (ABC) l trung im E ca AB v SE = 2a. Gi I, J ln lt l trung im ca EC, SC; M l im di ng trn tia i ca tia BA sao cho 0ECM = ( < 90 ) v H l hnh chiu vung gc ca S trn MC. Tnh th tch ca khi t din EHIJ theo a, v tm th tch ln nht.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
39
S 37 PHN CHUNG CHO TT C CC TH SINH Cu I (2im). Cho hm s y = x4 - 8x2 + 7 (1) 1. Kho st s bin thin v v th hm s (1). 2. Tm cc gi tr thc ca tham s m ng thng y = mx - 9 tip xc vi th hm s (1). Cu II (2im)
1. Gii phng trnh 2sin 2x - = sin x - + 4 4 2
.
2. Gii phng trnh 2 21 31
1 1x
x x
.
Cu III (2im) Trong khng gian Oxyz cho mt phng (P): 2x + 3y - 3z + 1 = 0, ng thng d:
3 52 9 1
x y z v 3 im A(4; 0; 3), B(- 1; - 1; 3), C(3; 2; 6).
1. Vit phng trnh mt cu (S) i qua ba im A, B, C v c tm thuc mf(P). 2. Vit phng trnh mt phng (Q) cha ng thng d v ct mt cu (S) theo mt ng trn c bn knh ln nht.
Cu IV (2im). 1. Tnh tch phn
2
0
sin2xdxI = 3 + 4sinx - cos2x .
2. Chng minh phng trnh 4x(4x2 + 1) = 1c ng ba nghim thc phn bit. PHN RING - TH SINH CH C LM 1 TRONG 2 CU: V.a HOC V.b Cu V.a. Theo chng trnh KHNG phn ban (2 im)
1. Tm h s ca s hng cha x5 trong khai trin nh thc Newton ca (1 + 3x)2n , bit rng 3 2n nA + 2A = 100 (n l s nguyn dng) 2. Trong mt phng Oxy cho ng trn (C): x2 + y2 = 1. Tm tt c cc gi tr thc m trn ng thng y = m tn ti ng hai im m t mi im c th k c 2 tip tuyn vi (C) sao cho gc gia hai tip tuyn bng 60o. Cu V.b. Theo chng trnh phn ban (2 im)
1. Gii phng trnh 3
1 63 log 9log x
xx x
.
2. Cho hnh chp S.ABC m mi mt bn l mt tam gic vung, SA = SB = SC = a. Gi M, N, E ln lt l trung im ca cc cnh AB, AC, BC; D l im i xng ca S qua E; I l giao im ca ng thng AD vi mf(SMN). Chng minh rng AD vung gc vi SI v tnh theo a th tch ca khi t din MBSI.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
40
S 38 PHN CHUNG CHO TT C CC TH SINH Cu I (2im). Cho hm s 3 2y = x - 3x - 3m(m + 2)x - 1 (1) , m l tham s thc 1. Kho st s bin thin v v th ca hm s (1) khi m = 0. 2. Tm cc gi tr ca m hm s (1) c hai cc tr cng du. Cu II (2im)
1. Gii phng trnh 12sin x + - sin 2x - = 3 6 2
.
2. Gii phng trnh 10x + 1 + 3x - 5 = 9x + 4 + 2x - 2 . Cu III (2im)
Trong khng gian Oxyz cho ng thng d 1 c phng trnh : x - 3 y z + 5 = =
2 9 1
v hai im A(5; 4; 3), B(6; 7; 2) 1. Vit phng trnh ng thng d 2 qua 2 im A, B. Chng minh rng hai ng thng d 1 v d 2 cho nhau. 2. Tm im C thuc d 1 sao cho tam gic ABC c din tch nh nht. Tnh gi tr nh nht . Cu IV (2im)
1. Tnh 2
0
x + 1I = dx4x + 1 .
2. Cho 3 s dng x, y, z tho mn h thc yzx + y + z = 3x
. Chng minh rng
2 3 - 3x (y + z)6
.
PHN RING - TH SINH CH C LM 1 TRONG 2 CU: V.a HOC V.b Cu V.a. Theo chng trnh KHNG phn ban (2 im)
1. Cho s nguyn n tho mn ng thc 3 3n nA + C = 35
(n - 1)(n - 2)( n 3 ). Tnh tng
2 2 2 3 n 2 nn n nS =2 C - 3 C + ... + (-1) n C 2. Trong mt phng Oxy, cho tam gic ABC vi AB = 5, C(- 1; - 1), ng thng AB c phng trnh x + 2y - 3 = 0 v trng tm ca tam gic ABC thuc ng thng x + y - 2 = 0. Hy tm to cc nh A v B. Cu V.b. Theo chng trnh phn ban (2 im)
1. Gii phng trnh 2 12
2log 2 2 log 9 1 1x x
2. Cho hnh chp ABCD c y ABCD l hnh vung c cnh bng a, SA = a 3 v SA vung gc vi mt phng y. Tnh theo a th tch khi t din S.ACD v tnh cosin ca gc gia hai ng thng SB, AC.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
41
S 39 PHN CHUNG CHO TT C CC TH SINH
Cu I (2im). Cho hm s 2x + (3m - 2)x + 1 - 2my =
x + 2 (1), m l tham s thc
1. Kho st s bin thin v v th ca hm s (1) khi m = 1. 2. Tm cc gi tr m hm s (1) ng bin trn tng khong xc nh ca n. Cu II (2im)
1. Gii phng trnh 2 x3sinx + cos2x + sin2x = 4sinxcos2
2. Gii h phng trnh
3
4
x - 1 - y = 8 - x
x - 1 = y
Cu III (2im). Trong khng gian Oxyz cho 3 im A(1; 0; - 1), B(2; 3; - 1);,
C(1; 3; 1) v ng thng d: 1 0
4x yx y z
1. Tm to im D thuc ng thng d sao cho th tch khi t din ABCD bng 1. 2. Vit phng trnh tham s ca ng thng i qua trc tm H ca tam gic ABc v vung gc vi mf(ABC). Cu IV (2im)
1. Tnh tch phn 1 3
20
x dxI = 4 - x
.
2. Cho s nguyn n( n 2) v hai s thc khng m x, y. Chng minh:
n n n + 1 n + 1n n + 1x + y x + y .
PHN RING - TH SINH CH C LM 1 TRONG 2 CU: V.a HOC V.b Cu V.a. Theo chng trnh KHNG phn ban (2 im)
1. Chng minh rng vi n l s nguyn dng
n 0 n - 1 1 0 n n + 1
n n n2 C 2 C 2 C 3 - 1 + + ... + = n + 1 n 1 2(n + 1)
.
2. Trong mt phng Oxycho hai im A(3; 0), B(0; 4). Chng minh rng ng trn ni tip tam gic OAB tip xc vi ng trn i qua trung im cc cnh ca tam gic OAB. Cu V.b. Theo chng trnh phn ban (2 im)
1. Gii phng trnh 2x + 1 2x + 1 x3 - 2 - 5.6 0 . 2. Cho t din ABCD c cc mt ABC v ABD l cc tam gic u cnh a, cc m ACD v BCD vung gc vi nhau. Hy tnh theo a th tch ca khi t din ABCD v tnh s o ca gc gia hai nng thng AD v BC.
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S 40 PHN CHUNG CHO TT C CC TH SINH
Cu I (2im). Cho hm s 3 11
xyx
(1)
1. Kho st s bin thin v v th ca hm s 3 11
xyx
(1)
2. Tnh din tch ca tam gic to bi cc trc to v tip tuyn vi th hm s (1) ti im M(-2; 5). Cu II (2im) 1. Gii phng trnh 4 44(sin x + cos x) + cos4x + sin2x = 0 . 2. Gii phng trnh 2 2( 1)( 3) 2 3 2 ( 1)x x x x x . Cu III (2im). Trong khng gian Oxyz cho mt phng ( ) : 2x - y + 2z + 1 = 0 v ng thng
d: x - 1 y - 1 z= = 1 2 - 2
1. Tm to giao im ca d vi ( ) ; tnh sin ca gc gia d v ( ) . 2. Vit phng trnh mt cu c tm thuc d v tip xc vi hai mt phng ( ) v Oxy. Cu IV (2im)
1. Tnh 1
2x
20
xI = xe - dx4 - x
2. Cho cc s thc x, y tha 0 ,3
x y . Chng minh rng:
cosx + cosy 1 + cos(xy) PHN RING - TH SINH CH C LM 1 TRONG 2 CU: V.a HOC V.b Cu V.a. Theo chng trnh KHNG phn ban (2 im)
1. Chng minh rng vi n l s nguyn dng, ta c: n - 1 2 n - 2 n n n - 1n n n2C + 2 C + ... + n.2 C = 2n.3 2. Trong mt phng Oxy cho ng trn (C): (x - 4)2 + y2 = 4 v im E(4; 1). Tm to im M trn trc tung sao cho t M k c hai tip tuyn MA, MB n ng trn (C) vi A, B l cc tip im sao cho ng thng AB qua E. Cu V.b. Theo chng trnh phn ban (2 im)
1. Gii bt phng trnh 2 22x - 4x - 2 2x - x - 12 - 1 6 .2 - 2 0 . 2. Cho t din ABCD v cc im M, N, P ln lt thuc cc cnh BC, BD, AC sao cho BC = 4BM, AC = 3AP, BD = 2BN. Mt phng (MNP) ct AD ti Q. Tnh t s AQAD v t s th tch hai phn ca khi t din ABCD c phn chia bi mt phng
(MNP).
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S 41 I. PHN CHUNG CHO TT C CC TH SINH (7, 0 im) Cu I. (2 im) Cho hm s y = - x3 - 3x2 + mx - 4, trong m l tham s thc, (1). 1. Kho st s bin thin v v th ca hm s (1) cho, vi m = 0. 2. Tm tt c cc gi tr ca tham s m hm s (1) cho ng bin trn khong (0; 2) Cu II. (2 im)
1. Gii phng trnh 2
2
tan t anx 2 sintan 1 2 4
x xx
2. Tm tt c cc gi tr ca tham s m phng trnh 24 2 4 1x x x m c ng mt nghim thc. Cu III. (2 im) Trong khng gian Oxyz, cho im A(5; 5; 0) v ng thng d:
1 1 72 3 4
x y z
1. Tm ta im A' i xng vi im A qua ng thng d. 2. Tm ta cc im B, C thuc d sao cho tam gic ABC vung ti C v BC = 29 Cu IV. (2 im)
1. Tnh tch phn 1
2
0
( 1) .xI x x e dx
2. Gii h phng trnh
2 2
2 2
2 2
36 60 25 036 60 25 036 60 25 0
x y x yy z y zz y z x
II. PHN RING. Th sinh ch c lm 1 trong 2 cu: V.a hoc V.b Cu V.a. Theo chng trnh KHNG phn ban (2 im) 1. C bao nhiu s t nhin gm 4 ch s khc nhau m mi s u ln hn 2500. 2. Trong mt phng Oxy, tm ta cc nh ca tam gic ABC bit rng ng thng AB, ng cao k t A v trung tuyn k t B ln lt c phng trnh l x + 4y - 2 = 0, 2x - 3y + 7 = 0, 2x + 3y - 9 = 0. Cu V.b. Theo chng trnh phn ban (2 im)
1. Gii phng trnh 5 1 2 5 1 3.2 .x x x . 2. Cho hnh chp S.ABC c y ABC l tam gic vung cn nh B, AB = a, SA = 2a v SA vung gc vi mt phng y. Mt phng qua A vung gc SC ct SB, SC ln lt ti H, K. Tnh theo a th tch khi t din SAHK.
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PHN TH HAI
HNG DN GII
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S 1
Cu I: 1. Kho st s bin thin v v th ca hm s khi m = 8. (Hc sinh t gii). 2. Xc nh, th hm s ct trc honh ti 4 im phn bit. Bi ton quy v xc nh m phng trnh x4 - mx2 + m - 1 = 0 c 4 nghim phn bit. t2 - mt + m - 1 = 0 (t = x2) c 2 nghim t1, t2 dng phn bit.
22 4 1 2 0m m m
2 24( 1) 0 ( 2) 02
0 01
1 0 1
m m mm
S m S mm
P m P m
Cu II: 1. Bt phng trnh 2 11 1
2 2
log 4 4 log 2 3.2x x x
2 12 3.2 4 4 2.4 3.2 4 4 4 3.2 4 0
2 12
2 4
x x x x x x x x
x
xx
2. 2(sin4x + cos4x) + cos4x + 2sin2x + m = 0 (1) 2(1 - 2sin2xcos2x) + cos4x + 2sin2x + m = 0 3 + m - 3sin22x + 2sin2x = 0 3t2 - 2t - (m + 3) = 0; t = sin2x (2)
Vi 0 0 2 0 12
x x t , nhn mi gi tr thuc on
[0;1]. phng trnh (1) c t nht mt nghim thuc
on 0;2
, iu kin cn v l phng trnh (2)
c nht mt nghim t [0;1] 3t2 - 2t = m + 3 c t nht mt nghim t [0; 1]. t f(t) = 3t2 - 2t (0 t 1)
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V d th (C) ca hm y = f(t) (0 t 1).
f 1 13 3
T th suy ra, phng trnh 3t2 - 2t = m + 3 c t nht mt nghim t [0; 1], iu kin cn v l:
1 103 1 23 3
m m
Cu III: 1. K AH BC, AO SH. V SA (ABC), BC AH nn BC SH. T BC (SAH), suy ra BC AO. Do AO (ABC). V vy AO l khong cch t A
n (SAB). ABC u nn AH = 32
a . SA (ABC) nn SA (ABC) nn SA
AH. Do SAH vung ti A. Ta c:
2 2 21 1 1
AO AH SA 2 2 2
4 2 23 3a a a
22
aAO
2. Tnh tch phn 31
20 1xI dx
x
Cch 1. t u = x2 + 1 du = 2xdx 31
20 1xI dx
x
=
2 2
1 1
1 (u - 1) 1 1 1du 1 - du 1 ln 22 u 2 u 2
Cch 2. t x = tant dx = 2osdt
c t
34 4 4
32 2 2
0 0 0
tan t d t 1I = . = tan t.d t = tan t - 1 d t1 + tan t co s t co s t
24 4 4
40
0 0 0
sin td t tg t= tan t.d tan t - = + ln co stcost 2
1 1 1 1 1 1= + 1n - 1n1 = + 1n = 1 - 1n22 2 2 2 22
.
Cu IV:
y
1
y = m+3 23
13
x
O 1
B H
C
O A
S
a
Hnh 13
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1. To giao im ca (C1) v (C2) l nghim ca h:
2 2
2 22 2
7 10 010 010 04 2 20 0
x yx y xx y xx y x y
Rt y t phng trnh u, th vo phng trnh th 2 ta c. x2 + (7x - 10)2 - 10x = 0 x2 - 3x + 2 = 0 x = 1, x = 2. Vi x = 1 ta c y = -3 Vi x = 2 ta c y = 4 Vy 2 giao im ca (C1) v (C2) l A1 (1;-3) A2 (2;4).
Trung im A ca A1A2 c to A3 1;2 2
. Ta c 1 2 1;7A A
. ng thng
qua A vung gc vi A1A2 c phng trnh
3 11.3 7 02 2
x y
hay: x + 7y - 5 = 0 To tm I ca ng trn cn tm l nghim ca h:
7 5 06 6 0
x yx y
(12; 1)I
ng trn cn tm c bn knh:
R = IA2 = 2 22 12 4 1 125
Vy ng trn cn tm c phng trnh: (x- 12)2 + (y + 1)2 = 125 2. Ta c: (C1): (x - 5)2 + y2 = 52 (C2): (x + 2)2 + (y - 1)2 = 52 (C1) c tm I1(5; 0) bn knh 5 (C2) c tm I2(-2; 1) bn knh 5 (C1) v (C2) ct nhau nn ch c hai tip tuyn ngoi.
Ta c 1 2 7;1I I
. Do d ng thng I1I2 c phng trnh:
1. (x + 2) + 7 (y - 1) = 0 hay: x + 7y - 5 = 0 Do tip tuyn chung ca (C1) v (C2) c phng trnh dng: x + 7y + d = 0 Khong cch t I1 n tip tuyn l
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2 2
55 5 25 2 5 25 2 5 25 2
1 7
dd d d
Vy cc tip tuyn phi tm l: x + 7y - 5 + 25 2 = 0 x + 7y - 5 - 25 2 = 0 Cu V:
1. Gii phng trnh 24 4 2 12 2 16x x x x iu kin: x - 4 0 x 4. Vi iu kin x 4, phng trnh tng ng vi
4 4 4 4 12 2 4. 4x x x x x x
24 4 4 4 12x x x x (1) t 4 4 0x x t t
Phng trnh (1) t2 - t - 12 = 0 4
3tt
t = 4 4 4 4x x (1)
22 2 16 16
4x x
x
2 16 8x x
2 2
4 816 64 16x
x x x
x = 5
2. Tng s cch chn 8 hc sinh t 18 em ca i tuyn l
81818.17.16.15.14.13.12.11 43758
8.7.6.5.4.3.2.C
Tng s cch trn c phn lm hai b phn ri nhau: B phn I gm cc cch chn t i tuyn ra 8 em sao cho mi khi u c em c chn (s cch phi tm). B phn II gm cc cch chn t i tuyn ra 8 em ch gm hai khi (lu l s em thuc mi khi u t hn 8 nn khng c cch chn no c 8 em thuc cng cng mt khi). Ring b phn II c th phn tch thnh 3 loi:
8 em c chn t khi 12 hoc khi 11: C 813C cch chn 513C 8 em c chn t khi 12 hoc khi 10: C 812C cch chn 412C
(loi)
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8 em c chn t khi 11 hoc khi 10: C 811C cch chn 311C S cch phi tm s l:
8 5 4 318 13 12 11 43758 1947 41811C C C C (cch). Cu VI. Ta c:
2 2 2
2 2 2 2a b c a b ca b c
R R R R
= a sinA + b sinB + c sinC = 2 2 2S S Sa b cbc ca ab
= 2Sa b cbc ca ab
Mt khc 2S = ax + by + xy + cz. Do :
2 2 2
ax2
a b c a b cby czR bc ca ab
Ta c, theo bt ng thc Bunhicpski:
2
1 1 1 1 1 1ax ax+by+cz2 2 2
b c c a a bby cza c b b a c c b a a b c
x y z
Suy ra: 2 2 2
2a b cx y z
R
(1)
Theo di phn chng minh trn, ta thy du "=" trong (1) xy ra khi v ch khi:
2b c c a a b
c b a c b aa x b y c z
a b cx y z
ABC u M trng vi trng tm G ca tam gic ABC.
Nhn xt: Cng c th c lng: 1 1 1a b cbc ca ab a b c
nh p dng
bt ng thc Csi cho mi cp s hng v tri:
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2 2 2
1 1 1 1 1 1 1 1.2 .2 .22 2 2 2 2
1 1 1
a b c a c c abc ca ab bc ab ab bc c a b
a b c
C
ch 2: C th lm cch khc nh sau:
1 1 1. ax .x y z by cza b c
2 2 2
1 1 1 1 1 1 1 1 1(ax ) .22
2 2
abcby cz Sa b c a b c a b c R
ab bc ca a b cR R
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S 2 Cu I: 1. Tm n N* tho mn bt phng trnh 3 22 9nn nA C n
iu kin: n N*, n 3. Bt phng trnh n(n - 1) (n - 2) + 22 9nC n
n(n - 1) (n - 2) + n(n - 1) 9 (n - 1)(n - 2) + n - 1 9
2 2 8 03 4 ( *)
n nn n N
Kt hp iu kin n 3 suy ra n = 3 hoc n = 4.
2. Phng trnh 84 221 1log 3 log 1 log (4 )2 4
x x x
iu kin:01
xx
Phng trnh 84 221 1log ( 3) log ( 1) log (4 )2 4
x x x
2 2 2log ( 3) log 1 log (4 )
( 3). 1 4
x x x
x x x
i. Nu x > 1, phng trnh (x + 3)( x -1) - 4x = 0
2 2 3 0
31
x xx
x
ii. Nu 0 < x < 1, phng trnh (x + 3)( 1 - x) - 4x = 0
2 6 3 0
3 2 20 1x x
xx
p s: phng trnh c 2 nghim 3
3 2 3
x
x
Cu II:
1.2 2
2 2x x m my x
x x
2
2 2
4 4' 1( 2) ( 2)
m x x myx x
hm s nghch bin trn on 1;0 , iu kin cn v l
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' 0 [ 1;0y x
2
[ 1;0 ]
( ) 4 4 , 1; 0max ( ) ( 1) 9g x x x m x
g x m g m m
2. kho st v v th hm s khi m = 1. ( Hc sinh t gii) 3. Phng trnh.
2 2
2
1 1 1 1
2
1 1
9 ( 2)3 2 1 0
2 1 ( 2) (1)
3
t t
t
a a
X X a X
X
Do 21 1 1 2,t t m 1 - t2 0. Suy ra min gi tr ca 21 13 tX l on 3;9 .
2 2 1
(1) 23 9
X X aXX
T th v cu 2, hn ch trong on 3;9 suy ra, phng trnh c nghim,
iu kin cn v l: 6447
a
* Ch : Bn c th thy th v hn, nu tm a phng trnh sau c nghim: 2 2t + 1 - t t + 1 - t9 - (a + 2).3 + 2a + 1 = 0 Cu III:
1. Gii phng trnh 4 4s in os 1 1cot 25sin 2 2 8sin 2
x c x xx x
iu kin: sin2x 0 Vi iu phng trnh:
2 21 2in cos 1 1os25 2 8x x c x
2 9os 2 5 os2 04
9os2 12 2 6 21 3cos 22
c x c x
c xx k
x
2. trong tam gic ABC h thc: bsinC (bcosC + ccosB) = 20
(loi)
6x k
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4R2sinBsinC (sinBcosC + sinCcosB) = 20 4R2sinBsinCsinA = 20
M 3
28 sin Asin sin 2 sin Asin sin4 4abc RS B C R B C
R R
Vy ta c: S = 10 (n v din tch) Cu IV: 1. cho gn, k hiu BC = a, AC = b, AC = c, OA = x, OB = y, OC = z. Theo gi thit ta c:
2 2 2
2 2 2
2 2 2
2 2 2
2 2 2
2 2 2
a y zb x zc x y
a b cb c ac a b
Suy ra, ABC l tam gic nhn, nn trc tm H nm trong tam gic. Gi AM, BK, CN l 3 ng cao ca tam gic ABC. Theo gi thit OA (OBC)
.BC OA
BC OMBC AM
Suy ra OMA = , OKB = ,ONC =
Ta c: cos = OMAM
.
Trong tam gic vung BOC ta c: 2 2 2 2 21 1 1 1 1
OM OB OC y z
Suy ra: OM2 = 2 2
2 2
y zy z
OM = 2 2
yzy z
Trong tam gic vung AOM:
AM2 = OA2 + OM2 = x2 + 2 2
2 2
y zy z
B
C
O
A
N
A K
H
Hnh 15
M
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2 2 2 2 2 2 2 2 2 2 2 2
22 2 2 2
x y y z x z x y y z z xAM AMy z y z
Vy, cos = 2 2 2 2 2 2
OM yzAM x y y z z x
Lp lun tng t ta cng c:
cos = 2 2 2 2 2 2
xzx y y z z x
cos = 2 2 2 2 2 2
xyx y y z z x
Suy ra, cos + cos + cos =
2 2 2 2 2 2 2
313
xy yz xz xy yz xzx y y z z x xy yz xz
(pcm)
Cch khc: cos + cos + cos = 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 2 2 2
3( )3
x y y z z xxy yz xzx y y z z x x y y z z x
2. a) (P) c vect php 1; 1;1n
. ng thng (d) qua A v (P) c phng
trnh:
1 3 21 1 1
x y z
Gi I l giao im ca (d) v (P), khi c to ca I l nghim ca h:
3 0
1 3 21 1 1
x y zx y z
Suy ra I(-2; -2; -3). A' l im i xng ca A qua (P) (d) v I l trung im ca AA'. Do A' c to (xo; yo; zo)
01 22
x , 03 2
2y
, 02 32
z
x0 = -3; y 0 = -1, zo = 4. Vy A' (-3; -1; -4). b) 'BA
= (2; -8; -16) = 2(1; -4; -8). BA' c phng trnh:
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3 1 41 4 8
x y z
Gi M l giao im ca BA' vi (P) th to ca M l nghim ca h
3 03 1 4 4;3;4
1 4 8
x y zx y z M
Ta c: MA + MB = MA' + MB A'B. Do MA + MB t gi tr nh nht khi MA' + MB t gi tr nh nht bng A'B. Vy gi tr nh nht ca biu thc MA +
MB l A'B = 2 2 22 8 16 18 , khi M l giao im ca ng thng A'B vi mt phng (P). Cu V:
Ta c: I =
1 3 1 3 1 3 3
233 20 0 0
11 1
1 1
xn n nxx x
x
d ee dx e d eex e
= 1
1 32 1 11 3 02 2
0
12[ 1 1 ]1
2
x nxn
x
ee e
= 1 12 2 1 12 4 2 2 2 1
2 2
.
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S 3
Cu I:
1. a) Kho st v v th hm s khi m = 12
.
(Hc sinh t gii). Gi (C) l th hm s. b) Vit phng trnh tip tuyn ca (C) bit tip tuyn y song song vi ng thng y = 4x + 2.
Ta c hm s y = 3 21 1 423 2 3
x x x
y' = x2 + x - 2 Theo gi thit tip tuyn th phi tm c h s gc k =4. Vy c:
2 22 4 6 0xx x x 1 1
2 2
22;313;6
x y
x y
Vy, c 2 tip tuyn tho mn iu kin bi.
Tip tuyn (d1): 2 264 2 43 3
y x y x
Tip tuyn (d2): 1 734 3 46 6
y x y x
2. Do 0 < m < 56
nn:
1 1(0) 2 03 3
y m
8 1 5(2) 2 4 2 03 3 3
y m m
Li c: y' = x2 + 2mx - 2 y" = 2x + 2m > 0, x [0; 2]
Suy ra th hm s 3 21 12 22 3
y x mx x m lm trn on [0; 2]. Kt hp
vi y(0) < 0; y(2) < 0 suy ra y < 0, x [0; 2]. Do :
2 2 2
2 3
0 0 0
1 12 23 3
S y dx ydx x mx x m dx
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4 3 2
23 12 012 3 xx xm x m x
4 8 2 4 104 43 3 3 3 3
mm m
Theo gi thit S = 4 m = 12
(tho mn iu kin 0 < m < 56
)
Ch : Khng cn dng tnh "lm" ca th hm s trn [0; 2] chng minh y < 0, x [0; 2] nh sau:
1 1(0) 2 03 3
y m
8 1 5(2) 2 4 2 03 3 3
y m m
Li c: y' = x2 + 2mx - 2 y" = 2x + 2m > 0, x [0; 2] Suy ra y' ng bin, lin tc trn [0; 2], vi tp gi tr [-2; 2 + 2m], nn i du t m sang dng trn [0; 2]. Do hm s cho nghch bin ri chuyn sang ng bin, lin tc trn [0; ]. ng thi vi g(0) < 0 v g(2) < 0, ta c pcm. Cu II: Gii h phng trnh:
4 2
4 3 (1)
log log 0 (2)
x y
x y
iu kin: 42
log 0 1log 0 1
x xy y
Vi iu kin phng trnh (2) log4x = log2y log2x = log2y2 x = y2 Phng trnh (1) y2 - 4y + 3 = 0 (do y 1) y = 1; y = 3 x = 1; x = 9 Vy, phng trnh c hai nghim (1;1) v (9;3)
2. Phng trnh tg4x + 1 = 2
4
2 sin 2 sin 3os
x xc x
iu kin: cos 0 sinx 1 Phng trnh sin4x + cos4x = (2 - sin22x) sin3x
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
58
2 211 sin 2 2 sin 2 sin 32 x x x
2 22 sin 2 2 sin 2 .2sin 3x x x
sin3x = 12
(d thy tho mn iu kin)
3 2
653 26
x k
x k
218 35 218 3
kx
kx
(k Z)
Cu III: 1. K AH BE. Do SA (ABC) nn BE SH. Do SH l khong cch t S
n BE. Ko di BE ct AD ti M. E l trung im ca CD nn ED = 2 2a AB D
l trung im ca AM AM = 2a ABM vung ti A ta c: SAH vung ti A ta c:
SH = 2
2 2 2 4 3 55 5a aSA AH a
a B C
E
D M
S
A
a
H
Hnh 16
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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59
2. (d) l hnh chiu vung gc ca trn (P) th (d) l giao tuyn ca (P) v mt phng (Q) cha v vung gc vi (P). rng, mt phng (Q) cha c phng trnh dng: 2 1 2 0x y z x y z , 2 2 0 (1)
Tht vy, tt c cc im M(x; y) thuc u c ta tha: 2 1 0
2 0x y z
x y z
v do tha (1).
(1) 2 2x y z , 2 2 0
suy ra (Q) c vct php tuyn 2 ; ;Qn
Do (Q) (P) nn . 0 4. 2 2 1 0 7 3 0Q Pn n
Chn 3 ta c 7 Vy, (Q) c phng trnh x + 4y + 4z + 11 = 0 Nh th, (d) l giao ca hai mt phng:
4 2 1 0
4 4 11 0x y z
x y z
(d) c vc t ch phng [ , ]P Qa n n
= (4; 5; -6), trong , Pn
=(4; - 2; 1), Qn
=(1; 4;
4) v i qua im (1; 0; - 3).
T , suy ra phng trnh (d): 1 34 5 6
x y z
Cch 2. (d) l hnh chiu vung gc ca trn (P) th (d) l giao tuyn ca (P) v mt phng (Q) cha v vung gc vi (P). Mt phng (Q) i qua Mo (1; -3; 0) l mt im thuc v c cp ch phng gm mt vc t l vc t ch phng ca , mt vc t l vc t php tuyn ca (P).
1 2 1 2(2;1;1), (1;1;1) [ , ] (0; 1;1)n n n n
Vc t ch phng ca l (0, 1;1)a
.
Suy ra, vc t php tuyn ca (Q) l [ , ]Q Pn a n
=(1; 4; 4).
Vy phng trnh (Q): 1(x - 1) + 4(y + 3) + 4z = 0 x + 4y + 4z + 11 = 0. Cch 3. Trn () chn im Mo no chng hn Mo (1; -3; 0). Gi H l hnh chiu ca M0 trn (P). Gi I l giao im ca () vi (P). Suy ra phng trnh ng thng IH chnh l phng trnh hnh chiu ca () ln mt phng (P):
1 3
4 5 6x y z
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60
Cu IV:
1. Tm 3
0
1 11x
x ximx
3 3
0 00
1 1 1 1 1 11 lim limx xx
x x x ximx x x
Xt tng s hng:
00
11 lim21 1 1 1xx
x ximx x x
3
00 23 3
1 11 11 lim[1 1 1 ]
xx
xximx x x x
20 3 3
1 1131 1x
imx x x
3
0
1 1 1 112 3x
x ximx
5
6
2. (C1): x2 + y2 - 4y - 5 = 0 x2 (y-2)2 = 9 (C2): x2 + y2 - 6x + 8y + 16 = 0 (x-3)2 + (y + 4)2 = 9 (C1) c tm I1 (0; 2) bn knh R1 = 3 (C2) c tm I2 (3; -4); bn knh R2 = 3
V I1I2 = 22
1 23 6 45 3 5 R R nn (C1) v (C2) nm ngoi nhau, do
c 4 tip tuyn chung. V R1 = R2 = 3 nn d1 // d2 // I1I2
Phng trnh ng thng I1I2 = 0 2 2 2 2 0
3 6x y x y
Phng trnh d1, d2 c dng 2x + y + c = 0
Khong cch t I1 n d1, d2 bng 2 22
32 1
c
1
2
3 5 2 2 3 5 2 02 3 5
3 5 2 2 3 5 2 0
c x yc
c x y
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
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61
Hnh 17
Do tnh i xng, d3 v d4 ct nhai ti trung im I ca on I1I2 c to (3/2;1).
Phng trnh d3, d4 c dng y + 1 = k 3 31 02 2
kx kx y
Khong cch t I1 ti d3, d4 bng 2
32 12 3
1
k
k
Gii ra ta c : 1
2
0 144 333
k y
y xk
Cu V:
Cho , 0
54
x y
x y
Tm min S vi S = 4 14x y
Cch 1: S = 5
1 1 1 1 1 5 5.5 25 54 4 5. . . .4x x x x y x x x x yx x x x y
min S = 5
1 14
454
x yx y
x y
114
x
y
Cch 2: S = 4 1 ( )5 4
f xx x
0 < x < 5
4
I1 I2 I
d1
d2
d3
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
62
f(x) = 22
4 4 05 4x x
22 5 4
1504
x xx
x
Lp bng du f '(x) suy ra min S = 5
Cch 3: 1 2 1 4 12 . . .2 42
x y x yx yx y
(3)
Du "=" (3) khi
54
2 1 4 1. 2 . 5 1
4 4
x y xx x y y
x y yx y
(3) 25 5 4 1 4 1 5
2 4 4 4x y x y
Vy min S = 5.
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
63
S 4
Cu I:
1. Gii bt phng trnh: 12312 xxx iu kin: x 3.
Bt phng trnh 12312 xxx
12312 2 xxx (v x +12 > x-3 0)
2
2 9 2 12 3 2 1
12 . 3 4
9 52 0 13 4
x x x x
x x
x x x
Do iu kin x 3, suy ra 3 4x 2. Gii phng trnh
2tan cos cos sin (1 tan tan )2xx x x x x
iu kin:
1cos0cos
02
cos
0cos
xx
xx
Ta c: sin sin
21 tan tan 12 cos cos
2
xxxx xx
xxx
xx
xx
xxxx
cos1
2coscos
2cos
2coscos
2sinsin
2coscos
Phng trnh 2 sintan cos coscos
xx x xx
cosx(1 - cosx) = 0. Do iu kin cosx 0 nn phng trnh cosx = 1
x = 2k, (k Z) Cu II: 1. y = (x - m)3 - 3x y' = 3(x-m)2 - 3 = 3[(x - m)2 - 1]
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Trn Xun Bang - Trng THPT Chuyn Qung Bnh
D b thi i Hc 2002 - 2008 ra v Hng dn gii. 6/2010
64
y'(0) = 3(m2 -1) y" = 6(x - m) y"(0) = -6m iu kin cn v hm s t cc tiu ti x = 0 l y'(0) = 0 m = 1 hoc m = -1. Vi m = 1 th y"(0) = -6 < 0, hm t cc i ti x = 0 Vi m = -1 th y"(0) = 6> 0, hm t cc tiu ti x = 0 p s: m = -1 . 2. Kho st v v th hm s khi x = 1( Hc sinh t gii) th hm s y = (x - 1)3 - 3x c cho trn hnh 18. 3. Tm k h bt phng trnh c nghim
)2(1)1(log31log
21
)1(031
32
22
3
xx
kxx
iu kin: (x - 1)3 > 0 x > 1 Khi x > 1, bt phng trnh (1) (x - 3)3 - 3k < k (1') Bt phng trnh (2) log2x + log2(x - 1) 1 (x > 1). x(x - 1) 2
1022
xxx
1< x 2 Bi ton quy v xc nh k bt phng trnh (1') c nghim tha iu kin 1< x 2. Da vo th hm s v cu 2, xt ch trn khong 1< x 2, ta suy ra tp mi tr s k cn tm l k > -5 (k >
(1;2]min ( ) (2) 5f x f )
Cu III: 1. Gi H l trung im BC. Do ABC vung cn ti A nn AH BC; AB = AC SB = SC SH BC. Do AHS = 600.
Hnh 18
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Ta c AH = .22aBC
Trong SAH vung ti A nn:
SA = 0 3tan 602 2a a
.
2. a) d1 l giao tuyn ca hai mt phng x - az - a = 0 v y - z + 1 = 0 nn i qua M1(a; - 1; 0) v vung gc vi cc vc t
1u
= (1; 0; - a), 1v
= ( 0; 1; - 1).
Suy ra mt vc t ch phng ca d1 l 1a
= [ 1u
, 1v
] = (a; 1; 1).
d2 l giao tuyn ca hai mt phng ax + 3y - 3 = 0 v x - 3z - 6 = 0 nn i qua M2(0; 1; - 2) v vung gc cc vc t 2u
= (a; 3; 0), 2v
= ( 1; 0; - 3). Suy ra mt vc
t ch phng ca d2 l 2a
= [ 2u
, 2v
] = (3; - a; 1).
1 1M M
= (- a; 2; - 2), 1 2[ ; ]a a
= (1 + a; 3 - a; - a2 - 3)
Ta c 1 2[ ; ]a a
. 1 1M M
= a2 - 3a + 12 > 0, a. Suy ra, khng c a d1 v d2 ct nhau.
Ta cng c kt qu l d1 v d2 cho nhau, vi a. Cch 2. Phng trnh tham s ca cc ng thng d1 v d2:
1 2
'x = a + atd : y = - 1 + t d : 1 '
3z = t 12 '
3
x tay t
z t
Xt h phng trnh:
a + at ' (1)
1 1 ' (2)3
12 ' (3)3
tat t
t t
T (2) v (3) suy ra: (1 ) ' 12a t
a = - 1 khng tha
Hnh 19
H
S
C
B
A
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