Mô tả:
SCC GD VA DT PHU YEN
TR
G THPT CHUYEN
DE CHINH THUC
DE THI THPT QUOC GIA 2015 — 2016 LAN 1
Mfin: TOAN
Théi gran lém béi.’ 180 phiit, khong ké thc i gian phét
dé
Ngay thi 09/10/2015
Cau 1(1,0 diém) Khao sat su bién thién va ve dfi thi ctia ham
—2.x— 2)
b) Giai phuong trinh 4‘—3.2” " "—4" " " '= 0.
.
'
Cau 4 (1,0 diém). Tinh tich phan I ——
Cau 5 (1,0 diem). Trong khfing gian vcii he toa do Oxyz, cho hai mat phang co phuong trinh
(PQ : 2.x — 3y + 4z + 20 = 0 va (QQ : 4s — 13y— 6z + 40 = 0 . Chung minh (P) cat (Q) theo
giao tuyén 1:i duéng thang d. Viet phuong trinh duéng thang d.
Cau 6 (1,0 diém).
a) Giai phuong trinh sin4 + cos4 i+—
4
b) Trong mat phang toa do On. Ci gé›c phan tu thu nhat ta lay 2 diem phan biét; cu the ‹ cac
go) c phan tu thu hai, thu ba, thu tu ta lan lupt lay 3,4,5 diem phan biet (cac diem
khong nam trén cac true tpa do). Trong 14 diém dfi ta lay 2 diem bat ky. Tinh xac suat
de doan thang noi hai diem dfi cat hai tru.c tpa dfi.
Cau 7 (1,0 diém). Cho hinh chop 5.ABCD ct day ABCD la hinh chu nh(at v‹ i
AB —— a,AD —— at. Canh bén SA vufing goc vc›i day, canh SC tao vcii day goc 30‘. Gpi K la
hinh chiéu vufing gfic cua A trén SD. Tinh thé tich khoi chop 5.ABCD va khoang each giua hai
duéng
thang AK, SC.
Cau 8 (1,0 diém). Trong mat phang v‹ i he toa dfi Ory, cho hinh vufing ABCD cc› dinh C(2;—5)
v:i noi tiép ducrng tron tain I. Trén cung nho BC cua ducing tron (I) lay diem E, trén tia doi cua
tia EA lay diem M sao cho EM = EC. Tim tpa do dinh A, biét dinh B thupc duéing thang d: y —
2
= 0 va diem M(8;—3).
Cau 9 (1,0 diém). Giai he phuong trinh
Tim gia tri nhfi nhat cua biéu thuc P ——" + ” + '
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DAP AN
Cau 1
Ta co y —— .x’— 3.x' + 2
TXD: D = 111.
*Su bién thién:
—Chiéu bién thién:
y '= 3s'— 6s ; y = 0 w x = 0 hoac x =
2
C:ic khoiing dong bién: (—m;0) vii (2;+in); kho:ing nghich bién (0;2)
—Cum tri: Ham so dat cue dai tai
x
—Giéri han
tai
“ cue: lim
Bang bién thién
dat cue tiéu tai x = 2; ycT '—2
Giii tr| lén nhat va gia tri nho nhat ciia f(x j trén doan [0;e] lan lupt 1:i 3 + 1n2 va 3.
2
×—2, ta cé›:
i)
b) 4‘— 3.2“ " " '— 4" " “ '— 0
x'= 4s'— 8x — 12
—1+
(1)
3x'— 8s— 12 = 0
4+2
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3
Vay phuong trinh da cho co nghiem duy nhat x
dx
4+
3
IQ ³ = In 4 — In 2 = In 2.
dx
du ———,
x
3
=
=
=
2
— In 3
4
dx
3
— In 3
4
+
— In 3
4
31n 3
+
— In
31n 3
4
'
4
Gia sit (P) song song ho)ac triing (Q), thi ton tai so thuc k sao cho:
(vo li)
Vay (P) cat (Q) theo mot giao tuyén lii ducing thang d .
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5
Vi d la giao tuyén ciia (P) va (Q) nén
nhan
1
14
(5; 2;—1) lam vecto chi phuong.
Mat khac diém M(0;4;—2) dong thcri thuoc (P) va (Q) nén M C d.
Phuong trinh (d):
Cau 6
.x
y—4
5
2
z +2
1
a)
4
2x + —
2
2x ——
4
2x ——
4
8 2
in
= —a + k2u
x = ——— +
8 2
4
8
ver o = arccos
2
b) Gpi A la bién co “Duéing thang noi hai diém dupe chpn cat hai true tpa do”.
Tinh so phan th cua khong gian mau:
So c:ich chon 2 trong 14 diém da cho la Cf4 - 91
Tinh so két qu:i thuan loi cho A:
De doan thang noi hai diém cat hai true toa do thi chung phai nam ci hai gfic phan tu doi xfmg
nhau qua goc tpa dp O (moi diém nam ci mot goc phiin tu)
—THI : Hai diém nam c› hai goc phan tu (I) va (III):
So cach chpn diém nam trong goc (I): co 2 cach
So c:ich chon diém nam trong gfic (III): co 4
cach.
Theo quy tac nhan, co 2.4 = 8 (cap diém) thfia man TH nay
—TH2: Hai diém nam ci hai gfic phan tu (II) va (IV):
So cach chpn diém nam trong goc (II): co 3 cach
So each chpn diém nam trong gfic (IV): co 5 each.
Theo quy tae nhan, co 3.5 = 15 (cap diém) thfia man TH
nay. Theo quy tac cong, so két qua co lpi cho A la 8
= 23
23
91
15
Tinh the tich
Vi SA vufing gfic v‹ i day nén goc gifra SC v:i (ABCD) la SCA —— 30‘
ABCD la hinh chfr nhat, tain giac ABD vufing tai A nén:
AC —— BD ——
B + AD’—— a
Tann giac SAC vufing tai A:
SA —— FC. tan30‘= a
The tich khoi chop:
"S.ABCD
3
ABCD
1
—=——— ..aa.. a.a
3
Vé AI ± SC tai I.
Vi SA ± CD, AD ± CD nén (SAD) ± CD
Suy ra AK ± CD. Ma AK ± SD nén AK 1(SCD)
Suy ra AK ± IK va AK 1SC.
AK ± SC, AI ± SC nén (AKI) ± SC in SC ± IK.
IK lit doan vuong goc chung cua AK v:i SC =› d(AK,SC) = IK.
Tann giac SAD vufing tai A:
1
1
1
AK’“ SA’ + AD’
Tain giac SAC vufing tai A:
AI ² ——
Tann giac AIK vuong tai K:
IK
——
AII’ AK’—’ —
6
BE cat CM tai F.
AC la ducing kinh ciia (I) nén AEC —— 90‘ in CEM ——
90‘ Suy ra tain giac ECM vuong can tai E
ECF
—— 45‘
ABEC la tu giac noi tiép nén CEF —— CAB —— 45‘ (A CAB vufing can)
Suy ra A ECF vufing can tai F
EF la duéng cao cua tain giac can ECM in F la trung diém CM.
»
5; —4)
Duéing thang BF di qua F , nh a) n vecto — CM —— (3; TQ liim vecto phap tuyén.
2
=› Phuong trinh BF . 3.x + y — 11= 0
Tpa do ctia diém B thfia man h(:
Ta co: CB —— (1; 7) . Do dfi duéng thang BC qua B va nh(an vecto n = $7;—TQ lain vecto phap
tuyén.
Phuong trinh BC . 7s— y — 19 = 0.
AB qua B va nhan CB —— $1; 7) lain vecto phap tuyén.
Phuong trinh M : x + 7y — 17 = 0.
DK:
2
Suy ra f(ti dcing bién trén 111.
2
in 2.x — 2 = 2
Thay vao phuong trinh (2) ta dupe:
>1
ttaa
co: 6(2.x —1)( x—
Ap dqng bat dang thiic Co—si cho ba so khfing am, ta co:
— 10×+ 4 = 8
5
2
H( co nghiem duy nhat 2; —
2
Tit (*) suy ra
5
2
4
Két hpp vcii (*) ta co:
4
Tu (*) suy ra‘²
—z
4 thi P ——
I’
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nhat!
3
+—+
Ly — Hfia — Sinh — Van — Anh tot
1
— + 1 trén
Suy ra f(I j dc›ng bién va lién tuc trén [4;+in)
71
4
71
4
Dau bang xiiy ra khi x — y — 2z, chang han x
71
4
2,
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