This math test contains 6 questions about calculus, geometry, and equations. Question 1 involves finding the derivative of a function and finding critical points. Question 2 involves solving equations. Question 3 involves finding integrals and values that satisfy an equation. Question 4 is about finding properties of a square pyramid. Question 5 involves finding points on lines and writing equations of circles. Question 6 is about finding the minimum value of an expression involving integrals.

1. TRU,ONG EHSP HA NOI
TRTIONG THPT CUUVNTV - EHSP
DE THI rnu DAI HQC t AN I Narr ZOrt
Mdn thi : TOAN
Thdi gian tdm bdi : IB0 phtit, kh6ng ke thdi gian phdt di
Ciu 1. ( 2,0 dilm )
Cho hAm s6 .y: Irt - 1**' + (*2 -3)x, trong do m littham s6.
32
I . I(hAo s6t sg bi€n thiOn vi v€ d6 thi cta hdm s6 khi z = l.
diri cdrc canh g6c vudng cira mQt tam giirc vudng c6 dQ ddi c4nh huy€n birg F .
"{2
Cdu 2. (2,0 diem)
1. Giii phuong trinh :
2. Gidi phuo-ng trinh :
CAu 3. ( 2,0 di1m )
1. Tim nguyOn hdm cria hdm sO f1x; : tanx.tan(x + ];.tan1x - l;.3" 3',
2. Tim c6c gi6 tri cta m d6 phuong trinh sau c6 nghiQm duy nh6t :
g-lz-xl _ 43-|2-xl= p.
Ciu 4. ( 1,0 didm,
,
Cho, hinh ch6p tu gi6c ttdu S ABCD
tt6
UU ddi canh ddy bing a, cgnh b6n bing t' ttnn goc tao boi'2
mdt b€n v6'i mdt d6y vi the tictr t<hOi cau ngopi ti6p hinh ch6p d6.
Cfru 5. ( 2,0 diem )
.3
Trong mdt phdng Oxy :
L Chohaidi€m A(2;1), B(-l;-3)vdhai cluongthing d1: x+y+3 =0; cl2: x-5y- 16:0.
Tim tga dQ c6c cti6m C, D lin luE thuQc clr vd dzsao cho t6'gi6c ABCD ld hinh binh hdnh.
2. Vi6tphuongtrinh tluo'ngthingtiOpxircv6'i duongtron (C), xt+ y?-zr] x+4y+4=0vdtao
v6'i trr,rc tung mdt g6c bing 60".
CAu 6. ( 1,0 diem )
Xdtcdctam thirc bgc hai f(x):ax2 + bx+ c,trotlgd6 a< bvnf(x)> 0 vd.i moi x € R.
Hay tiin gi6 tri nho nh6t cira biriu thric M =
TT
Het
2. Tim t6t cd cdc gi6 tr! oba m ae ham s6 c6 cuc d4i t4i x6p, c!.rc ti6u t4i xcr ddng thd'i xss, .rcr lh dO
z+^[s-x, +12-.,[s-*,1
sinal3x*|l * sinal3x -?=:
x2
-l
Dy kiAn ki ttti thrb Egi hgc l6n 2 sE ihrqc t6 chrbc vito ngdy 26,27/2/2011
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2. nAp AN - THANG omirn
rru fiillon IAN rntl Nnar nAu zotl
Ciu DAP AN DIEM
I
e aidm)
(t,O a$ni. Hoc sinh tu ei6i.t.
2. (1,0 itiiinr) . Tim c6c gi|trim
- y'=x'-mx+m2-3
. Hdm sti c6 CD, CT vd Xco ) 0, xcr > 0 e pt y' = 0 c6 hai nghiQrn duo'ng phdn biQt
0,25
(L:m2-4(m2-3)>0 ._^
(+ l" -i'= '"';^; * [";l!'r'
o o
t ;=m2-3>o
( m>o
.,13 < m <2. 0,25
ld dO dei c6c cqnh g6c vudng cta tam gi6c vu6ng c6 canh f,ufAn Uine
f
".,=1
€(xco+xcr)2-z*"u'*.t: i (*)
' Xcor xct
e xzru+
0,25
. Theo dinh l)t Vi-et, thi (*) trd thdnh : m' - 21m'- 3) =
.l;
t **=J;. (fi,2) 0,25
il
(2 iliim)
l. ( 1,0 iti€m . Gi6i phuong trinh .
a EiAukiQn:-3< x<3,
Phuong trinh trd thenh
xl0. Dat 1=1f$Qz +xt= 9-t2,t>0,1+3
g-tz
-+-=l
3+t 4(3-t)
a
0,50
(+ 4Q-q2 -4Q-t)+ I : 0 e 2(3-t)= I <+
*:9 - E o *=*E( th6a mdn diOu kiOn).
5
t- -2
e
0,50
2. (1,0 iti€nt). Gidi phuong oinf .
Phuongtrinh tuongduongv6i phuongtrinh: cosa(:*-l ) +sin4(3x-T l:i' 4 4' 2
0,25
e [cos2(3x -l I *sin213x-i lt' -2 cos2(3x -[ ).rin'{:* -on) = } 0,25
(+ t-]sin'{ox-
sin26x = 0 <+
?I 1
-)=-z' z
kn
x: -6
<+ sin2(6x -
, kez.
lt
-)=l <+
2' cos26x = I
(+
0,50
ilI
(2diim)
l. (1,0 iliAm) . Tim nguyOn hdm
a f(x):
sinx.sin(x +
l).sin (x - T)
.or*..or(* +
i).costx - i)
slnx.(cosf - coszx)
_
sinx.(zsin'x - i) sin3x:--
cos3x.orr.(.orf + .orz*) .os*.(zcoszx - ])
0,50
. Jf(x)dx=-JHo* = lJffi = llnlcos3xl + c. 0,50
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3. 2. (1,0
. DAt t: t- x , phuong trinh tr6'thdnh 9-ltl - 4. 3-ltl : m (l )
Phuong trinh dd cho c6 nghiQm duy dr6t khi vd chi khi pt (l) c6 nghiQm dr-ry nh6t.
+ ) EiAu kien cAn : Gid sft (l ) c6 nghiQm duy nhAt t" . khi d6 - to cfrng ld nghi€m.
Suy ra to = 0, thay vdo (l ) ta cluo.'c m = - 3.
+) DiAukigndrl:Khirn=-3,thi (l)trdthdnh: 9-ltl -4.3-ltl :-3 (2).
Suy ra 3-ltl = 1 a:r t = 0, d0y ld nghiQnr duy nh6t cria (2).
. E6ps6: m:-3.
t (0,5 eti6@. Gqi H ld t6m cta d6y ABCD, ta c6 SH I (ABCD); M td trung di6m cfia BC thi
BC I (SHM), do c6c m{t ben t4o v6i mqt d6y ctrng mQt g6c, n€n 5fr8 bing g6c tao bdi
m[t b€n voi cl6y.
Ta c6 : 311: r/ffi - 4112
+ tan sffE= ffi= vr +fliliH=60o.
o (0,5 diA@. Hinh ch6p S.ABCD ddu, nOn tdm I cria
t<trOi cAu ngo4i tii5p hinh ch6p ld giao <ti6m c0a club'ng
thing SH v6'i m{t phing trung trqc cria mQt c4nli b€n
ndo d6 cta hinh ch6p.
GqiN ld trung cli6m ctra SC, thi IN ld trung trr,rc
cria SC. Suy ra: ASNI - ASHC
SN.SC 5a
=rR=SI=-;fr-=m
*l
4 . rzsa3{1i A
V0y, V: -tR' :
4n .
/::
rl
!1
,l
tl
ll
I
I
I
rl'I
I
,_-. 1
t
I
tr I
rl
-H'
I
I
I
I
t
t
I
IV
Q iri6m)
t. (1,0
V
Q di6m
Gi6 sri ABCD ld hinh binh hdnh, ta c6 Cd = ffi = 13; +; (+
[}3 - ;: = i
Vi D ed2 ndn xe-5yo- 16=0 + (xc+3)-5(ys+4)=
ViC€d1,n€n xc+yc*3=0.
Tu d6 ta c6 hQ phuong trinh : fT l?It-=j=
(x' = 3 (xo = 6
- (xc+yc=-3 <+
tv.=-o ttvl=-z
Ta c6 : el = (3; a) ve Be = (4; - 3 ) n6n hai vectcv Bf, Be khdng ctrng phuong, tric ld 4
dit5m A, B, C, D kh6ng thdng hing, hay tri gi6c ABCD ld hinh binh hdnh.
Edp sd : C(3; -6) vi D(6; -2)
a? r-l'f
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4. 2.(1,0
@2)vib6nkinhR:J3;
Tir gi6 thii5t suy ra ti6p tuyiin cAn tim t?o v6i trr,rc hodnh mQt g6c bing 30o , n€n ti6p tuy6n c6
h0s6g6cbing ** t€rrptti€ptuy6nc6d4ng:y=*rt x*m e +*-V3y +"13m=0'
Khoang cich htir t6m l(r/3; -2) den ti6p tuy€n bing b5n kinh R = v3'
ItJs +ztr + VSml r;
TfclA ir=-7=5
rfm = -5
* |ilill =22
* lfn=-lL[m=1
V{y c6 4 tiiip tuy€n th6a mdn y6u cAu bii to6n : x-rfgy-5r/3=o;
"+r/3y+3V3=0;
*-fsy-V3=o;
x+y'3y-J3=0.
Tim gi6 tri nh6 n
w
Q diAm) Tir giAthi6t suy ra a> 0 vd A = b2 - 4ac<0 + c >
-yza+b+-
M># b-a
Dat t=b-a>0+MZ
+a2 + +a(t+a)+(t+a)z
-
gaz + 6at+ tz
4at 4at
3 gaz + t2 3 T''lw7
=-a-/-a-=J
2 4at -Z 4at
(^_b'
Dingthftcxiyrakhi vdchi kf
ti = A
(+ b=c=4a =+ f(x):a(x+2)2 vhminM = 3'
- 3t
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