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thi D tr khi D-nm 2007 II

Cu I: Cho hm s1x

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 :

=+

=1x yx

0myx2c nghim duy nht

Cu III: Cho mt phng (P): x 2y + 2z 1 = 0 v cc ng thng

2

z

3

3y

2

1x:d1 =

=

v5

5z

4

y

6

5x: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 khongbng 2.

Cu IV:

1. Tnh

=2

0

2 xdxcosxI

2. Gii phng trnh:x

x

2 2x1x

12log +=

.

Cu Va (cho chng trnh THPT khng phn ban):1. T cc ch s 0, 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s t nhinchn 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 ngthng: d1: (m 1)x + (m 2)y + 2 m = 0d2: (2 m)x + (m 1)y + 3m 5 = 0

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Chng minh d1 v d2 lun ct nhau. Gi P = d1 d2. Tm m sao choPBPA + ln nht

Cu Vb (cho chng trnh THPT phn ban):1. Gii phng trnh: 022.72.72 xx21x3 =++ .

2. Cho lng tr ng ABCA1B1C1 c tt c cc cnh u bng a. M l trungim ca on AA1. Chng minh BM B1C v tnh d(BM, B1C).

Bi giiCu I:1. Kho st hm s (Bn c t gii)

2. Ta c ( )2

1y ' 0, x 1

x 1= <

T th ta thy tip tuyn to vi hai tim cn mt tam gic vungcn ta phi c h s gc ca tip tuyn l 1 tc l:

( )( ) 2x,0x11x1

1x

121

2

2====

. Ti x1 = 0 y1 = 0 phng trnh tip tuyn l y = x

. Ti x2 = 2 y2 = 2 phng trnh tip tuyn l y = x + 4

Cu II :1. Gii phng trnh: (1 tgx)(1 + sin2x) = 1 + tgx (1)

t: t = tgx 2t1

t2x2sin

+

= . Pt (1) thnh

( ) 22t

1 t 1 1 t1 t

+ = + + ( ) ( )

2 2

1 t t 1 (t 1)(1 t ) + = + +

( ) ( )2t 1 0 hay 1 t t 1 (1 t ) + = + = +

t 1 hay t 0 = =

Do (1) tgx = 0 hay tgx = 1

x = k hay x =4

+ k , k

Cch khc

(1) (cosx sinx)(cosx + sinx)2 = cosx + sinx

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(hin nhin cosx = 0 khng l nghim)

cosx + sinx = 0 hay (cosx sinx)(cosx + sinx) = 1

tgx = -1 hay cos2x = 1 x =4

+ k hay x = k , k

2. Tm m h sau c nghim duy nht

(I)

=

=

=+

=

x1x y

0myx2

1x yx

0myx2

Vi iu kin:

1x

0x yta c

(I) ( )

( )( )

22

y 2x my 2x m1 x

xy 1 x y x 1x

= = = =

( )( )

2

21 x

2x m x 2 m x 1 0x

= + = ()

( hin nhin x = 0 khng l nghim ca () )

t ( )2f (x) x 2 m x 1= + , ( a = 1 )

ycbt tm m phng trnh () c ng 1 nghim tha x 1

af(1) < 0 hayf (1) 0 0(vn,do ac 0)c bhay1 1(VN) 1a 2a

= =

2 m < 0 m > 2

Cu III :1. d1 i qua A(1, 3, 0), VTCP ( )2,3,2a =

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Mt phng (P) c PVT ( )2,2,1nP =

M/phng (Q) cha d1 v (P) nn (Q) c PVT( )1,2,2n,an PQ ==

Vy (Q) qua A c PVT ( )1,2,2nQ = nn phng trnh (Q):

2(x 1) 2(y 3) 1(z 0) = 0 2x + 2y + z 8 = 0

2. P/trnh tham s d1:x 1 2ty 3 3tz 2t

= + = =

( )1M d M 1 2t,3 3t, 2t +

P/trnh tham s d2:

x 5 6t '

y 4t 'z 5 5t '

= += =

( )2M d N 5 6t ', 4t ', 5 5t ' +

Vy ( )5t2't5,3t3't4,4t2't6MN ++=

Mt phng (P) c PVT ( )2,2,1nP =

V MN // (P) 0n.MN P =

( ) ( ) ( )1 6t ' 2t 4 2 4t ' 3t 3 2 5t ' 2t 5 0 t t ' + + + = = . Ta li c khong cch t MN n (P) bng d(M, P) v MN // (P)

( ) ( )2

441

1t22t332t21=

++

++

6 12t 6 6 12t 6 hay 6 12t 6 t 1hay t 0 + = + = + = = =

. t = 1 t' = 1 M1(3, 0, 2) N1(1, 4, 0)

. t = 0 t' = 0 M2(1, 3, 0) N2(5, 0, 5)

Cu IV :

1. Tnh

=2

0

2 xdxcosxI

t: u = x2du = 2xdx ; dv = cosxdx , chn v = sinx

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Vy I =

= 2 2

2 2 20

0 0

x cosxdx x sinx 2 xsinxdx

Ta c

=2

2 20

x sinx4

I1 =2

0

xsinxdx

; t u = x du = dx

dv = sinxdx, chn v = cosx

I1 =

= + 2 2

20

0 0

xsinxdx xcosx cosxdx

= [ ] 20xcosx sinx 1

+ =

Vy : I = 2 22

0

x cosxdx 24

=

2. Gii phng trnhx

x2

2 1log 1 x 2 (*)

x

= +

iu kin

x x 02 1 0 2 1 2

x 0x 0 x 0

> > =

>

(*) = +x

x2

2 1log 1 2 x

xv x > 0

= +x x2 2log (2 1) log x 1 2 x v x > 0

(2x 1) + log2(2x1) = x + log2x (**)

Xt hm f(t) = t + log2t ng bin nghim cch khi t > 0Do f(u) = f(v) u = v, vi u > 0, v > 0Vy t (**) 2x1 = x 2xx 1 = 0 (***)

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Li xt hm g(x) = 2xx 1 khi x > 0

g'(x) = 2xln2 1 , g'(x) = 0 = = >x 21

2 log e 1ln2

2 2x log (log e) 0= >

Ta c g//(x) > 0 vi mi x nn g'(x) l hm tng trn R/

2 2g (x) 0, x log (log e) < < v/

2 2g (x) 0, x log (log e)> >

g gim nghim cch trn ( ]2 2; log (log e)

v g tng nghim cch trn [ )2 2log (log e);+

g(x) 0 = c ti a l 1 nghim trn ( ]2 2; log (log e) , v c ti a l 1

nghim trn [ )2 2log (log e);+ .bng cch th nghim ta c pt g(x) 0= (***) c 2 nghim lx = 0 v x = 1 . V x > 0 nn (*) x = 1.

Cu Va :

1/ Gi n = 1 2 3 4a a a a l s cn tm. V n chn a4 chn.

* TH1 : a4 = 0 Ta c 1 cch chn a4

6 cch chn a15 cch chn a24 cch chn a3

Vy ta c 1.6.5.4 = 120 s n

* TH2: a4 0. Ta c 3 cch chn a45 cch chn a1

5 cch chn a24 cch chn a4Vy ta c 3.5.5.4 = 300 s n .Tng cng hai trng hp ta c : 120 + 300 = 420 s n

2. Ta giao im P ca d1, d2 l nghim ca h phng trnh

(m 1)x (m 2)y m 2

(2 m)x (m 1)y 3m 5

+ = + = +

Ta c2

2m 1 m 2 3 1D 2m 6m 5 2 m 0 m2 m m 1 2 2

= = + = + >

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V2

3 1D 2 m 0 m

2 2

= + >

nn d1, d2 lun lun ct nhau.

Ta d thy A(0,1) d1 ; B(2,1) d2 v d1 d2 APB vung ti P P nm trn ng trn ng knh AB.

Ta c (PA + PB)2 2(PA2 + PB2) = 2AB2 = 2 2(2 2) 16=

PA + PB 4. Du "=" xy ra PA = PB P l trung im cacung AB

Vy Max (PA + PB) = 4 khi P l trung im ca cung ABP nm trn ng thng y = x 1 qua trung im I (1 ;0) ca AB

v IP = 2 P (2 ; 1 ) hay P (0 ;- 1)Vy ycbt m = 1 v m = 2

Cu Vb :

1. Gii phng trnh : 23x+17.22x + 7.2x2 = 0 2.23x7.22x + 7.2x2 = 0

t t = 2x > 0 th (1) thnh

2t

3

7t2

+ 7t 2 =0 (t 1)(2t25t + 2) = 0 t = 1 hay t = 2 hay t =

1

2

Do pt cho tng ng

= = =x x x1

2 1hay2 2hay22

x = 0 hay x = 1 hay x = 1

2. Chn h trc Oxyz sao cho

ta c A(0 ;0 ;0); A1(0,0,a); C ( - a ;0 ;0 ) B

a a 3, ,0

2 2;

B1

a a 3, ,a

2 2;M

a0,0,

2

= =

uuuur uuuur

1

a a 3 a a a 3

BM , , ;CB , ,a2 2 2 2 2

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= + =uuuur uuuur 2 2 2

1a 3a a

BM.CB 04 4 2

BM B1C

Ta c 1B.B (0,0,a)= = =uuuur uuuur1 1

1

1

[BM.B C].BB a 30d(BM,B C)

10[BM.B C]

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H VN CHNG - PHM HNG DANH(Trung tm Bi dng vn ha v Luyn thi i hc Vnh Vin)

x

CB