Đề thi Đại học chính thức môn toán - khối a,a1 - năm học 2013

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s0 ctAo DUc vA pAo rAo Ari DE TIII TUYEN SINH DAI HQc NAtvt eoTT Mon: ToAn; Khdi A vi kh6'i Al Thdi gian lhrn bd,i: 180 phtlt, kh,6ng kd thdi gian phdt dd pf; culuH rrl0c t. prrAN cHIJNG cHo r{r cA rn( sINH e,0 itidm) Ctul 12,Adi€my. Chohlmsdy= *r3+ 3u2 +}mt- 1 (1), v6imli thamsdthgc. -E a) Kh6o sdt str bidn thiOn vh vE dd thi cfia hlm sd (t) k*ri nr :0. &rO* m dd him sd (1) nghlch bidn r6n kho6ng (0; +oo). \J/ -LC;u 2 (1,0 ilidm), Gif,i phtrong trlnh I * tan r = Ztft,sin (o .r l) cnu 3 (1,0 iri&n). ciai hQ phrrdng rrinh {n"; ffi: I (r,s e R)" 2 Cau 4 (1,0 itidd. Ttnh tlch phan , : I#ln r dr, 1 C0u 5 (1,0 ifiild. Cho htnh ch6p S.ABC c6 tl{y l} tam gi6c vuOng tqi A, ffi - B0:, ,gBC h tam gi6c ddu canh a vi m{t-b0n ,SBC vuOng- g6c vdi ddy. Tfnh theo a thd dch cria khdi cnOp S.ABC vh ktroflng cdch tU didm C 6€n m{t phing (SAB). C0u 6 $,A ilidd. Cho cdc sd thr,fc dddng a,b,c th6a m\n di€u kiQn (a + c)(b * c) : 4c2. Tlm gi6 d nh6 nhdt cfia bidu thrftc r : ${ * , t'b!'*- Vffi - (b*3c)s ' (a+3c1s c It. pHAN RIENG $,0 ilifid; Thi skh chl iIWc thm mfit trons lwi phd @tdn A ho(c phdn B) A. Theo chrfdng trinh Chudn Cflu 7.a (1,0 ilidn). Trong m{t ph8ng vdi hS tqa tl$ Oay, cho htnh ch8 $bilt ABCD cd rfidm C thu0e dudng thfrng d,:2o*A*5:0 vi A(-4;8). Gqi M h didm 6di x8ng crla B qua C, N H htnhchidu w6ng g6c crla .B tr€n dddng th8ng MD. TIm tga itQ c6c didm B vA, C, bidt rilng N(5;- ). ; C&u 8.a{Lr0iti€m). TrongkhOnggianvdihQtqa 6SOayz,chodr/dngthEngA,'-=6:'11 :'!2 e-3-21 vi didm A(t;7;3). Vidt phrrong trlnh m{t ph8ng (P) di qua / vi vu6ng g6c vdi A. fim tga dS ttidm M thuOc A sao cho AM :2\,m. C8u 9.a QrA ilidd. Ggi S li t0p hgp tdt cL cic sd tU nhi6n g6m ba chit sd phtn bigt ttrrgc chgn til cdc ch8 sd t;Z;3;4;5; 6;7. Xie tlinh sd pMn tt? cfa S. Chgn ng6u nhitn m$t sd til .9, tlnh xdc sudt dd sd tluqc chgn ld sd chfrn. B. Theo chddng trinh Ntng cao Cflu7.b (lr0ilidd, Trong @ph8ng vdi h$ to.a dO Ouy,cho thfOng th8ng A: r *A:0. Drrdng trOn (C) c6 bdn kinh fi: /16 c{t A tai hai didm A vh B sao cho AB : qrfr. ti€p tuydn crla (C) tpi A vil B c{t nhau tai mQt ttidm thuQc tia Oy. Vi6t phrrong trlnh drrdng trOn (C). C0u 8.b {1r0 didd. Trong khOng gian vdi h$ tga d$ Oayz, cho m{t phfing (P):2r* 3y + z - 1.1 = 0 vl m{t cdu (,S): rz +a2 + z2 -2s*4y-22 -$:0. Chrlng minh (P) ti6p xric vdi (S). fim tga rl0 tidp tfidm o&a (P) vn (fl. Ctu 9,b $rA iridd. Cho sdphfc z :1+{5t. Vidt d4ng lugng gi6c cria z. Tlm phdn th{c vi phdn 6o cfia sd phrlc ur : (L + i,)26. -s$1 Tht sinh khhng thtgc sfr dVng tti li|u. Cdn bQ coi thi khAng gihi thtch gi thdm. Hqvitenthtsinh:.,0^rir0 fLl,.lfl: ...;sdb6odanh: .Q.t.[,1.A.l.al.,ti.

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Page 1: Đề thi Đại học chính thức môn Toán - Khối A,A1 - Năm học 2013

s0 ctAo DUc vA pAo rAo AriDE TIII TUYEN SINH DAI HQc NAtvt eoTT

Mon: ToAn; Khdi A vi kh6'i AlThdi gian lhrn bd,i: 180 phtlt, kh,6ng kd thdi gian phdt ddpf; culuH rrl0c

t. prrAN cHIJNG cHo r{r cA rn( sINH e,0 itidm)

Ctul 12,Adi€my. Chohlmsdy= *r3+ 3u2 +}mt- 1 (1), v6imli thamsdthgc.

-E a) Kh6o sdt str bidn thiOn vh vE dd thi cfia hlm sd (t) k*ri nr :0.&rO* m dd him sd (1) nghlch bidn r6n kho6ng (0; +oo).\J/

-LC;u 2 (1,0 ilidm), Gif,i phtrong trlnh I * tan r = Ztft,sin (o .r l)cnu 3 (1,0 iri&n). ciai hQ phrrdng rrinh {n"; ffi: I (r,s e R)"

2

Cau 4 (1,0 itidd. Ttnh tlch phan , : I#ln r dr,1

C0u 5 (1,0 ifiild. Cho htnh ch6p S.ABC c6 tl{y l} tam gi6c vuOng tqi A, ffi - B0:, ,gBC htam gi6c ddu canh a vi m{t-b0n ,SBC vuOng- g6c vdi ddy. Tfnh theo a thd dch cria khdi cnOpS.ABC vh ktroflng cdch tU didm C 6€n m{t phing (SAB).

C0u 6 $,A ilidd. Cho cdc sd thr,fc dddng a,b,c th6a m\n di€u kiQn (a + c)(b * c) : 4c2. Tlm gi6 dnh6 nhdt cfia bidu thrftc r : ${ * , t'b!'*- Vffi- (b*3c)s ' (a+3c1s c

It. pHAN RIENG $,0 ilifid; Thi skh chl iIWc thm mfit trons lwi phd @tdn A ho(c phdn B)

A. Theo chrfdng trinh Chudn

Cflu 7.a (1,0 ilidn). Trong m{t ph8ng vdi hS tqa tl$ Oay, cho htnh ch8 $bilt ABCD cd rfidm C thu0edudng thfrng d,:2o*A*5:0 vi A(-4;8). Gqi M h didm 6di x8ng crla B qua C, N H htnhchiduw6ng g6c crla .B tr€n dddng th8ng MD. TIm tga itQ c6c didm B vA, C, bidt rilng N(5;- ).

; C&u 8.a{Lr0iti€m). TrongkhOnggianvdihQtqa 6SOayz,chodr/dngthEngA,'-=6:'11 :'!2e-3-21vi didm A(t;7;3). Vidt phrrong trlnh m{t ph8ng (P) di qua / vi vu6ng g6c vdi A. fim tga dS ttidmM thuOc A sao cho AM :2\,m.

C8u 9.a QrA ilidd. Ggi S li t0p hgp tdt cL cic sd tU nhi6n g6m ba chit sd phtn bigt ttrrgc chgn tilcdc ch8 sd t;Z;3;4;5; 6;7. Xie tlinh sd pMn tt? cfa S. Chgn ng6u nhitn m$t sd til .9, tlnh xdc sudtdd sd tluqc chgn ld sd chfrn.

B. Theo chddng trinh Ntng cao

Cflu7.b (lr0ilidd, Trong @ph8ng vdi h$ to.a dO Ouy,cho thfOng th8ng A: r *A:0. DrrdngtrOn (C) c6 bdn kinh fi: /16 c{t A tai hai didm A vh B sao cho AB : qrfr. ti€p tuydn crla (C)tpi A vil B c{t nhau tai mQt ttidm thuQc tia Oy. Vi6t phrrong trlnh drrdng trOn (C).

C0u 8.b {1r0 didd. Trong khOng gian vdi h$ tga d$ Oayz, cho m{t phfing (P):2r* 3y + z - 1.1 = 0vl m{t cdu (,S): rz +a2 + z2 -2s*4y-22 -$:0. Chrlng minh (P) ti6p xric vdi (S). fim tga rl0tidp tfidm o&a (P) vn (fl.

Ctu 9,b $rA iridd. Cho sdphfc z :1+{5t. Vidt d4ng lugng gi6c cria z. Tlm phdn th{c vi phdn 6ocfia sd phrlc ur : (L + i,)26.

-s$1Tht sinh khhng thtgc sfr dVng tti li|u. Cdn bQ coi thi khAng gihi thtch gi thdm.

Hqvitenthtsinh:.,0^rir0 fLl,.lfl: ...;sdb6odanh: .Q.t.[,1.A.l.al.,ti.