Luyện giải đề trước kỳ thi đại học tuyển chọn và giới thiệu đề thi toán học phần 1

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Luyện giải đề trước kỳ thi đại học   tuyển chọn và giới thiệu đề thi toán học phần 1

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Cty TNHH MTVDWHTOiaitg V^ DETHITHUfSOl I P H A N C H U N G C H O T A T C A C A C T H I S I N H (5^c em H Q C sinh than men! "Luyen gidi de truoc ky thi dai hgc - Tuyen chon vd giai thieu de thi Todn hgc" la mpt nhOng cuon thupc bp sach "On luy$n thi Dai hgc", nhom tac gia chuyen toan THPT bien soan Voi each viet khoa hpc va sinh dpng giiip ban dpc tiep can voi mon toan mpt each t y nhien, khong ap luc, ban dpc tro nen t y tin va nang dpng hon; hieu ro ban chat, biet each phan tich de tim tarn ciia van de va biet giai thich, lap luan cho tirng bai toan Sy da dang ciia h^ thong bai tap va tinh huong giiip ban dpc luon thii giai toan Tac gia chii trpng bien soan nhung cau hoi mo, npi dung co ban bam sat sach giao khoa va cau true de thi Dai hpc, dong thai phan bai tap eac dang toan co lai giai chi tiet Hi^n de thi Dai hpc khong kho, to hop eua nhieu van de dan gian, nhung chua nhieu cau hoi mo neu khong nam chae ly thuye't se lung tiing vifc tim 16i giai bai toan Voi mpt bai toan, khong nen thoa man voi mpt lai giai minh vira tim dupe ma phai co' gang tim nhieu each giai nhat cho bai toan do, moi mpt each giai se eo them phan kien thue mai on tap Mon Toan la mpt mon rat ua phong each tai tu, nhung phai la tai tit mpt each sang tao va thong minh Khi giai mpt bai toan, thay v i dung thoi gian de luc Ipi tri nho, thi ta can phai suy nghT phan tich de tim phuong phap giai quyet bai toan Do'i voi Toan hpc, khong eo trang sach nao la thua Tung trang, tung dong deu phai hieu Mon Toan doi hoi phai kien nhan va ben bi t u nhirng bai tap don gian nhat, nhiing kien thiic co ban nhat V i chinh nhiing kien thue co ban moi giiip ban dpc hieu dupe nhij'ng kien thuc nang cao sau Cau 1: Cho ham so y = ^ (C) X a) Khao sat sy bien thien va ve thj (C) b) Gpi I la giao diem eua hai duong ti^m can Tim diem A thupc thj ( C ) , biet tam giac OIA co di?n tich bang i , voi O la goc tpa dp 6sl2x-3+-j^ Vx+1 x+1 Cau 4: Tinh tich phan: I = 7t X + — xdx Cau 5: Cho hinh chop S.ABC eo day ABC la tam giae vuong can tai B, AC = 2a Tam giae ASC vuong tai S va nkm mat phSng vuong goc voi day, SA = a Tinh theo a the tich khoi chop S.ABC va khoang each tix C den mat phiing (SAB) Cau 6: Cho cac so thye khong am a,b,e thoa a + b + e = l va khong co hai so nao dong thoi bang Tim gia trj nho nha't ciia bieu thuc: P =1 + ^ r + (e + l ) ( + a + b ) (a + b)(b + e) (e + a)(a + b) ^ ' II P H A N R I E N G Thi sinh chi dxxtfc chpn lam mpt hai phan (phan A hoac B) A Theo chUorng trinh chuan Cau 7a: Trong mat phang Oxy cho tam giae ABC npi tiep duong tron (C) ec Mac du tac gia da danh nhieu tam huyet cho cuon sach, xong sy sai sot la dieu kho tranh khoi Chung toi rat mong nhan dupe sy phan bi^n va gop y quy bau eua quy dpc gia de nhirng Ian tai ban sau cuon sach dupe hoan thi^n hon phuong trinh: (x + 4)^ + y^ =25, H ( - ; - ) la trye tam tam giac ABC; M ( - ; -2 la trung diem canh BC Xae djnh tpa dp cac dinh A , B , C Cau 8a: Viet phuong trinh m|it cau (S) co tam nam tren duong than^ Thay rnat nhom bien soan Tac gid: Nguyen Phu Khanh d:2iz2 = yzi = £zi ~3 2 ( Q ) : x + y - z + = va tiep xuc voi hai m^t phSng (P):x + y - z - = v Tuyen chgn & Giai thifu dethi Todu hqc - Nguyen Phii Khdnh , Nguyen Tat Thu Cau 9a: Chung minh dang thuc sau: Vay CO diem thoa yeu cau bai toan: A,(2;3), A ( ; l ) , A - ; u , „2n-l_22"-l 2n ^" C t y TNHH MTV DWH Khang Viet 2n + l Cau 2: Dieu ki|n: • (n la so nguyen duong, CJ^ la so to hop chap k ciia n phan tu) B Theo chUorng trinh nang cao Cau 7b: Trong mat phang Oxy cho elip (E) C6 hai tieu diem I^(W3;0); I^(V3;0) va di qua diem A sfS;- Lap phuong trinh chinh t5c cua ( E ) va voi moi V ^/ diem M tren elip, hay tinh bieu thuc: P = F^M^ + FjM^ - 30M^ - F1M.F2M , X— z +1 V Cau 8b: Trong khong gian Oxyz cho duong thang A: — ^ ~ ^ — ] ~ phang ( a ) : x + y - z + = Chung minh rang A va (a) cat tai A Lap sinx + cosx ^ X* — cosx sin2x o-^r-— + s i n x sinx + cosx _ 2cosx = o + kjt sm X cosx 2cos^x sinx smx + cosx = X = — + k7C cosx = + cos x = 2\/2 sinx cosx = sin2x sm X = — + k7l n ,^ n X = - + k2K, X = - 4 kin + — Ket hop dieu kien ta c6 nghiem cua phuong trinh la: + z^ = Cau 9b: Tim cac so phuc z, w thoa X ?t k n Phuong trinh phuong trinh mat cau (S) c6 tarn nMm tren A, di qua A va (S) cat m p ( a ) theo mgt duong tron c6 ban kinh bang sinx ^ 7t W^Z-5=:1 X = H\i(}m DAN GIAI llTt — + nn, X= „ + 1771 2nn, x = „ ^ + Inn, n&Z Cau 3: Dieu kiC^n: x > - I P H A N C H U N G C H O T A T C A C A C T H I S I N H Ba't phuong trinh c=> V x - + 3x^7+1 = j ( x - 3)(x + l ) + Cau 1: I a) Dpc gia t u lam o ( V x - - l ) + 3^/^(^-2^/2x-3)>0 b) Ta CO l ( l ; ) = > O I = (l;2)=r>IO = \/5 va phuong trinh O I : x - y = GQi A e ( C ) = ^ A a; 2a-l ,a^l a-1 h = d(A,IO) = - ^ 2a- Dodo Nen S^,OA4''°^4 2a-l a-1 13 >0-0 (8x-13)(7-9x) Cau 4: Ta c6: I - V'^^'"^dx - 'fxe^^dx = A - B 23-^ - 5a + = 2a^-3a = a = 2,a = a = 0,a = , A4 Dat t = V l + n x ^ l n x = l ( t - l ) : ^ — = ^ t d t 3^ / X Doi can x = l = > t = l , x = e=:>t = -;4 U J TuySii chgn & Giai thifu dethi Toan hqc - Nguyen Phu Khdnh , Nguyen Tat Thu » Suy A = Jt-tdt = - t ^ c=0 14 du = dx D|t I I PHAN R I E N G Thi sinh chi dupe chpn lam mpt hai phan (phan A 2x dx = e2''dx Suy = ^ X ^ " ho?c B) A Theo chUerng trinh chuan Cau 7a: Duong tron (C) c6 tarn I ( - ; O ) , ban kinh R = 2x , 2J 2 ^ 14 e2«-e2 Vgy I = — ^ = ^1;_2) laVTPTcua BC nen phuong trinh BC la: x - y - l = Do tpa dp B, C la nghi^m ciia h?: Cau 5: Ta c6 A B = B C = ^ - n^^A v Tpa dp A la nghi^m cua he: aVs Uijc Ve H E I A B ^ S E I A B va ™ Suy SE = VsH^ + H E ^ = ^ 3V.S A B C = AC _ = 1BC = — ^ V a y A { - ; - ) hoac A ( - ; ) Cau 8a: Vi mat cau (S) c6 tarn I e d -t vol x, y > x+ y 1 •+a + bl^b + c c + a, (a + b)(a + b + 2c) - t | « > t = l = > l ( - l ; ; ) va R = l Vay p h u o n g t r i n h mat cau ( S ) : (x + i f + ( y - 3^ + (z - 3^ = Cau 9a : Ta c6: ( l + x f " = C°„ + xC^„ + + x^^C^jJ {l-xf".C^„-xC^„ x^"Ci^ ( l - c ) ( l + c) i-c^ D o d o : P > — i - + (c + l ) ( - c ) = — ^ + + 3c-c2 l-c^ ^ ' l-c^ = - J _ + ( l _ c ) + 3c2+3c>2, l { - t ; l + 2t;l + 2t) d(l,(P)) = d ( l , ( Q ) ) R 6-3t (c + a)(a + b) (x + 4)^ + y = Mat cau (S) tiep xuc voi hai mat phang (P) va ( Q ) nen 27213 y 2x + y + 13 = Giai h$ ta tim dupe ( x ; y ) = (-4;-5),(-8;3) ^ S , , „ = isE.AB = X (a + b)(b + c) -7- AC^ 1 Cau 6: A p dung bat dang thuc - + — > Ta c6: a^>/3 =M= ^ A ^ ^ l ^ ^ E H BC Vay d ( C { S A B ) ) = phuong trinh A H : 2x + y +13 = = i ^ Do Vs.ABC = S H S , i A B C = 3- ^ - ^ (x + 4)^ + y = Do B ( l ; ) , C ( - ; - ) A H // I M AC x-2y-l =0 Giai h?nay ta dupe cac cap nghifm (x;y) = { l ; ) , ( - ; - ) a72, suy S ^ g c = ^ ( ^ ^ j = Gpi H la chan duang cao t u S ciia tarn giac S A C ri> S H ( A B C ) AC = a7i::.SH = ^ I Vay minP = a=b = - DSng thuc xay xf" - _ , ) n ( x C L - ^ C L ^^(Uxf"-(l-xrdx -i_.4fl-c2]=8 ! ~ x^-^C^ir') (1) ( l + x p ' - ( l - x )2n+l ,2n+l 2n + l 2n + l (2) Cty TNHH MTV DWH Khang Viet Tuyen chqn & Gi&i thifu dethi Todn hgc - Nguyen Phu Khdnh, Nsuuen Tat Thu Ma: j(xC2„ + Tu ( l ) suy w'' =-z^ x3cL+ + x2"-i w Suy (2) » w^.|z|^° = z^ ^2 v2n -2n 2n Tu (1), (2) va (3) suy ra: fz = -2n s Z Tu {2) suy = z^ => + w W = 2n-l -2n (3) ^2n-l ic^„ i c ^ -'-Cl +- + ^1 C ^^ 0: z= l • w= -l: v6 nghiem Z , = !=> w = z =1 = w = 0, w = - z5 =l (z) = Thu lai ta thay cap (w,z) = (-1,1) thoa yeu cau bai toan 22"-l OETHITHllfSOZ B Theo chUorng trinh nang cao / - ^ =ic^„+ici,+ +^c 2n 2n V^ Cau7b:Giasu (E): — + ^ = voi a,b>0 a^ I PHAN CHUNG CHO TAT CA CAC THI SINH Cau 1: Cho ham so y = x^ - 3x2 - 3m (m +1) x - a2=b2 + Theo gia thiet bai toan ta c6 h^ la^ ^c:>a2=4,b2=l a) Khao sat sy bien thien va ve thi ham so m = 0, b) Tim tat ca cac gia tri cua tham so m de ham so (l) c6 hai cue tri ciing dau 4b2 Suyra(E):^ + I - = l Cau 2: Giai phuang trinh : (l + tanx)(2cos2x-l) rr '- = 2V2 cos3x ' sm x + 71 ' XetM(xo;yo)e(E)^^ + y ^ = l = ^ y = i - i '(x2+l)y4+l = 2xy2(y3-l) Suy P = (a + exg )^ + (a - exp )^ - 2(x^ + y2 j _ (a^ - e^x^) x= -l x-l_y_z+l Cau 8b: Xet h^ phuong trinh : < ~ T ~ ~ ^ y = - l ^ A ( - l ; - l ; ) 2x+y-2z+3=0 z=0 Goi I la tarn cua mat cau, suy I (l + 2t; t; -1 -1) Theo gia thiet bai toan ta c6 • t = l r : > l ( ; l ; - ) , R = IA = = > ( S ) : ( x - f + ( y - l f + ( z + f =24 • t = -3t:>l(-5;-3;2),R = IA = 2V6=>(S):(x + f + (y + f + ( z - f =24 Cau 9b: Tim cac so phiic w,z thoa: w''' + z^=0 (1) w^z-5=l (2)' Cau 3: Giai h$ phuong trinh: Cau 4: Tinh tich phan: I = xy2 |3xy'* - 2j = xy"* (x + 2y) +1 (voi x,y e '^l (x-l)sin(lnx) + xcos(lnx) ^ ^ ^dx Cau 5: Cho hinh chop S.ABCD c6 day ABCD la hinh thoi canh a, BAD = 60° va SA = SB = SD Mat cau ngoai tiep hinh chop S.ABCD c6 ban kinh bang va SA > a Tinh the tich khoi chop S.ABCD Cau 6: Cho cac sothuc duong a,b,c thoa man a + b + c = 3bc 2ca ^ ;— + > — c + ab a + bc b + ca I I PHAN RIENG Thi sinh chi du(?c chpn lam mpt hai phan (phan A hoac B) A Theo chUorng trinh chuan ^, , , I Chung mmh rang: 2ab + Tuyen chiftt b Giai thifu dethi Todn HQC - Nguyen Phu Kh,\nh , Nguyen Cau 7a: Trong mat phSng Oxy cho tam giac ABC npi tie'p duong tron (C): (x-1)^ +{y-lf =10 Diem M(0;2) la trung diem canh BC va di^n tich tam giac ABC bang 12 Tim tpa dp cac dinh cua tam giac ABC Cau 8a: Trong khong gian Oxyz cho hai duong th^ng: fx = l + t y = -2 + t , : v —4 z - l • /\ == vam|itph5ng ( a ) : x + y + z - l l = X — z=l Viet phuong trinh duong thang A c3t hai duong thang A,, A j va mat phang (a) lanluqrttai A , B , M thoa man A M = 2MB dong thoi A l A j Cau 9a: Gpi zi la nghi^m phuc c6 phan ao am cua phuong trinh z^ - 2z + = 2z-z^+l = Tim tap hp-p cac diem Mcbieu dien so phuc z thoa: z + zf+2 M ( ; ) ; N { ; - ) ; P(2;0); Q ( l ; ) Ian lupt thupc c^inh AB, BC CD, A D Hay lap phuong trinh cac canh ciia hinh vuong Cau 8b: Trong khong gian Oxyz cho diem A{3; 2; 3) va hai duong th3ng , x-2 y-3 z-3 « x-1 y-4 z - ^, , • ^ - ^.i dj : — — = — = — — va d2 : — ^ = ^ = — Chung minh duong thang di, d2cva = m ( m + l ) x j - m ( m + l ) - x j - m ( m + l ) x j - = |m^ + m + l j ( - x j - l ) Tuongty y2=(m^ + m + l j ( - x - l ) Do yiy2 > (2xj + l)(2x2 +1) > o 4x,X2 + 2{xj + X2) +1 > o -4m (m + l ) + > < = > m ^ + m - < o — — — < m < V ; Cau 2: Dieu kien: sin 7t X + — 4j I PHAN CHUNG CHO TAT CA CAC THI SINH Cau 1: a) Ban dpc ty lam b) Tap xac djnh D = M Taco: y ' = x ^ - x - m ( m + l ) => y' = O o x ^ - x - m ( m + l)=:0 Ham so CO hai eye trj va chi ( l ) c6 hai nghifm phan bi^t x,,X2 o A'= 1+ m ( m + l ) = m^ + m + > dung voi Vm 10 x?.tJ + k K l ^ x) o 2cos2x - = 2cos3xcosx = cos4x + cos2x o 2cos2 2x -cos2x = kn 7t cos2x = cos2x = x=±- + k7t Ket hpp voi dieu ki^n ta c6 nghi^m cua phuong trinh da cho la: X = tuyen C M ciia tam giac ABC HMGDANGIAI ^0; c o s x t O o x = - + k7i; COS X (sin X + COS va C ciia tam giac ABC biet di chua duong cao BH va d2 chua duong trung — tie'p xiic voi Parabol y = x + m Vift yj = x ^ - x j - m ( m + l ) x i - l = x , ( x ^ - ) » i ) - ( x j - x j j - x j - m ( m + l ) x i - l diem A ciing n^m mpt mat phang Xac dinh toa dp cac dinh B Cau 9b: Tim m de thj ham so' y = Khang V i X , langhiemciia ( l ) nen X j - x j = m ( m + l ) Suy ra: B Theo chiToTng trinh nang cao Cau 7b: Trong m^t phSng voi h^ toa dp Oxy cho hinh vuong ABCD biet DWH CtyTNHHMTV Tat Thu Cau 3: H? o - + nT[, X = ± - + nn, n G Z x V + 2xy2+l + y*-2xy-'=0 3xV-2xy2-xV-2xy-'-l =0 x2+24 + ^-2xy y x+ =- l y x V - ^ - x - x y - ^ =0 y' y 3x2y^-2xy- (do y = khong la nghi^m ciia h?) D l t a = x + ^ , b = xy,tac6he: -2xy = - l a2-2b = - l a2-3b2+2b = f I ' =0 x + —• y J a2=2b-l W-4b +l =0 b=l a = ±l Tuyen chgn & Giai thieu dethi Tomi h^c - Nguyen huu Khdnh , Nguyen a=l b=l -7- X =• X = — y xy = y= hoac Cty TNIIU Af IV DWH Khang Viet TatThu^ X = 1+ = -1 y 'X- =y - l = he v6 nghiem b=l y y^ + y + l = xy = Vay nghifm ciia he da cho la: (x;y) = -1±V5 i + Vs^ 71 e'2 c2 „: Cau4:Tac6I= j sin(lnx) + cos(lnx) d x - | ^'"^'"'^)(jx 71 o2 e2 I x'sin(lnx) + x.(sin(lnx)) dx - [sin(lnx)d(lnx) -I o2 " = (xsin(lnx) + cos(lnx)) ^ = e - l Cau 5: Tu gia thiet, suy ABD la tarn giac deu nen SABD la hinh chop deu Goi H, O Ian luot la tarn ciia tarn giac ABD va hinh thoi ABCD Suy S H I ( A B C D ) Mat phSng trung true canh SA cat SH tai I, ta c6 I la tarn mat cau ngoai tie'p hinh chop S.ABD Vi ASFI - ASHA, suy — = — =^ SA^ = 2SI.SH SH SA Ma A H = - A O = ^ ^ S H = S A - ^ 3 Nen ta c6 phuang trinh 2\ 2^ 12a' S A ^ - ^ SA^=4Sl2 SA^-^ SA^2 = 2a' (loai) SA^ = 2a2 => SA = aV2 12 SH = Mat khac: S^BCD = ^S^^BD = ,,2 Vay the tich khoi chop S.ABCD la: V = |SH.SABCD = ^ ^ ' ^ = Cau 6: Bat d3ng thuc can chung minh tuong duong voi 2ab 3bc 2ca ^5 (c + a)(c + b)^(a + b)(a + c)^(b + c)(b + c ) ~ ' ~ o 2ab(l - c ) + 3bc(l - a) + 2ca(l - b) > | ( l - a)(l - b)(l - c) 11 ab + 4bc + ca > 16abc - + — + ->16 a b c 11 Ap dung bat dang thuc - + — > ta c6: X y x+y 1 4 ^ 16 , - +—+- > - + > = 16 (dpcm) a b c a b+c a + b + c Dang thuc xay a = i , b = c = II PHAN RIENG T h i sinh chi dirg-c chpn lam mgt hai phan (phan A hoac B) {x-l)%(y-lf =10^ y=x+ ^ A Thee chUorng trinh chuan x2=4 Cau 7a: Duong tron (C) c6 tam l(l;l)/ suy MI = (l;-l) ViBCdiquaM va vuonggoc voi MI n e n B C : x - y + = Toa dp B, C la nghiem ciia he: "x = 2,y==24| a - b + 2| Taco: d(A,BC) = l ^ — B C = 4V2 =>SAABC Nen[ xta- CyO+!T2-=b0 + 2| = a =[x'=4 b + 4,a = b -Lx8.= -2,y = Suyra • a = bB(2;4),C(-2;0) + thay vao (l)hoac ta c6:B(-2;0),C(2;4) Gpi(bA(a;b), a - l f + ( b - l f= 0

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