Luyện giải đề trước kỳ thi đại học tuyển chọn và giới thiệu đề thi toán học phần 2

Tuyen chgn & Giai thifu dethi Todn hgc - Nguyen Phu Khdnh, Nxuyht Tai Thu Cau 9.^: Tinh modun cua so'phiic z, biet: z = (2 - i)"^ + ( l + i)'* - ^ — i - OETHITHUfSdzi Theo chUomg trinh nang cao Cau 7.b: Trong mSt phSng tpa dp Oxy, cho hinh vuong ABCD eo phuong trinh duong thing AB: 2x + y - = 0, va C, D Ian lupt thupc dupng thing d j : 3x - y - = 0, d j : x + y - = Tinh di|n tich hinh vuong 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 Cau 1: Cho ham so : y = x'^ - 3x^ + mx +1 c6 thi la (C^^) a) Khao sat sy bien thien va ve thi (C) cua ham so m = x = -t Cau 8.b: Trong khong gian Oxyz, cho duong thang (d): y = + 2t va mSt cau [z = - - t b) Tim m de ham so c6 cue dai, cxfc tieu Gpi ( A ) la duong thang di qua hai diem eye dai, cue tieu Tim gia trj ion nhat khoang each tir diem I - ; — den U 4j duong thSng ( A ) cosx + yfz sin f I X 7t — (S): x^ + y^ + z^ - 2x - 6y + 4z -11 = Viet phuong trinh mat phing (p) vuong goc duong thang (d), cat mat cau (S) theo giao tuyen la mpt duong tron c6 ban kinh r = \ Cau 9.b: Tim so phue z thoa man ( l - 3i) z la so thuc va z - + 5i = 4j Cau 3: Giai phuong trinh: sVZx + l + 2x = loVx-3 +13 HMGDANGIAI ixe" (e^+lj + l Cau 4: Tinh tich phan sau: I = f ^ —dx 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 Caul: a) Danh cho ban dpc I e^+l ^ Cau 5: Cho hinh hpp dung ABCD.A'B'C'D' eo day la hinh thoi e^nh a, BAD=a voi cosa=-, canh ben AA' = 2a Gpi M la diem thoa man DM = k.DA va N la trung diem cua canh A'B' Tinh the tich khoi tu dien C'MD'N theo a va tim kde C ' M I D ' N b) Taco y' = 3x^-6x + m Ham so c6 eye dai, eye tieu phuong trinh y' = c6 hai nghi^m phan bi?t.Tuclaeanc6: A ' - - m > o m < Chia da thiic y cho y ' , ta dupe: y = y' x _ l _ 3 Cau 6: Cho so thuc khong am a, b, c thoa man a + b + c = Tim gia tri nho nhai cua bieu thuc: P = a + b^ + c^ II P H A N R I E N G Thi sinh chi dupe chpn lam mpt hai phan (phan A hoac B) A Theo chUorng trinh chuan Cau 7.a: Trong mat phang tpa dp Oxy, cho tam giac ABC vuong tai A va diem B(1;1) Phuong trinh duong thSng AC: 4x + 3y - 32 = Tia BC lay M cho (d;): ^ = = ~^ \ \g (P): x + y - z + = Lap phuong trinh duong thSng (d) song song voi m^t phSng (P) va cat (d^), ( d j ) Ian lupt tai A, B cho dp dai doan AB nho nhat 140 m , -2 x + — + Gia sir ham so c6 eye d^i, eye tieu t^ii cae diem (xi;yi),(x2;y2) • Vi y'(xj) = 0,y'(x2) = nen phuong trinh duong thing (A)qua hai diem eye dgi, eye tieu la: y = r2m_2^ + ^ + hay y = —(2x + l ) - x + l ( 5>/2 BM.BC = 75 Tim C biet ban kinh duong tron ngoai tiep tam giac AMC la — ^ Cau 8.a: Trong khong gian Oxyz, cho hai duong thSng ( d j ) : 2m \ Ta thay, duong thang (A) luon di qua diem co'djnh A — ; so'goc |*a duong thing lA la k = | Ke IH ( A ) ta thay d ( l ; A ) = I H ^ I A = | I Ding thuc xay I A ± ( A ) o ^ - = - i = - - < » m = l V^y, max d ( l ; A ) = | k h i m = 141 Cty TNHH MTV DWH Khang Vi$t Tuye'n chgn b Giai thi?u dethi Todu hgc - Nguyett Phu Khanh , Nguyen Tat Thu Cau 2: Phuang trinh cho tuong duong voi phuang trinh: = id(M,(A'B'C'D')).isABCD sin3x + 3sinx = 4sin^ x.cosx + 2cosx + s i n x - c o s x sin 3x + s inx = sin x.sin 2x + cos x - s inx o sin 2x.cos X - sin 2x.s inx = cos x - s inx (cosx-sinx)(2sin2x-l) = o c o s x = sinx hoac s i n x - l = X = ^ + k27t — = —.2a.—.a.a.sina = — , / l = 3 V 16 12 • Dat AB = x, A D = y, A A ' = z Taco D ' C'M = C'D' + D'D + D M = - x - z - k y Voi cosx = sinx x = — + Voi sin2x = — o c + k27t 12 57t X = /''>• - D ' N = D ' A ' + A ' N = - y + ^x — + kTl Khido C ' M D ' N < : > C ' M D ' N = , — '12 + kn • |x + ky + z j / A- \ \ \ \ D' =0 Cau 3: Dieu ki#n: x > Phuong trinh cho tuong duong: ( V x - - N/2X + I ) = 2X - Nhan ve voi bieu thiic lien hg-p va dat thua so chung: (2X-13)(5-2N/X^-N/2X 2N/)r^ fk xy = ~a^ - k a ^ + - - D^t P(b) = - b - c + b2+c^ voi b e [ U , ' , fx m i n P = c ' ' - c + — 13 Vgy, phuong trinh cho c6 nghiem la: x = - y , x = Cau 4:1= f x e M x + f dx e^+l l2= \— — Ta c6: P'(c) = 3c^ - va P'(c) = o c = - Dat u = e" + du - dx -Y = ln +I2 = l + ln Ii=xe'' du = e^dx u-1 +e I = Ii Xet P(c) = c - c + J voi ce[0;3" ^ +^ fu = X at k = — Cau6:a + b + c = 3=>a = - b - c , k h i d ( S P = - b - c + b^+c^ N/2X +1 = + I ) = c^x = y hoac -k eMx = (xe''-e'') = ^ v3 Tir do, ta duoc P > P(c) = H _ ^ 3V3 Vay, (a;b;c) = 1 thi minP = 373 tt- PHAN RIENG Thi sinh chi dug-c chpn lam mpt hai phan (phan A ko^c B) Theo chi/orng trinh chuan 7.a: Gpi I la tarn duong tron ngoai tiep tarn giac A M C Vi B n i m ngoai duong tron ( l ) nen ta c6: BM.BC = BM.BC ( l ) Taco V^MDN = | d K ( A ' ^ ' C ' ^ ' ) ) S c N D 142 143 Ta c6: P^g^^j^j = B M B C = B I ^ - R ^ vaiR = ^ AB = ^(a - 5)^ + {-a - ) ^ + (-3)^ = Vza^ - a + 35 = ^2(a - ) ^ + 27 > sVs (2) ^ ' Suy ra, m i n A B = 3V3 A(1;2;2), AB = (-3;-3;-3) Tir ( l ) va (2) suy V^y, phuong trinh duong t h i n g (d) la: = = B I ^ - R ^ =75