Hướng dẫn giải bài tập Đại số và Giải tích 11: Phần 1

v u TUAN (Chu bien) - TRAN VAN HAO OAO NGOC NAM - LE VAN TIEN -IVU VIET YEN BAI TAP y ,»;p7X*"^' ,•• * • • • \ ;»vr*»« ' ' • • • • • • Ơ ằ.ã ã ã ã T' ai'' a NHA XUAT BAN GIAO DUC VIET NAM VU TUAN (Chu bien) TRAN VAN HAO - BAG NGOC NAM LEVANTI^N-VUVI^TYEN BAITAP DAIS6 VAGIAI TICH (Tdi bdn ldn thd tu) r NHA XUAT BAN GIAO DUC VIET NAM Ban quy^n thu6c Nha xu^t ban Giao due Vi6t Nam 01 - 201 l/CXB/824 - 1235/GD Ma s6': CB103T1 m.' huang L HAM SO Ll/ONG GIAC PHUONG TRINH Ll/ONG GIAC §1 Ham so laong giac A KIEN THCTC CAN NHd Ham so sin Ham s6' j = sinx co tap xae dinh la M va -1 < sinjc < 1, Vx G R y = sin X la ham s6' le y = sinx la ham s6' tu^n hoan v6i chu ki 2jt Ham s6 y = sinx nhan cae gia tri dac bi6t: • sinx = x = kn, k e Z n • sm X = x = — + k2n, k G Z • sinx = -1 x = -— + k2n, k e Z D6 thi ham s6 y = sinx (H.l) : Hinh Ham so cosin Ham s6' y = cosx eo tap xae dinh la R va -1 < cosx < 1, Vx G y = cosx la ham so ehSn y = cosx la ham so tu^n hoan vdi chu ki 2n Ham s6' y = cosx nhan cac gia tri dac bi6t: • cosx = X = — + kn, k eZ • cos X = X = k2n, k e Z • cosx = -1 X = {2k + l)7i, k e It D6 thi ham s6' y = cosx (H.2) : Hinfi Ham so tang Ham sd V = tanx = eo tap xae dinh la cosx D = R\{^ + kn,ke y = tanx la ham s6 le y = tanx la ham sd tu5n hoan vdi chu ki n Ham sd y = tar v nhan eae gia tri dae biet: • tanx = x =kn, k e Z • tanx = X = n— + kn, k e.Z • tanx = -1 x = -— + kn, k G D6 thi ham sd 3^ = tanx (H.3): -37t Hinh Ham so cotang COSX Ham s6 y = coix = —— c6 tap xae dinh la smx D= R\{kTi,keZ] y = cotx la ham sd le y = coix la ham sd tuSn hoan vdi chu ki % Ham sd y = cot x nhan cac gia tri dac bi6t: 71 • cot X = X = — + kn, k e Z 71 • cot X = X = — + ^71, k eZ It, • cotx = -1 X = —— + ^7r, )t G Z D6 thi ham sd j = cotx (H.4): O -27t ]£2 Hinh B Vi DU • Vidul Tim tap xae dinh cua eae ham sd a) y = sin3x ; b) y = cos— ; X c) y = cosVx ; d) y = sin 1+X 1-x" Gidi a) Dat t = 3x, ta duoc ham sd y = sin r co tap xae dinh la D = R Mat khae, rGRx = - G R nfen tap xae dinh eua ham s6 y = sin3x la R ' • b) Ta CO — e R X ;^ Vay tap xae dinh eiia ham sd y = cos— la X ^ D = R\{0} e) Ta CO Vx G R o x > Vay tap xae dinh cua ham s6 y = cosVx la D = [0 ; +00) d ) T a CO + ^ 1-X ir» l + ^ ,^ G R 0 « 1-x 1+X vay tap xae dinh eua ham sd j = sin J-j 1^ - < X < la D = [-1 ; 1) • Vidul Tim tap xae dinh eua cae ham sd a) y = ; ^ 2cosx b) y = cot 2x - — , , ' ^ y A)' cotx ,^ sinx+ Gidi , K a) Ham sd y = x^c dinh va ehi cosx ^ hay x ?t — + kn, k G ' ^ • 2cosx • • vay tap x^e dinh cua ham sd la D = R \ { | + itTi, A: G I 71 I \ Aj 7C b) Ham sd y = cot 2x - — xae dinh va chi 2x - — ^t kn, k G • , hay x * — + k—, k e Z o vay tap xae dinh cua ham sd y = cot 2x - — la D = R \ { | + ^|,A:G e) Ham sd y = cotx ^ , [sinx 9^0 xae dmh < cosx-1 • lcosx?tl lx^kn,keZ < Ix^t A:27i,;tGZ Tap {^27:, k &Z] la tap eua tap [kn, k eZ} (umg vdd cac gia tri k cot X chan) vay tap xae dinh cua ham sd la cosx-1 R\{kn,k€Z] D= sinx + d) Bieu thiie ludn khdng am va no eo nghla cosx + 15«t 0, hay cosx + " cosx 9t - vay ta phai c6 x ^ (2k + l)n, it G Z, do tap xae dinh cua ^ smx+ ham so y = J la ^'cosx + D = R\{(2A: + l)7i, A;GZ} • Vi dn ? Tim gia tri ldn nhS^t va gia tri nho nha't cua cac h£im sd : b) y = - sin X cos x ; a) y = + 3eosx ; c)y= l + 4cos^x ; d) y = 2sin x - cos2x Gidi a) Vl -1 < cosx < ndn -3 < 3eosx < 3, do - < + 3cosx < vay gia tri ldn nha't eua ham sd' la 5, dat duoc cosx = o X = 2kn, keZ Gia tri nho nha't cua ham sd la - , dat duoc cos x = -1 d' x = {2k + l)7t, keZ b) y = - 4sin^ xcos^ x = - (2sinxcosx)^ = - sin^ 2x Ta ed < sin^ 2x < nen -1 < -sin^ 2x < vay 2