Đại số và giải tích 11 cơ bản

vu TUAN (Chu bien) - TRAN VAN HAO OAO NGOC NAM - LE VAN TIEN -IVU VIET YEN BAI TAP y ,»;p7X*"^' ,•• * • • • ;»v<»*?firFJ^ • •• .• • • • 1 ,j»VTIJ»>r*»« '' ¥ ».• • • • \ 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) 9 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 1. 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 = 0 khi x = kn, k e Z. n • sm X = 1 khi x = — + k2n, k G Z. • sinx = -1 khi x = -— + k2n, k e Z. D6 thi ham s6 y = sinx (H.l) : Hinh 1 2. 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 = 0 khi X = — + kn, k eZ. • cos X = 1 khi X = k2n, k e Z. • cosx = -1 khi X = {2k + l)7i, k e It. D6 thi ham s6' y = cosx (H.2) : Hinfi 2 3. 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 = 0 khi x =kn, k e Z. n • tanx = 1 khi X = — + kn, k e.Z. 4 • tanx = -1 khi x = -— + kn, k G D6 thi ham sd 3^ = tanx (H.3): -37t 2 4. Ham so cotang Hinh 3 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 = 0 khi X = — + kn, k e Z. 71 • cot X = 1 khi X = — + ^71, k eZ. 4 It, • cotx = -1 khi X = —— + ^7r, )t G Z. D6 thi ham sd j = cotx (H.4): -27t O ]£- 2 Hinh 4 B. Vi DU • Vidul Tim tap xae dinh cua eae ham sd a) y = sin3x ; c) y = cosVx ; b) y = cos— ; X 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, rGR<=>x = - GR nfen tap xae dinh eua ham s6 y = sin3x la R. 2 ' • 2 b) Ta CO — e R <=> X ;^ 0. Vay tap xae dinh eiia ham sd y = cos— la X ^ D = R\{0}. e) Ta CO Vx G R o x > 0. Vay tap xae dinh cua ham s6 y = cosVx la D = [0 ; +00). d) Ta CO 1 + .^ ir» l + ^ ,^ 1^ G R <^ > 0 « -1 < X < 1. 1-X 1-x 1 + X vay tap xae dinh eua ham sd j = sin J-j la D = [-1 ; 1). • Vidul. Tim tap xae dinh eua cae ham sd a) y = ; b) y = cot 2x - — , , ^ 2cosx ' ^ y A)' cotx ,^ sinx+ 2 Gidi 3 , K a) Ham sd y = x^c dinh khi va ehi khi cosx ^ 0 hay x ?t — + kn, k G ' ^ • 2cosx • •2 vay tap x^e dinh cua ham sd la D = R \ {| + itTi, A: G I 71 I 7C b) Ham sd y = cot 2x - — xae dinh khi va chi khi 2x - — ^t kn, k G \ Aj • ,4 hay x * — + k—, k e Z. o 2 vay tap xae dinh cua ham sd y = cot 2x - — la D = R\{| + ^|,A:G cotx . ^. , [sinx 9^0 lx^kn,keZ e) Ham sd y = xae dmh <:> < <:> < cosx-1 • lcosx?tl Ix^t A:27i,;tGZ. Tap {^27:, k &Z] la tap con eua tap [kn, k eZ} (umg vdd cac gia tri k cot X chan). vay tap xae dinh cua ham sd la cosx-1 D = R\{kn,k€Z]. sinx + 2 d) Bieu thiie ludn khdng am va no eo nghla khi cosx + 15«t 0, hay cosx + 1 " cosx 9t -1. vay ta phai c6 x ^ (2k + l)n, it G Z, do do tap xae dinh cua ^ smx+ 2 ham so y = J la ^'cosx + 1 D = R\{(2A: + l)7i, A;GZ}. • Vi dn .? Tim gia tri ldn nhS^t va a) y = 2 + 3eosx ; l + 4cos^x c)y= 3 ; gia tri nho nha't cua cac h£im sd : b) y = 3 - 4 sin X cos x ; d) y = 2sin x - cos2x. Gidi a) Vl -1 < cosx < 1 ndn -3 < 3eosx < 3, do do -1 < 2 + 3cosx < 5. vay gia tri ldn nha't eua ham sd' la 5, dat duoc khi cosx = 1 o X = 2kn, keZ. Gia tri nho nha't cua ham sd la -1, dat duoc khi cos x = -1 d' x = {2k + l)7t, keZ. b) y = 3 - 4sin^ xcos^ x = 3 - (2sinxcosx)^ = 3 - sin^ 2x. Ta ed 0 < sin^ 2x < 1 nen -1 < -sin^ 2x < 0. vay 2<y<3. [...]... cos— la ham sd chSn, nen ta chi eSn ve dd thi ham sd dp tren doan [0 ; 27i] rdi la'y ddi xiing qua true tiing, se duge dd thi ham sd tren doan [-27t; 27r] Dd thi ham sd duoc bidu dien tren hinh 5 Hinh 5 11 X X cos—, ndu cos— > 0 2 2 X b) Ta cd cos— 2 X X -cos—, ne'u cos— < 0 2 2 Vi vay, tit dd thi ham sd y = cos— ta giii nguyen nhflng phSn dd thi nam phia tren true hoanh va l^y dd'i xiing qua true hoanh... cos(2x + 50°) = cos60° » 2x + 50° = ±60° + it360°, it G 2 2x = - 5 0 ° + 6 0 ° + i t 3 6 0 ° , i t G «• 2x = - 5 0 ° - 60° + it360°, it G X = 5° + /:180°, it G Z . 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 . 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) 9 r NHA XUAT BAN GIAO DUC VIET NAM Ban. bidu dien tren hinh 5. Hinh 5 11 b) Ta cd X cos— 2 X X cos—, ndu cos— > 0 2 2 X X -cos—, ne'u cos— < 0. 2 2 Vi vay, tit dd thi ham sd y = cos— ta giii nguyen nhflng