MỘT SỐ KÍ HIỆU THÔNG DỤNG

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Tên gọi Số các hoán vị của n phần tử Số các chỉnh hợp chập k của n phần tử Số các tổ hợp chập k của n phần tử Xác suất của biến cố A Giới hạn của dãy số (un) Giới hạn của hàm số f(x) khi x dần tới x0 Giới hạn của hàm số f(x) khi x dần tới âm vô cực Giới hạn của hàm số f(x) khi x dần tới dương vô cực Giới hạn bên phải của hàm số f(x) khi x dần tới x0 Giới hạn bên trái của hàm số f(x)...

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Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 MOÄT SOÁ KÍ HIEÄU THOÂNG DUÏNG Kí hieäu Teân goïi Dieãn giaûi Pn Soá caùc hoaùn vò cuûa n phaàn töû Permutation Soá caùc chænh hôïp chaäp k cuûa n phaàn töû k An Soá caùc toå hôïp chaäp k cuûa n phaàn töû Combinatory k Cn P(A) Xaùc suaát cuûa bieán coá A Probability lim un Giôùi haïn cuûa daõy soá (un) Limit lim f ( x ) Giôùi haïn cuûa haøm soá f(x) khi x daàn tôùi x0 x  x0 lim f ( x ) Giôùi haïn cuûa haøm soá f(x) khi x daàn tôùi aâm voâ cöïc x  lim f ( x ) Giôùi haïn cuûa haøm soá f(x) khi x daàn tôùi döông voâ cöïc x  lim f ( x ) Giôùi haïn beân phaûi cuûa haøm soá f(x) khi x daàn tôùi x0  x  x0 lim f ( x ) Giôùi haïn beân traùi cuûa haøm soá f(x) khi x daàn tôùi x0  x  x0 y' hoaëc f'(x) Ñaïo haøm cuûa haøm soá y = f(x) y'' hoaëc f''(x) Ñaïo haøm caáp hai cuûa haøm soá y = f(x) y(n) hoaëc f(n)(x) Ñaïo haøm caáp n cuûa haøm soá y = f(x) dy hoaëc df(x) Vi phaân cuûa haøm soá y = f(x) Differenttial n(A) hoaëc A Soá phaàn töû höõu haïn cuûa taäp A ----- Voõ Thanh Huøng - THPT Traàn Quoác Toaûn - Ñoàng Thaùp ----- 1 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 CHÖÔNG I. ÖÙNG DUÏNG ÑAÏO HAØM ÑEÅ KHAÛO SAÙT HAØM SOÁ ----- oOo -----  CHUAÅN BÒ KIEÁN THÖÙC: 1. Caùc giaù trò löôïng giaùc cuûa cung (goùc) :  sin luoân xaùc ñònh  R vaø sin( + k2) = sin cos luoân xaùc ñònh  R vaø cos( + k2) = cos  - 1  sin  1 (sin 1). - 1  cos  1 (cos  1).   tan xaùc ñònh khi    k vaø tan(k) = tan; 2 cot xaùc ñònh khi   k vaø cot( + k) = cot.  Daáu cuûa caùc giaù trò löôïng giaùc cuûa goùc  Phaàn tö I II III IV Giaù trò löôïng giaùc sin + + - - cos + - - + tan + - + - cot + - + - 2. Baûng caùc giaù trò löôïng giaùc ñaëc bieät:     (300) (450) (600) (900) 0 (00)  6 4 3 2 1 2 3 sin 0 1 2 2 2 1 3 2 cos 1 0 2 2 2 1 3 tan 0 1 kxñ 3 1 3 cot kxñ 1 0 3 3. Coâng thöùc löôïng giaùc cô baûn: 1   sin2 + cos2 = 1  1  tan 2   (   k , k  Z). 2 cos  2 1   1  cot 2   (  k, k  Z).  tan.cot = 1 (   k , k  Z). 2 2 sin  4. Giaù trò löôïng giaùc cuûa caùc cung coù lieân quan ñaëc bieät: Cung buø:( - ) vaø  Cung phuï:(  - ) vaø  Cung ñoái:(-) vaø  Cung hôn keùm : ( + ) vaø  2 sin(-) = -sin sin( - ) = sin sin( + ) = -sin  sin( - ) = cos cos(-) = cos cos( - ) = -cos cos( + ) = -cos 2  cos( - ) = sin tan(-) = -tan tan( - ) = -tan tan( + ) = tan 2  cot(-) = -cot cot( - ) = -cot cot( + ) = cot tan( - ) = cot 2  cot( - ) = tan 2 2 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 5. Caùc coâng thöùc löôïn giaùc thöôøng söû duïng: Coâng thöùc coäng: Coâng thöùc nhaân ñoâi: Coâng thöùc haï baäc: 1  cos 2a cos(a - b) = cosacosb + sinasinb sin2a = 2sinacosa cos 2 a  2 2 2 cos(a + b) = cosacosb - sinasinb cos2a = cos a - sin a 1  cos 2a 2 sin(a - b) = sinacosb - cosasinb = 2 cos a - 1 sin 2 a  2 2 sin(a + b) = sinacosb + cosasinb = 1 - 2sin a 1  cos 2a tan a  tan b 2tana tan 2 a  tan( a  b)  tan 2a  1  cos 2a 1  tan a tan b 2 1  tan a tan a  tan b tan( a  b)  1  tan a tan b Coâng thöùc bieán tích thaønh toång: Coâng thöùc bieán ñoåi toång thaønh tích: 1 uv uv cosacosb = [cos(a + b) + cos(a - b)] cosu + cosv = 2cos cos 2 2 2 1 uv uv sinasinb =- [cos(a + b) - cos(a - b)] cosu - cosv = -2sin sin 2 2 2 1 uv uv sinacosb = [sin(a + b) + sin(a - b)] sinu + sinv = 2sin cos 2 2 2 uv uv sinu - sinu = 2cos sin 2 2  Coâng thöùc nhaân ba: sin3a = 3sina - 4sin3a cos3a = 4cos3a - 3cosa  Coâng thöùc sina + cosa:   2 sin(a + 2 sin(a - sina + cosa = ) sina - cosa = ) 4 4   2 cos(a - ) sina - cosa = - 2 cos(a + ) sina + cosa = 4 4  Ghi chuù: ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... 3 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 §1. HAØM SOÁ LÖÔÏNG GIAÙC I- ÑÒNH NGHÓA: 1. Haøm soá sin vaø haøm soá coâsin: a) Haøm soá sin: y y B M' sinx sinx M x A' x O A x x O B'  Quy taéc ñaët töông öùng moãi soá thöïc x vôùi soá thöïc sinx sin: R  R x  y = sinx ñöôïc goïi laø haøm soá sin, kí hieäu laø y = sinx  Taäp xaùc ñònh cuûa haøm soá sin laø: D = R. b) Haøm soá coâsin: y y B M'' cosx M x A' O x cosx A x O x B'  Quy taéc ñaët töông öùng moãi soá thöïc x vôùi soá thöïc cosx cos: R  R x  y = cosx ñöôïc goïi laø haøm soá coâsin, kí hieäu laø y = cosx  Taäp xaùc ñònh cuûa haøm soá coâsin laø: D = R. 2. Haøm soá tang vaø haøm soá coâtang: a) Haøm soá tang: sin x  Haøm soá tang laø haøm soá ñöôïc xaùc ñònh bôûi coâng thöùc y = (cosx ≠ 0), kí hieäu laø y = tanx. cos x   Taäp xaùc ñònh cuûa haøm soá y = tanx laø: D = R\{ + k, k  Z}. 2 b) Haøm soá coâtang: cos x  Haøm soá coâtang laø haøm soá ñöôïc xaùc ñònh bôûi coâng thöùc y = (sinx ≠ 0), kí hieäu laø y = cotx. sin x  Taäp xaùc ñònh cuûa haøm soá y = cotx laø: D = R\{k, k  Z}.  Nhaéc laïi ñònh nghóa haøm soá chaün, haøm soá leû. Xeùt tính chaün, leû cuûa caùc haøm soá y = sin(x), y = cos(x), y = tan(x) vaø y = cot(x). 4 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 * Nhaän xeùt: Haøm soá y = sinx laø haøm soá leû, haøm soá y = cosx laø haøm soá chaün, töø ñoù suy ra caùc haøm soá y = tanx vaø y = cotx ñeàu laø nhöõng haøm soá leû. II- TÍNH TUAÀN HOAØN CUÛA HAØM SOÁ LÖÔÏNG GIAÙC:  Giaûi nghóa töø tuaàn hoaøn, laáy ví duï thöïc teá ñôøi soáng. Tìm nhöõng soá T sao cho f(x + T) = f(x) vôùi moïi x thuoäc taäp xaùc ñònh cuûa caùc haøm soá: a) y = sinx; b) y = tanx.  Haøm soá y = sinx laø haøm soá tuaàn hoaøn vôùi chu kì 2.  Haøm soá y = cosx laø haøm soá tuaàn hoaøn vôùi chu kì 2.  Haøm soá y = tanx vaø y = cotx cuõng laø haøm soá tuaàn hoaøn, vôùi chu kì . III- SÖÏ BIEÁN THIEÂN VAØ ÑOÀ THÒ CUÛA HAØM SOÁ LÖÔÏNG GIAÙC: 1. Haøm soá y = sinx:  Haøm soá y = sinx xaùc ñònh vôùi moïi x  R vaø -1  sinx  1;  Laø haøm soá leû;  Laø haøm soá tuaàn hoaøn vôùi chu kì 2. a) Söï bieán thieân vaø ñoà thò haøm soá y = sinx treân ñoaïn [0; ]: y y 1 B x3 x2 sinx2 sinx2 x1 x4 sinx1 sinx1  x x3 x2 x4 x1 Ax  O A' O 2 B'   Haøm soá y = sinx ñoàng bieán treân [0; ] vaø nghòch bieán treân [ ; ]. 2 2 Baûng bieán thieân:  0  x 2 1 y = sinx 0 0 y 1 * Chuù yù: Vì haøm soá y = sinx laø haøm soá leû neân laáy ñoái  - xöùng ñoà thò haøm soá treân ñoaïn [0; ] qua goác toïa ñoä O, - 2 x  O  ta ñöôïc ñoà thò haøm soá treân ñoaïn [-; 0]. 2 -1 b) Ñoà thò haøm soá y = sinx treân R: y 1 5 3  - - -2 -  2 2 2 x 3 O  2 5 - 2 2 2 -1 2 5 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 c) Taäp giaù trò cuûa haøm soá y = sinx: Taäp giaù trò cuûa haøm soá y = sinx laø T = [-1; 1]. 2. Haøm soá y = cosx:  Haøm soá y = cosx xaùc ñònh vôùi moïi x  R vaø -1  cosx  1;  Laø haøm soá chaün;  Laø haøm soá tuaàn hoaøn vôùi chu kì 2;  Haøm soá y = cosx ñoàng bieán treân [-; 0] vaø nghòch bieán treân [0; ].  Baûng bieán thieân: x - 0  1 y = cosx -1 -1  Ñoà thò haøm soá y = cosx: y 1 5 3  - - -  -2 2 2 2 x O 3  2 5 - 2 2 2 -1  Taäp giaù trò cuûa haøm soá y = cosx laø T = [-1; 1]. Ñoà thò haøm soá y = sinx, y = cosx ñöôïc goïi chung laø caùc ñöôøng hình sin. 3. Haøm soá y = tanx:   Taäp xaùc ñònh: D = R\{  k , k  Z}; 2  Laø haøm soá leû;  Laø haøm soá tuaàn hoaøn vôùi chu kì ;  a) Söï bieán thieân cuûa haøm soá y = tanx treân nöûa khoaûng [0; ): 2 y T2 tanx2 B M2 tanx1 T1 M1 x  A x1 x2 A' O O 2 B'  Haøm soá y = tanx ñoàng bieán treân nöûa khoaûng [0; ). 2 6 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 Baûng bieán thieân:   - x 4 2 + y = tanx 1 0   * Nhaän xeùt: Khi x caøng gaàn thì ñoà thò haøm soá y = tanx caøng gaàn ñöôøng thaúng x = . 2 2 b) Ñoà thò haøm soá y = tanx treân D:   Ñoà thò haøm soá y = tanx treân ( ; ) : 22 y x O   - 2 2  Ñoà thò haøm soá y = tanx treân D: y 3 -3  - - 2 2 2 x  O  2  Taäp giaù trò cuûa haøm soá y = tanx laø T = (-; +). 4. Haøm soá y = cotx:  Taäp xaùc ñònh: D = R\{k, k  Z};  Laø haøm soá chaün;  Laø haøm soá tuaàn hoaøn vôùi chu kì ; a) Söï bieán thieân vaø ñoà thò haøm soá y = cotx treân khoaûng (0; ): Haøm soá y = cotx nghòch bieán treân khoaûng (0; ).  0  x 2 + y = tanx 0 - 7 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 y  x O  2 b) Ñoà thò haøm soá y = cotx treân D: y O x  2 - -2 -3   3 - 2 2 2 2  Taäp giaù trò cuûa haøm soá y = cotx laø T = (-; +).  Ghi chuù: ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... 8 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... BAØI TAÄP REØN LUYEÄN 1. Baøi taäp cô baûn: 3 Baøi 1: Haõy xaùc ñònh caùc giaù trò cuûa x treân ñoaïn [-; ] ñeå haøm soá y = tanx: 2 a) Nhaän giaù trò baèng 0; b) Nhaän giaù trò baèng 1; c) Nhaän giaù trò döông; d) Nhaän giaù trò aâm. Baøi 2: Tìm taäp xaùc ñònh cuûa caùc haøm soá: 1  cos x 1  cos x   c) y = tan( x  ) ; d) y = cot( x  ). a) y = ; b) y = ; 1  cos x sin x 3 6 1 Baøi 3: Döïa vaøo ñoà thò haøm soá y = cosx, tìm caùc giaù trò cuûa x ñeå cosx = . 2 Baøi 4: Döïa vaøo ñoà thò haøm soá y = sinx, tìm caùc khoaûng giaù trò cuûa x ñeå haøm soá ñoù nhaän giaù trò döông. Baøi 5: Döïa vaøo ñoà thò haøm soá y = cosx, tìm caùc khoaûng giaù trò cuûa x ñeå haøm soá ñoù nhaän giaù trò aâm. Baøi 6: Tìm giaù trò lôùn nhaát cuûa caùc haøm soá: a) y = 2 cos x + 1; b) y = 3 - 2sinx. Baøi 7: Döïa vaøo ñoà thò cuûa haøm soá y = sinx, haõy veõ ñoà thò cuûa haøm soá y = sinx. Baøi 8: Chöùng minh raèng sin2(x + k) = sin2x vôùi moïi soá nguyeân k. Töø ñoù veõ ñoà thò haøm soá y = sin2x. 2. Baøi taäp naâng cao: Baøi 1: Xeùt tính chaün - leû cuûa moãi haøm soá sau: d) y = sinxcos2x + tanx. a) y = -2sinx; b) y = 3sinx - 2; c) y = sinx - cosx; Baøi 2: Tìm giaù trò lôùn nhaát vaø giaù trò nhoû nhaát cuûa moãi haøm soá sau:  c) y = 4sin x . a) y = 2cos(x + ) + 3; b) y = 1  sin( x 2 ) - 1; 3 CAÂU HOÛI CHUAÅN BÒ BAØI ............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. 9 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 §2. PHÖÔNG TRÌNH LÖÔÏNG GIAÙC CÔ BAÛN 1. Phöông trình sinx = a: Xeùt phöông trình sinx = a (a  R) (1) Tröôøng hôïp a > 1: phöông trình (1) voâ nghieäm sin B Tröôøng hôïp a  1: 1 M M' aK  x    k 2 sinx = sin   (k  Z )  x      k 2 A' -1 1A coâsin O sinx = a  x  arcsin a  k 2 sinx = a   (k  Z ) -1  x    arcsin a  k 2 B' * Chuù yù: sin u( x )  sin   u( x )    k 2 [ 0  k 360 0 ] (k  Z )   [sin u( x )  sin  0 ] 0 0 0 u( x )      k 2 [180    k 360 ]  sinu(x) = a (-1  a  1) sin u( x )  a  u( x )  arcsin a  k 2 [arcsin a  k 360 0 ] (k  Z )   (sin u( x )  a) 0 0 u( x )    arcsin a  k 2 [180  arcsin a  k 360 ]  f ( x )  g( x )  k 2 (k  Z )  Toång quaùt: sin[f(x)] = sin[g(x)]    f ( x )    g( x )  k 2   Ñaëc bieät: sin[f(x)] = 1  f(x) = + k2, k  Z 2  sin[f(x)] = -1  f(x) = - + k2, k  Z 2 sin[f(x)] = 0  f(x) = k, k  Z. Ví duï: Giaûi caùc phöông trình sau: 1 1 2 d) sin(x + 450) = - a) sinx = ; b) sinx = ; c) sin2x = 1; . 2 5 2 Giaûi: ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... 10 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 2. Phöông trình cosx = a: Xeùt phöông trình cosx = a (a  R) (2) Tröôøng hôïp a > 1: phöông trình (2) voâ nghieäm sin B Tröôøng hôïp a  1: 1 M  x    k 2 cosx = cos   (k  Z )  x    k 2 A' -1 1A a coâsin O H cosx = a  x  arccos a  k 2 cosx = a   (k  Z ) M' -1  x   arccos a  k 2 B' * Chuù yù: cos u( x )  cos  u( x )    k 2 [ 0  k 360 0 ] (k  Z )   [cos u( x )  cos  0 ] 0 0  u( x )    k 2 [   k 360 ]  cosu(x) = a (-1  a  1) cos u( x )  a u( x )  arccos a  k 2 [arccos a  k 360 0 ] (k  Z )   [cos u( x )  a] 0  u( x )   arccos a  k 2 [ arccos a  k 360 ]  f ( x )  g( x )  k 2  Toång quaùt: cos[f(x)] = cos[g(x)]   (k  Z )  f ( x )   g( x )  k 2  Ñaëc bieät: cos[f(x)] = 1  f(x) = k2, k  Z cos[f(x)] = -1  f(x) =  + k2, k  Z  cos[f(x)] = 0  f(x) = + k, k  Z. 2 Ví duï: Giaûi caùc phöông trình sau: 1  2 2 d) cos(x + 600) = a) cosx = cos ; b) cos3x = - ; c) cosx = ; . 6 3 2 2 Giaûi: ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... 11 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 3. Phöông trình tanx = a: tanx = tan  x =  + k, k  Z [x = 0 + k1800, k  Z] tanx = a tanx = a x = arctana + k, k  Z [x = arctana + k1800, k  Z] * Chuù yù: tan[u(x)] = tan  u(x) =  + k, k  Z [ux) = 0 + k1800, k  Z]  tan[u(x)] = a tan[u(x)] = a ux) = arctana + k, k  Z [ux) = arctana + k1800, k  Z]  Toång quaùt: tan[f(x)] = tan[g(x)]  f(x) = g(x) + k, k  Z.  Ñaëc bieät: tan[u(x)] = 0  u(x) = k, k  Z. Ví duï: Giaûi caùc phöông trình sau: 1  c) tan(3x + 150) = 3 . a) tanx = tan ; b) tan2x = - ; 5 3 Giaûi: ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... 4. Phöông trình cotx = a: cotx = cot  x =  + k, k  Z [x = 0 + k1800, k  Z] cotx = a cotx = a x = acrcota + k, k  Z [x = acrcota + k1800, k  Z] * Chuù yù: cot[u(x)] = cot  u(x) =  + k, k  Z [ux) = 0 + k1800, k  Z]  cot[u(x)] = a cot[u(x)] = a ux) = acrcota + k, k  Z [ux) = acrcota + k1800, k  Z]  Toång quaùt: cot[f(x)] = cot[g(x)]  f(x) = g(x) + k, k  Z.   Ñaëc bieät: cot[u(x)] = 0  u(x) = + k, k  Z. 2 Ví duï: Giaûi caùc phöông trình sau: 2 1 c) cot(2x - 100) = a) cot4x = cot ; b) cot3x = -2; . 7 3 12 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 Giaûi: ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... 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Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 BAØI TAÄP REØN LUYEÄN 1. Baøi taäp cô baûn: Baøi 1: Giaûi caùc phöông trình sau: 1 a) sin(x + 2) = ; b) sin3x = 1; 3 2x  1 3 d) sin(x + 200) = - c) sin(  )=- ; . 33 2 2 Baøi 2: Vôùi nhöõng giaù trò naøo cuûa x thì giaù trò cuûa caùc haøm soá y = sin3x vaø y = sinx baèng nhau? Baøi 3: Giaûi caùc phöông trình sau: 2 b) cos3x = cos120; a) cos(x - 1) = ; 3 3x  1 1 d) cos22x = . c) cos(  )=- ; 24 2 4 2 cos 2 x  0. Baøi 4: Giaûi phöông trình 1  sin 2 x Baøi 5: Giaûi caùc phöông trình sau: 3 a) tan(x - 150) = b) cot(3x - 1) = - 3 ; ; 3 c) cos2x.tanx = 0; d) sin3xcotx = 0.  Baøi 6: Vôùi nhöõng giaù trò naøo cuûa x thì giaù trò cuûa caùc haøm soá y = tan( - x) vaø y = tan2x baèng nhau? 4 Baøi 7: Giaûi caùc phöông trình sau: a) sin3x - cos5x = 0; b) tan3x.tanx = 1. 2. Baøi taäp naâng cao: Baøi 1: Tìm nghieäm cuûa caùc phöông trình sau trong khoaûng ñaõ cho: 1 3 a) sin2x =  vôùi 0 < x < ; b) cos(x - 5) = vôùi - < x < ; 2 2 1  c) tan(2x - 150) = 1 vôùi -1800 < x< 900; d) cot3x = vôùi  < x < 0. 2 3 Baøi 2: Tìm taäp xaùc ñònh cuûa moãi haøm soá sau: 1  cos x sin( x  2) a) y = ; b) y = cos 2 x  cos x 2 sin x  2 tan x 1 c) y = ; d) y = . 1  tan x 3 cot 2 x  1 CAÂU HOÛI CHUAÅN BÒ BAØI ............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. 14 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 §3. MOÄT SOÁ PHÖÔNG TRÌNH LÖÔÏNG GIAÙC THÖÔØNG GAËP I- PHÖÔNG TRÌNH BAÄC NHAÁT ÑOÁI VÔÙI MOÄT HAØM SOÁ LÖÔÏNG GIAÙC Ñònh nghóa: Phöông trình baäc nhaát ñoái vôùi moät haøm soá löôïng giaùc laø phöông trình daïng: at + b = 0 trong ñoù a, b laø caùc haèng soá (a ≠ 0) vaø t laø moät trong caùc haøm soá löôïng giaùc. Caùch giaûi: Bieán ñoåi phöông trình ñaõ cho veà phöông trình löôïng giaùc cô baûn. Ví duï: Giaûi caùc phöông trình sau: c) 3 tanx + 1 = 0; d) 3 cotx - 3 = 0. a) 2sinx - 3 = 0; b) 5cosx + 3 = 0; Giaûi: ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... II- PHÖÔNG TRÌNH BAÄC HAI ÑOÁI VÔÙI MOÄT HAØM SOÁ LÖÔÏNG GIAÙC: Ñònh nghóa: Phöông trình baäc hai ñoái vôùi moät haøm soá löôïng giaùc laø phöông trình daïng: at2 + bt + c = 0 trong ñoù a, b, c laø caùc haèng soá (a ≠ 0) vaø t laø moät trong caùc haøm soá löôïng giaùc. Caùch giaûi: Ñaët aån phuï (ñieàu kieän cho aån phuï neáu coù), giaûi phöông trình theo aån phuï roài ñöa veà vieäc giaûi caùc phöông trình löôïng giaùc cô baûn. Ví duï: Giaûi phöông trình sau: x x 2 2 2 b) 3tan x - 2 3 tanx + 3 = 0; c) 2sin 2 + 2 sin - 2 = 0. a) 3cos x - 5cosx + 2 = 0; 2 Giaûi: ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... 15 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... III- PHÖÔNG TRÌNH BAÄC NHAÁT ÑOÁI VÔÙI sinx VAØ cosx: 1. Coâng thöùc bieán ñoåi bieåu thöùc asinx + bcosx: asinx + bcosx = a 2  b 2 sin(x + ) a b vôùi cos = vaø sin = a2  b2 a2  b2 2. Phöông trình daïng asinx + bcosx = c: Xeùt phöông trình asinx + bcosx = c (a2 + b2 ≠ 0) (1) Neáu a = 0, b ≠ 0 (hoaëc a ≠ 0, b = 0) thì (1) laø phöông trình baäc nhaát ñoái vôùi moät haøm soá löôïng giaùc. c Neáu a ≠ 0, b ≠ 0 thì (1)  a 2  b 2 sin(x + ) = c  sin(x + ) = a2  b2 Ví duï 1: Giaûi phöông trình sinx + 3 cosx = 1. Giaûi: ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... Ví duï 2: Giaûi phöông trình 3sin3x - 4cos3x = 5. Giaûi: ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... .....................................................................................................................  Ghi chuù: ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... 16 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 BAØI TAÄP REØN LUYEÄN 1. Baøi taäp cô baûn: Baøi 1: Giaûi caùc phöông trình sau: a) sin2x - sinx = 0; b) 2cos2x - 3cosx + 1 = 0; c) 2tan2x + 3tanx + 1 = 0. Baøi 2: Giaûi caùc phöông trình sau: a) cosx - 3 sinx = 2 ; b) 3 sin3x - cos3x = 2 ; c) 2sinx + 2cosx - 2 = 0; d) 5cos2x + 12sin2x - 13 = 0. Baøi 3: Giaûi caùc phöông trình sau: x x b) sin 2  2 cos  2  0 ; a) 2sin2x + 2 sin4x = 0; c) tanx - 2cotx + 1 = 0. 2 2 Baøi 4: Giaûi caùc phöông trình sau: a) 2sin2x + sinxcosx - 3cos2x = 0; b) 3sin2x - 4sinxcosx + 5cos2x = 2; 1 c) sin2x + sin2x - 2cos2x = ; d) 2cos2x - 3 3 sin2x - 4sin2x = -4. 2 Baøi 5: Giaûi caùc phöông trình sau:  a) tan(2x + 1)tan(3x - 1) = 1; b) tanx + tan(x + ) = 1. 4 Baøi 6: Giaûi caùc phöông trình sau: a) 2(sinx + cosx) + 6sinxcosx – 2 = 0; b) 2sin2x - 3 3 (sinx + cosx) + 8 = 0; c) (1 - 2 )(1 + sinx – cosx) = sin2x; d) cosx – sinx + 3sin2x – 1 = 0;  e) 5sin2x + sinx + cosx + 6 = 0; f) sin2x + 2 sin(x - ) = 1. 4 2. Baøi taäp naâng cao: Baøi 1: Giaûi caùc phöông trình sau: a) cosxcos5x = cos2xcos4x; b) sin2x + sin4x = sin6x; c) sin24x + sin23x = sin22x + sin2x; d) (sinx – cosx)2 – ( 2 + 1)(sinx – cosx) + 2 = 0. Baøi 2: Giaûi caùc phöông trình sau:  a) sinx + 3 cosx = 2sin(2x + ); b) 2sinx(cosx - 1) = 3 cos2x; 6 2 (sin3x - cos3x). b) cos3x - sinx = 3 (cosx - sin3x); c) 3 cosx - sinx = Baøi 3: Giaûi caùc phöông trình sau:    2 a) cos[ cos( x  )]  ; b) tan[ (cos x  sin x)]  1 . 4 2 4 2 CAÂU HOÛI CHUAÅN BÒ BAØI ............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. 17 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 * OÂN TAÄP CHÖÔNG I * ............................................................................................................................................................................................................................................................... ............................................................................................................................................................................................................................................................... ............................................................................................................................................................................................................................................................... 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............................................................................................................................................................................................................................................................... 18 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 ............................................................................................................................................................................................................................................................... ............................................................................................................................................................................................................................................................... ............................................................................................................................................................................................................................................................... ............................................................................................................................................................................................................................................................... ............................................................................................................................................................................................................................................................... ............................................................................................................................................................................................................................................................... BAØI TAÄP REØN LUYEÄN 1. Baøi taäp cô baûn: Baøi 1: a) Haøm soá y = cos3x coù phaûi laø haøm soá chaün khoâng? Taïi sao?  b) Haøm soá y = tan(x + ) coù phaûi laø haøm soá leû khoâng? Taïi sao? 5 3 Baøi 2: Caên cöù vaøo ñoà thò haøm soá y = sinx, tìm nhöõng giaù trò cuûa x treân ñoaïn [  ; 2] ñeå haøm soá ñoù: 2 a) Nhaän giaù trò baèng -1; b) Nhaän giaù trò aâm. Baøi 3: Giaûi caùc phöông trình sau: x1 2 1  b) sin22x = ; c) cot2 2 = 3 ; + 12x) = - 3 . a) sin(x + 1) = ; d) tan( 3 2 12 a) 2cos2x - 3cosx + 1 = 0; Baøi 4: Giaûi caùc phöông trình sau: b) 2sinx + cosx = 1. Baøi 5: Tìm giaù trò lôùn nhaát cuûa caùc haøm soá sau:  a) y = 2(1  cos x ) + 1; b) y = 3sin(x - ) - 2; 6 cos x . c) y = 3 - 4sinx; d) y = 2 - 2. Baøi taäp naâng cao: Baøi 1: Tìm taäp xaùc ñònh cuûa caùc haøm soá: 2  cos x tan x  cot x a) y = ; b) y = .  1  sin 2 x 1  tan( x  ) 3 Baøi 2: Tìm giaù trò lôùn nhaát vaø giaù trò nhoû nhaát cuûa moãi haøm soá sau:  c) y = 4sin x . a) y = 2cos( x  ) + 3; b) y = 1  sin( x 2 ) - 1; 3 CAÂU HOÛI CHUAÅN BÒ BAØI ............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................. 19 ----- Taøi lieäu löu haønh noäi boä -----
Taøi lieäu höôùng daãn töï hoïc moân Ñaïi soá vaø giaûi tích 11 CHÖÔNG II. TOÅ HÔÏP - XAÙC SUAÁT ----- oOo -----  CHUAÅN BÒ KIEÁN THÖÙC: 1. Taäp hôïp:  Taäp roãng:  laø taäp hôïp khoâng chöùa phaàn töû naøo.  A = B  A  B vaø B  A  Taäp con:  Tính chaát: a) A  A vôùi moïi taäp hôïp A. B b) Neáu A  B vaø B  C thì A  C. A c)   A vôùi moïi taäp hôïp A.  Kí hieäu: N*, Z*, Q*, R* laø caùc taäp hôïp soá A  B  x : x  A  x  B )  Soá taäp con cuûa taäp coù n phaàn töû laø 2n. khoâng coù phaàn töû 0. 2. Caùc pheùp toaùn treân taäp hôïp: Giao Hôïp Hieäu Phaàn buø B A A B A A B B Khi B  A thì A\B  A  B ={xx  A hoaëc x  B}  A  B ={xx  A vaø x  B}  A\ B ={xx  A vaø x  B} goïi laø phaàn buø cuûa B x  A x  A x  A  x A B    x A B   trong A, kí hieäu C A .  x A\ B   B x  B x  B x  B 3. Daáu hieäu chia heát:  Soá chia heát cho 2 laø nhöõng soá coù chöõ soá taän cuøng laø 0; 2; 4; 6; 8.  Soá chia heát cho 5 laø nhöõng soá coù chöõ soá taän cuøng laø 0 hoaëc 5.  Soá chia heát cho 3 laø nhöõng soá coù toång caùc chöõ soá chia heát cho 3.  Soá chia heát cho 9 laø nhöõng soá coù toång caùc chöõ soá chia heát cho 9. 4. Soá vaø chöõ soá: soá coù ba chöõ soá 128 chöõ soá  Ghi chuù: ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................... 20 ----- Taøi lieäu löu haønh noäi boä -----