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{"id":28013,"date":"2018-04-19T13:13:09","date_gmt":"2018-04-19T06:13:09","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=28013"},"modified":"2018-04-19T13:13:09","modified_gmt":"2018-04-19T06:13:09","slug":"chuong-1-bai-tap-trac-nghiem-giai-tich-12-cuc-tri-cua-ham-so-phan-5","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/chuong-1-bai-tap-trac-nghiem-giai-tich-12-cuc-tri-cua-ham-so-phan-5\/","title":{"rendered":"Ch\u01b0\u01a1ng 1 – B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: C\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 (Ph\u1ea7n 5)"},"content":{"rendered":"

C\u00e2u 13:<\/b>\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– 3mx2<\/sup>\u00a0+ m c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb B, C th\u1eb3ng h\u00e0ng v\u1edbi \u0111i\u1ec3m A(-1;3)?<\/p>\n

A. m = 0 \u00a0\u00a0\u00a0 B. m = 1 \u00a0\u00a0\u00a0 C. m = -3\/2 \u00a0\u00a0\u00a0 D. m = -3\/2 ho\u1eb7c m = 1<\/p>\n

C\u00e2u 14:<\/b>\u00a0Cho h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0– 6x + 8 (C). Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 (C) l\u00e0:<\/p>\n

A. y = 6x – 6 \u00a0\u00a0\u00a0 B. y = -6x – 6 \u00a0\u00a0\u00a0 C. y = 6x + 6 \u00a0\u00a0\u00a0 D. y = -6x + 6<\/p>\n

C\u00e2u 15:<\/b>\u00a0Cho h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0-3x2<\/sup>\u00a0– 9x + 4. Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 tr\u00ean l\u00e0:<\/p>\n

A. y = -8x – 17 \u00a0\u00a0\u00a0 B. y = x + 7 \u00a0\u00a0\u00a0 C. y = -x + 1 \u00a0\u00a0\u00a0 D. Kh\u00f4ng t\u1ed3n t\u1ea1i<\/p>\n

C\u00e2u 16:<\/b>\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 3mx + 1 – m t\u1ea1o v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng \u0394: 3x + y – 8 = 0 m\u1ed9t g\u00f3c 45o<\/sup>\u00a0?<\/p>\n

A. m = 0 \u00a0\u00a0\u00a0B. m = 2\u00a0\u00a0\u00a0 C.m = 3\/4 \u00a0\u00a0\u00a0D. m = 2 ho\u1eb7c m = 3\/4<\/p>\n

C\u00e2u 17:<\/b>\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ m2<\/sup>x + m c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb \u0111\u1ed1i x\u1ee9ng qua \u0111\u01b0\u1eddng th\u1eb3ng:<\/p>\n

\"\"<\/p>\n

A. m = 0 \u00a0\u00a0\u00a0B. m = 1 \u00a0\u00a0\u00a0 C. m = -1\u00a0\u00a0\u00a0 D. Kh\u00f4ng t\u1ed3n t\u1ea1i<\/p>\n

C\u00e2u 18:<\/b>\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x4<\/sup>\u00a0– 2mx2<\/sup>\u00a0+ m\u00a04<\/sup>\u00a0+ 2m c\u00f3 ba \u0111i\u1ec3m c\u1ef1c tr\u1ecb t\u1ea1o th\u00e0nh tam gi\u00e1c \u0111\u1ec1u?<\/p>\n

A. m = 0 \u00a0\u00a0\u00a0 B. m = \u221b3 \u00a0\u00a0\u00a0C.-\u221b3 \u00a0\u00a0\u00a0 D. Kh\u00f4ng t\u1ed3n t\u1ea1i<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
13-D<\/td>\n14-D<\/td>\n15-A<\/td>\n16-C<\/td>\n17-D<\/td>\n18-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 13:<\/b><\/p>\n

y\u2019=3x2<\/sup>-6mx=3x(x – 2m)<\/p>\n

H\u00e0m s\u1ed1 c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb => y\u2019=0 c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t <=> m \u2260 0 (*)<\/p>\n

T\u1ecda \u0111\u1ed9 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb l\u00e0 B(0;m) v\u00e0 C(2m;-4m3<\/sup>\u00a0+ m)<\/p>\n

AB\u2192<\/i>\u00a0=(1;m \u2013 3);\u00a0AC\u2192<\/i>\u00a0=(2m+1; -4m3<\/sup>\u00a0+ m-3)<\/p>\n

A, B, C th\u1eb3ng h\u00e0ng<\/p>\n

\"\"<\/p>\n

\u0110\u1ed1i chi\u1ebfu v\u1edbi \u0111i\u1ec1u ki\u1ec7n (*) c\u00f3 m \u2208 {-3\/2; 1}<\/p>\n

C\u00e2u 14:<\/b><\/p>\n

C\u00e1ch 1: Ta c\u00f3 y\u2019=3x2<\/sup>-6x-6 ; y\u201d=6x – 6<\/p>\n

\"\"<\/p>\n

Do \u0111\u00f3 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 \u0111i\u1ec3m c\u1ef1c tr\u1ecb l\u00e0 A(1 + \u221a3; -6\u221a3) v\u00e0 B(1 – \u221a3; 6\u221a3) .<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb l\u00e0:<\/p>\n

\"\"<\/p>\n

C\u00e1ch 2: Ta c\u00f3:<\/p>\n

\"\"<\/p>\n

G\u1ecdi x1<\/sub>, x2<\/sub>\u00a0l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh y\u2019(x)= 3x2<\/sup>-6x-6=0 . Khi \u0111\u00f3 ta c\u00f3 A(x1<\/sub>, y(x1<\/sub>)), BA(x2<\/sub>, y(x2<\/sub>)) l\u00e0 hai c\u1ef1c tr\u1ecb c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 C v\u1edbi y'(x1<\/sub>) = y'(x2<\/sub>) = 0 .<\/p>\n

Do \u0111\u00f3 ta c\u00f3:<\/p>\n

\"\"<\/p>\n

V\u1eady A, B thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng y= – 6x+6.<\/p>\n

C\u00e2u 15:<\/b><\/p>\n

y’ = 3x2<\/sup>\u00a0– 6x – 9, y” = 6x – 6<\/p>\n

\"\"<\/p>\n

Do \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb l\u00e0 A(-1;0) v\u00e0 B(3;-23).<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb l\u00e0:<\/p>\n

\"\"<\/p>\n

C\u00e2u 16:<\/b><\/p>\n

Ta c\u00f3 y’ = 3x2<\/sup>\u00a0– 6x + 3m. H\u00e0m s\u1ed1 c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb <=> y\u2019=0 c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t<\/p>\n

<=> \u0394’ = 32<\/sup>\u00a0-3.3m > 0 <=> m < 1 (*)<\/p>\n

Chia y cho y\u2019 ta \u0111\u01b0\u1ee3c:<\/p>\n

\"\"<\/p>\n

Gi\u1ea3 s\u1eed x1<\/sub>, x2<\/sub>\u00a0l\u00e0 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t c\u1ee7a y\u2019=0<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u00f3 d\u1ea1ng (d) : y= (2m-2)x+1<\/p>\n

(d) c\u00f3 vect\u01a1 ph\u00e1p tuy\u1ebfn l\u00e0\u00a0n1<\/sub>\u2192<\/i>\u00a0= (2m – 2; -1)<\/p>\n

(\u0394) : 3x+y-8=0 c\u00f3 vect\u01a1 ph\u00e1p tuy\u1ebfn l\u00e0\u00a0n2<\/sub>\u2192<\/i>(3; 1)<\/p>\n

V\u00ec g\u00f3c gi\u1eefa \u0111\u01b0\u1eddng th\u1eb3ng (d) v\u00e0 (\u0394) l\u00e0 45o<\/sup>\u00a0n\u00ean<\/p>\n

\"\"<\/p>\n

\"B\u00e0i<\/p>\n

\u0110\u1ed1i chi\u1ebfu \u0111i\u1ec1u ki\u1ec7n (*) c\u00f3 m = 3\/4<\/p>\n

C\u00e2u 17:<\/b><\/p>\n

y’ = 3x2<\/sup>\u00a0+ 6x + m2<\/sup>\u00a0. H\u00e0m s\u1ed1 c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb => y\u2019=0 c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t <=> \u0394’ = 32<\/sup>\u00a0– 3.m2<\/sup>\u00a0> 0 <=> -\u221a3 < m < \u221a3<\/p>\n

Chia y cho y\u2019 ta \u0111\u01b0\u1ee3c:<\/p>\n

\"\"<\/p>\n

Gi\u1ea3 s\u1eed x1<\/sub>, x2<\/sub>\u00a0l\u00e0 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t c\u1ee7a y\u2019=0.<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u00f3 d\u1ea1ng<\/p>\n

\"\"<\/p>\n

(d) c\u00f3 vect\u01a1 ph\u00e1p tuy\u1ebfn l\u00e0<\/p>\n

\"B\u00e0i\"\"<\/p>\n

V\u00ec hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb \u0111\u1ed1i x\u1ee9ng v\u1edbi nhau qua (\u0394) n\u00ean (d) \u22a5 (\u0394)<\/p>\n

\"\"<\/p>\n

Th\u1eed l\u1ea1i khi m=0 ta c\u00f3: y = x3<\/sup>\u00a0+ 3x2<\/sup>; y’ = 3x2<\/sup>\u00a0+ 6x; y” = 6x + 6<\/p>\n

\"\"<\/p>\n

y”(0) = 6 > 0; y”(-2) = -6 < 0<\/p>\n

T\u1ecda \u0111\u1ed9 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 l\u00e0 O(0;0), A(-2;4)<\/p>\n

Trung \u0111i\u1ec3m c\u1ee7a OA l\u00e0 I(-1;2).<\/p>\n

Ta th\u1ea5y I(-1,2) kh\u00f4ng thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng (\u0394) . V\u1eady kh\u00f4ng t\u1ed3n t\u1ea1i m.<\/p>\n

C\u00e2u 18:<\/b><\/p>\n

y’ = 4x3<\/sup>\u00a0– 4mx = 4x(x2<\/sup>\u00a0– m)<\/p>\n

H\u00e0m s\u1ed1 c\u00f3 ba \u0111i\u1ec3m c\u1ef1c tr\u1ecb => y\u2019=0 c\u00f3 ba nghi\u1ec7m ph\u00e2n bi\u1ec7t <=> m > 0.<\/p>\n

Khi \u0111\u00f3 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 ba \u0111i\u1ec3m c\u1ef1c tr\u1ecb l\u00e0 :<\/p>\n

A(0; m4<\/sup>\u00a0+ 2m), B(-\u221am; m4<\/sup>\u00a0– m2<\/sup>\u00a0+ 2m), C(\u221am; m4<\/sup>\u00a0– m2<\/sup>\u00a0+ 2m)<\/p>\n

\u0394ABC \u0111\u1ec1u khi AB=AC<\/p>\n

\"\"<\/p>\n

\u0110\u1ed1i chi\u1ebfu v\u1edbi \u0111i\u1ec1u ki\u1ec7n t\u1ed3n t\u1ea1i c\u1ef1c tr\u1ecb ta c\u00f3 m = \u221b3 l\u00e0 gi\u00e1 tr\u1ecb c\u1ea7n t\u00ecm.<\/p>\n","protected":false},"excerpt":{"rendered":"

C\u00e2u 13:\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x3\u00a0– 3mx2\u00a0+ m c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb B, C th\u1eb3ng h\u00e0ng v\u1edbi \u0111i\u1ec3m A(-1;3)? A. m = 0 \u00a0\u00a0\u00a0 B. m = 1 \u00a0\u00a0\u00a0 C. m = -3\/2 \u00a0\u00a0\u00a0 D. m = -3\/2 ho\u1eb7c m = 1 C\u00e2u 14:\u00a0Cho h\u00e0m s\u1ed1 […]<\/p>\n","protected":false},"author":3,"featured_media":28014,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[157],"tags":[1447,1446],"yoast_head":"\nCh\u01b0\u01a1ng 1 - B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: C\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 (Ph\u1ea7n 5)<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/chuong-1-bai-tap-trac-nghiem-giai-tich-12-cuc-tri-cua-ham-so-phan-5\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Ch\u01b0\u01a1ng 1 - B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: C\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 (Ph\u1ea7n 5)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 13:\u00a0V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a m, \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x3\u00a0– 3mx2\u00a0+ m c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb B, C th\u1eb3ng h\u00e0ng v\u1edbi \u0111i\u1ec3m A(-1;3)? 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