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{"id":26594,"date":"2018-04-13T13:11:17","date_gmt":"2018-04-13T06:11:17","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=26594"},"modified":"2018-04-13T13:11:17","modified_gmt":"2018-04-13T06:11:17","slug":"de-kiem-tra-giai-tich-12-cuoi-nam-phan-2","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-cuoi-nam-phan-2\/","title":{"rendered":"\u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 2)"},"content":{"rendered":"

C\u00e2u 8:<\/b>\u00a0S\u00f4\u0301 \u0111i\u00ea\u0309m c\u01b0\u0323c ti\u00ea\u0309u cu\u0309a ha\u0300m s\u00f4\u0301 y = x4<\/sup>\u00a0+ x2<\/sup>\u00a0+ 1 la\u0300<\/p>\n

A. 0 \u00a0\u00a0\u00a0B. 1 \u00a0\u00a0\u00a0 C. 2\u00a0\u00a0\u00a0D. 3.<\/p>\n

C\u00e2u 9:<\/b>\u00a0Cho ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 1. Ti\u0301ch ca\u0301c gia\u0301 tri\u0323 c\u01b0\u0323c \u0111a\u0323i cu\u0309a ha\u0300m s\u00f4\u0301 la\u0300<\/p>\n

A. 0 \u00a0\u00a0\u00a0B. \u20133 \u00a0\u00a0\u00a0C. 3\u00a0\u00a0\u00a0D. \u20136<\/p>\n

C\u00e2u 10:<\/b>\u00a0\u0110\u01b0\u01a1\u0300ng th\u0103\u0309ng \u0111i qua hai \u0111i\u00ea\u0309m c\u01b0\u0323c tri\u0323 cu\u0309a ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 2 co\u0301 ph\u01b0\u01a1ng tri\u0300nh la\u0300<\/p>\n

A. y = -x + 2 \u00a0\u00a0\u00a0B. y = x + 2\u00a0\u00a0\u00a0C. y = 2x + 2\u00a0\u00a0\u00a0D. y = -2x + 2<\/p>\n

C\u00e2u 11:<\/b>\u00a0Ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 6x2<\/sup>\u00a0co\u0301 gia\u0301 tri\u0323 nho\u0309 nh\u00e2\u0301t va\u0300 gia\u0301 tri\u0323 l\u01a1\u0301n nh\u00e2\u0301t tr\u00ean \u0111oa\u0323n [-1; 5] t\u01b0\u01a1ng \u01b0\u0301ng la\u0300<\/p>\n

A. \u201325 va\u0300 \u20137 \u00a0\u00a0\u00a0B. \u20137 va\u0300 0 \u00a0\u00a0\u00a0C. \u201332 va\u0300 0 \u00a0\u00a0\u00a0D. \u201332 va\u0300 \u20137.<\/p>\n

C\u00e2u 12:<\/b>\u00a0Ti\u00ea\u0301p tuy\u00ea\u0301n ta\u0323i \u0111i\u00ea\u0309m A(0; 2) cu\u0309a \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 3x + 2 co\u0301 ph\u01b0\u01a1ng tri\u0300nh la\u0300<\/p>\n

A. y = -3x + 2\u00a0\u00a0\u00a0B. y = 3x + 2\u00a0\u00a0\u00a0C. y = 2x + 2\u00a0\u00a0\u00a0D. y = x + 2<\/p>\n

C\u00e2u 13:<\/b>\u00a0Cho ha\u0300m s\u00f4\u0301 y = 2x4<\/sup>\u00a0– 5x2<\/sup>\u00a0– 7. S\u00f4\u0301 ti\u00ea\u0301p tuy\u00ea\u0301n \u0111i qua \u0111i\u00ea\u0309m M(0; -7) cu\u0309a \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 la\u0300<\/p>\n

A. 1 \u00a0\u00a0\u00a0B. 2\u00a0\u00a0\u00a0 C. 3 \u00a0\u00a0\u00a0D. 4<\/p>\n

C\u00e2u 14:<\/b>\u00a0S\u00f4\u0301 giao \u0111i\u00ea\u0309m cu\u0309a \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+ 2x + 1 v\u01a1\u0301i tru\u0323c hoa\u0300nh la\u0300<\/p>\n

A. 0\u00a0\u00a0\u00a0B. 1\u00a0\u00a0\u00a0C. 2\u00a0\u00a0\u00a0D. 4<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
8-B<\/td>\n9-B<\/td>\n10-D<\/td>\n11-C<\/td>\n12-A<\/td>\n13-C<\/td>\n14-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 8:<\/b><\/p>\n

Ta c\u00f3: y’ = 4x3<\/sup>\u00a0+ 2x = 2x(2x2<\/sup>\u00a0+ 1) . Do \u0111\u00f3: y’ = 0 <=> x = 0 L\u1eadp b\u1ea3ng bi\u1ebfn thi\u00ean ta suy ra h\u00e0m s\u1ed1 \u0111\u00e3 cho ch\u1ec9 c\u00f3 m\u1ed9t c\u1ef1c ti\u1ec3u x = 0<\/p>\n

C\u00e2u 9:<\/b><\/p>\n

Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 6x; y’ = 0 <=> x = 0, x = 2. C\u00e1c gi\u00e1 tr\u1ecb c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 l\u00e0:<\/p>\n

y1<\/sub>\u00a0= y(0) = 1, y2<\/sub>\u00a0= y(2) = -3<\/p>\n

V\u1eady t\u00edch c\u00e1c gi\u00e1 tr\u1ecb c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 y1<\/sub>y2<\/sub>\u00a0= -3<\/p>\n

C\u00e2u 10:<\/b><\/p>\n

Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 6x, y’ = 0 <=> x = 0, x = 2. C\u00e1c \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho l\u00e0 A(0; 2), B(2; -2)<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng AB l\u00e0<\/p>\n

\"\"<\/p>\n

C\u00e2u 11:<\/b><\/p>\n

Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 12x, y’ = 0 <=> x = 0, x = 4 So s\u00e1nh c\u00e1c gi\u00e1 tr\u1ecb:<\/p>\n

y(-1) = -7, y(0) = 0, y(4) = -32, y(5) = -25<\/p>\n

Ta c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho tr\u00ean \u0111o\u1ea1n [-1; 5] l\u00e0 0 v\u00e0 -32.<\/p>\n

C\u00e2u 12:<\/b><\/p>\n

Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 3. H\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a ti\u1ebfp tuy\u1ebfn l\u00e0: k = y'(0) = -3<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn l\u00e0 y – 2 = -3(x – 0) <=> y = -3x + 2<\/p>\n

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

X\u00e9t \u0111i\u1ec3m A(x0<\/sub>, y0<\/sub>) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. Ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn d t\u1ea1i A c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 l\u00e0<\/p>\n

\"\"<\/p>\n

Ti\u1ebfp tuy\u1ebfn \u0111i qua M(0;-7) khi v\u00e0 ch\u1ec9 khi<\/p>\n

\"\"<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean (\u1ea9n x0<\/sub>) c\u00f3 ba nghi\u1ec7m n\u00ean c\u00f3 ba ti\u1ebfp tuy\u1ebfn th\u1ecfa m\u00e3n y\u00eau c\u1ea7u b\u00e0i to\u00e1n.<\/p>\n

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

Ta c\u00f3: y’ = 3x2<\/sup>\u00a0– 6x + 2,<\/p>\n

\"\"<\/p>\n

Gi\u00e1 tr\u1ecb c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 l\u00e0:<\/p>\n

\"\"<\/p>\n

L\u1eadp b\u1ea3ng bi\u1ebfn thi\u00ean ta suy ra \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u1eaft tr\u1ee5c ho\u00e0nh t\u1ea1i 1 \u0111i\u1ec3m.<\/p>\n","protected":false},"excerpt":{"rendered":"

C\u00e2u 8:\u00a0S\u00f4\u0301 \u0111i\u00ea\u0309m c\u01b0\u0323c ti\u00ea\u0309u cu\u0309a ha\u0300m s\u00f4\u0301 y = x4\u00a0+ x2\u00a0+ 1 la\u0300 A. 0 \u00a0\u00a0\u00a0B. 1 \u00a0\u00a0\u00a0 C. 2\u00a0\u00a0\u00a0D. 3. C\u00e2u 9:\u00a0Cho ha\u0300m s\u00f4\u0301 y = x3\u00a0– 3×2\u00a0+ 1. Ti\u0301ch ca\u0301c gia\u0301 tri\u0323 c\u01b0\u0323c \u0111a\u0323i cu\u0309a ha\u0300m s\u00f4\u0301 la\u0300 A. 0 \u00a0\u00a0\u00a0B. \u20133 \u00a0\u00a0\u00a0C. 3\u00a0\u00a0\u00a0D. \u20136 C\u00e2u 10:\u00a0\u0110\u01b0\u01a1\u0300ng th\u0103\u0309ng \u0111i qua hai \u0111i\u00ea\u0309m […]<\/p>\n","protected":false},"author":3,"featured_media":26595,"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":"\n\u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 2)<\/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\/de-kiem-tra-giai-tich-12-cuoi-nam-phan-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"\u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 2)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 8:\u00a0S\u00f4\u0301 \u0111i\u00ea\u0309m c\u01b0\u0323c ti\u00ea\u0309u cu\u0309a ha\u0300m s\u00f4\u0301 y = x4\u00a0+ x2\u00a0+ 1 la\u0300 A. 0 \u00a0\u00a0\u00a0B. 1 \u00a0\u00a0\u00a0 C. 2\u00a0\u00a0\u00a0D. 3. C\u00e2u 9:\u00a0Cho ha\u0300m s\u00f4\u0301 y = x3\u00a0– 3x2\u00a0+ 1. Ti\u0301ch ca\u0301c gia\u0301 tri\u0323 c\u01b0\u0323c \u0111a\u0323i cu\u0309a ha\u0300m s\u00f4\u0301 la\u0300 A. 0 \u00a0\u00a0\u00a0B. \u20133 \u00a0\u00a0\u00a0C. 3\u00a0\u00a0\u00a0D. \u20136 C\u00e2u 10:\u00a0\u0110\u01b0\u01a1\u0300ng th\u0103\u0309ng \u0111i qua hai \u0111i\u00ea\u0309m […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-cuoi-nam-phan-2\/\" \/>\n<meta property=\"og:site_name\" content=\"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12 l\u1edbn nh\u1ea5t t\u1ea1i Vi\u1ec7t Nam\" \/>\n<meta property=\"article:published_time\" content=\"2018-04-13T06:11:17+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/1-172.png\" \/>\n\t<meta property=\"og:image:width\" content=\"242\" \/>\n\t<meta property=\"og:image:height\" content=\"56\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/png\" \/>\n<meta name=\"author\" content=\"H\u00e0 Trang\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"H\u00e0 Trang\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"3 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-cuoi-nam-phan-2\/\",\"url\":\"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-cuoi-nam-phan-2\/\",\"name\":\"\u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 2)\",\"isPartOf\":{\"@id\":\"https:\/\/lop12.edu.vn\/#website\"},\"datePublished\":\"2018-04-13T06:11:17+00:00\",\"dateModified\":\"2018-04-13T06:11:17+00:00\",\"author\":{\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/2da855e5d06b553ac0065157cc261a45\"},\"breadcrumb\":{\"@id\":\"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-cuoi-nam-phan-2\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-cuoi-nam-phan-2\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-cuoi-nam-phan-2\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/lop12.edu.vn\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"\u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 2)\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/lop12.edu.vn\/#website\",\"url\":\"https:\/\/lop12.edu.vn\/\",\"name\":\"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12\",\"description\":\"\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/lop12.edu.vn\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"Person\",\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/2da855e5d06b553ac0065157cc261a45\",\"name\":\"H\u00e0 Trang\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/545a8100db1147f314111301d261dc33?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/545a8100db1147f314111301d261dc33?s=96&d=mm&r=g\",\"caption\":\"H\u00e0 Trang\"},\"url\":\"https:\/\/lop12.edu.vn\/author\/trangvth2\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"\u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 2)","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/lop12.edu.vn\/de-kiem-tra-giai-tich-12-cuoi-nam-phan-2\/","og_locale":"en_US","og_type":"article","og_title":"\u0110\u1ec1 ki\u1ec3m tra Gi\u1ea3i t\u00edch 12 cu\u1ed1i n\u0103m (ph\u1ea7n 2)","og_description":"C\u00e2u 8:\u00a0S\u00f4\u0301 \u0111i\u00ea\u0309m c\u01b0\u0323c ti\u00ea\u0309u cu\u0309a ha\u0300m s\u00f4\u0301 y = x4\u00a0+ x2\u00a0+ 1 la\u0300 A. 0 \u00a0\u00a0\u00a0B. 1 \u00a0\u00a0\u00a0 C. 2\u00a0\u00a0\u00a0D. 3. 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