Đường tiệm cận của đồ thị hàm số

Home Lớp 12 Toán lớp 12 - Sách Kết nối tri thức Bài 3. Đường tiệm cận của đồ thị hàm sốBài tập nâng cao

{"common":{"save":0,"post_id":"7524","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5372","post_id":"7524","mon_id":"1159285","chapter_id":"1159288","question":"Cho hàm s\u1ed1 $y=f(x)$<\/span> xác \u0111\u1ecbnh, liên t\u1ee5c trên R và có b\u1ea3ng bi\u1ebfn thiên nh\u01b0 hình bên d\u01b0\u1edbi.<\/p><svg height=\"100\" width=\"320\" xmlns:xlink=\"http:\/\/www.w3.org\/1999\/xlink\"> <g><title><\/title><rect fill=\"#fff\" height=\"102\" id=\"canvas_background\" width=\"322\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" height=\"100%\" id=\"canvasGrid\" overflow=\"visible\" width=\"100%\" x=\"0\" y=\"0\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g><title><\/title><image height=\"95\" id=\"svg_1\" stroke=\"null\" width=\"319.00001\" x=\"-0.5\" 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y=\"-0.5\"><\/image> <\/g> <\/svg><\/span><\/p>T\u1ed5ng s\u1ed1 ti\u1ec7m c\u1eadn ngang và ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb hàm s\u1ed1 $y=\\dfrac{1}{2f(x)-1}$<\/span> là<\/p>","options":["A.<\/strong> 4","B.<\/strong> 3","C.<\/strong> 1","D.<\/strong> 2"],"correct":"1","level":"3","hint":"S\u1eed d\u1ee5ng \u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn ngang, ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/p>Cho hàm s\u1ed1 $y=\\dfrac{P(x)}{Q(x)}$<\/span> có t\u1eadp xác \u0111\u1ecbnh D.<\/p>\u0110i\u1ec1u ki\u1ec7n c\u1ea7n: Gi\u1ea3i Q(x) = 0 ⇔ x = $x_0$<\/span> là ti\u1ec7m c\u1eadn \u0111\u1ee9ng khi tho\u1ea3 mãn \u0111i\u1ec1u ki\u1ec7n \u0111\u1ee7.<\/p>\u0110i\u1ec1u ki\u1ec7n \u0111\u1ee7: $x_0$<\/span> không ph\u1ea3i là nghi\u1ec7m c\u1ee7a P(x) ⇒ x = $x_0$<\/span> là ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/p>","answer":"Ch\u1ecdn A.<\/strong> 4.<\/span><\/p>\u0110\u1eb7t h(x) = $\\dfrac{1}{2f(x)-1}$<\/span>.<\/p>*) Ti\u1ec7m c\u1eadn ngang<\/p>Ta có $\\displaystyle\\lim_{x\\to+\\infty}h(x)=\\displaystyle\\lim_{x\\to+\\infty}\\dfrac{1}{2f(x)-1}=0$<\/span>.<\/p>$\\displaystyle\\lim_{x\\to-\\infty}h(x)=\\displaystyle\\lim_{x\\to-\\infty}\\dfrac{1}{2f(x)-1}=0$<\/span>.<\/p>Suy ra \u0111\u1ed3 th\u1ecb hàm s\u1ed1 có m\u1ed9t \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn ngang y = 0. *) Ti\u1ec7m c\u1eadn \u0111\u1ee9ng<\/p>Xét ph\u01b0\u01a1ng trình: $2f(x)-1=0$<\/span> ⇔ $f(x)=\\dfrac{1}{2}$<\/span>.<\/p>D\u1ef1a vào b\u1ea3ng bi\u1ebfn thi\u1ec7n ta th\u1ea5y ph\u01b0\u01a1ng trình $f(x)=\\dfrac{1}{2}$<\/span> có ba nghi\u1ec7m phân bi\u1ec7t a, b, c tho\u1ea3 mãn a < 1 \u0110\u1ed3ng th\u1eddi $\\displaystyle\\lim_{x\\to a^+}h(x)=\\displaystyle\\lim_{x\\to b^-}h(x)=\\displaystyle\\lim_{x\\to c^+}h(x)=+\\infty$<\/span> nên \u0111\u1ed3 th\u1ecb hàm s\u1ed1 y = h(x) có ba \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng là x = a; x = b và x = c.<\/p>V\u1eady t\u1ed5ng s\u1ed1 ti\u1ec7m c\u1eadn ngang và ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb hàm s\u1ed1 y = h(x) = $\\dfrac{1}{2f(x)-1}$<\/span> là 4.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-06-28 00:54:54","option_type":"txt","len":0},{"id":"5373","post_id":"7524","mon_id":"1159285","chapter_id":"1159288","question":"Cho hàm s\u1ed1 y = f(x) liên t\u1ee5c trên R\\{1} và có b\u1ea3ng bi\u1ebfn thiên nh\u01b0 sau.<\/p><svg height=\"120\" width=\"320\" xmlns:xlink=\"http:\/\/www.w3.org\/1999\/xlink\"> <g><title><\/title><rect fill=\"#fff\" height=\"122\" id=\"canvas_background\" width=\"322\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" height=\"100%\" id=\"canvasGrid\" overflow=\"visible\" width=\"100%\" x=\"0\" y=\"0\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g><title><\/title><image height=\"119\" id=\"svg_1\" stroke=\"null\" width=\"318.99999\" x=\"1\" xlink:href=\"data:image\/png;base64,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\" y=\"-1.5\"><\/image> <\/g> <\/svg><\/span><\/p>\u0110\u1ed3 th\u1ecb $y=\\dfrac{1}{2f(x)+3} $<\/span> có bao nhiêu \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng?<\/p>","options":["A.<\/strong> 3","B.<\/strong> 0","C.<\/strong> 1","D.<\/strong> 2"],"correct":"4","level":"3","hint":"S\u1eed d\u1ee5ng \u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn ngang, ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/p>Cho hàm s\u1ed1 $y=\\dfrac{P(x)}{Q(x)}$<\/span> có t\u1eadp xác \u0111\u1ecbnh D.<\/p>\u0110i\u1ec1u ki\u1ec7n c\u1ea7n: Gi\u1ea3i Q(x) = 0 ⇔ x = $x_0$<\/span> là ti\u1ec7m c\u1eadn \u0111\u1ee9ng khi tho\u1ea3 mãn \u0111i\u1ec1u ki\u1ec7n \u0111\u1ee7.<\/p>\u0110i\u1ec1u ki\u1ec7n \u0111\u1ee7: $x_0$<\/span> không ph\u1ea3i là nghi\u1ec7m c\u1ee7a P(x) ⇒ x = $x_0$<\/span> là ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/p>","answer":"Ch\u1ecdn D.<\/strong> 2.<\/span><\/p>\u0110\u1eb7t $y=g(x)=\\dfrac{1}{2f(x)+3}$<\/span> có t\u1eed s\u1ed1 là 1 $\\ne$<\/span> 0, v\u1edbi m\u1ecdi x ∈ R.<\/p>Ta có $2f(x)+3=0$<\/span> ⇔ $f(x)=-\\dfrac{3}{2}$<\/span>.<\/p>D\u1ef1a vào b\u1ea3ng bi\u1ebfn thiên ta th\u1ea5y ph\u01b0\u01a1ng trình trên có 2 nghi\u1ec7m phân bi\u1ec7t $x_1 \\in (-\\infty;0)$<\/span> và $x_2 \\in (0 ; 1)$<\/span>.<\/p>Do \u0111ó \u0111\u1ed3 th\u1ecb hàm s\u1ed1 $y=g(x)=\\dfrac{1}{2f(x)+3}$<\/span> có 2 \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-06-28 01:07:42","option_type":"txt","len":0},{"id":"5374","post_id":"7524","mon_id":"1159285","chapter_id":"1159288","question":"Cho hàm s\u1ed1 y = f(x) liên t\u1ee5c trên R\\{1} và có b\u1ea3ng bi\u1ebfn thiên nh\u01b0 sau.<\/p><svg height=\"130\" width=\"320\" xmlns:xlink=\"http:\/\/www.w3.org\/1999\/xlink\"> <g><title><\/title><rect fill=\"#fff\" height=\"132\" id=\"canvas_background\" width=\"322\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" height=\"100%\" id=\"canvasGrid\" overflow=\"visible\" 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y=\"1.5\"><\/image> <\/g> <\/svg><\/span><\/p>\u0110\u1ed3 th\u1ecb $y=\\dfrac{1}{2f(x)-5} $<\/span> có bao nhiêu \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng?<\/p>","options":["A.<\/strong> 0","B.<\/strong> 4","C.<\/strong> 3","D.<\/strong> 2"],"correct":"2","level":"3","hint":"S\u1eed d\u1ee5ng \u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn ngang, ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/p>Cho hàm s\u1ed1 $y=\\dfrac{P(x)}{Q(x)}$<\/span> có t\u1eadp xác \u0111\u1ecbnh D.<\/p>\u0110i\u1ec1u ki\u1ec7n c\u1ea7n: Gi\u1ea3i Q(x) = 0 ⇔ x = $x_0$<\/span> là ti\u1ec7m c\u1eadn \u0111\u1ee9ng khi tho\u1ea3 mãn \u0111i\u1ec1u ki\u1ec7n \u0111\u1ee7.<\/p>\u0110i\u1ec1u ki\u1ec7n \u0111\u1ee7: $x_0$<\/span> không ph\u1ea3i là nghi\u1ec7m c\u1ee7a P(x) ⇒ x = $x_0$<\/span> là ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/p>","answer":"Ch\u1ecdn B.<\/strong> 4.<\/span><\/p>Ta có $2f(x)-5=0$<\/span> ⇔ $f(x)=\\dfrac{5}{2}$<\/span>.<\/p>Ph\u01b0\u01a1ng trình trên có 4 nghi\u1ec7m phân bi\u1ec7t $x_1, x_2, x_3, x_4$<\/span> $\\ne 1$<\/span> và gi\u1edbi h\u1ea1n c\u1ee7a hàm s\u1ed1 $y=\\dfrac{1}{2f(x)-5} $<\/span> t\u1ea1i các \u0111i\u1ec3m $x_1, x_2, x_3, x_4$<\/span> \u0111\u1ec1u b\u1eb1ng $\\pm \\infty$<\/span>.<\/p>M\u1eb7t khác $\\displaystyle\\lim_{x\\to1^\\pm}\\dfrac{1}{2f(x)-5}=0$<\/span> nên x = 1 không ph\u1ea3i ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/p>V\u1eady \u0111\u1ed3 th\u1ecb hàm s\u1ed1 $y=\\dfrac{1}{2f(x)-5} $<\/span> có 4 \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-06-28 01:25:25","option_type":"txt","len":0}]}