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":"7523","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5408","post_id":"7523","mon_id":"1159285","chapter_id":"1159288","question":"Cho hàm s\u1ed1 $y=\\dfrac{ax+1}{bx-2}$<\/span>. Tìm a, b \u0111\u1ec3 \u0111\u1ed3 th\u1ecb hàm s\u1ed1 có x = 1 là ti\u1ec7m c\u1eadn \u0111\u1ee9ng và y = $\\dfrac{1}{2}$<\/span> là ti\u1ec7m c\u1eadn ngang.<\/p>","options":["A.<\/strong> a = –1; b = 2","B.<\/strong> a = 4; b = 4","C.<\/strong> a = –1; b = –2","D.<\/strong> a = 1; b = 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>","answer":"Ch\u1ecdn D.<\/strong> a = 1; b = 2.<\/span><\/p>N\u1ebfu b = 0 thì \u0111\u1ed3 th\u1ecb hàm s\u1ed1 $y=\\dfrac{ax+1}{-2}$<\/span> không có ti\u1ec7m c\u1eadn.<\/p>N\u1ebfu b $\\ne$<\/span> 0 thì t\u1eadp xác \u0111\u1ecbnh c\u1ee7a hàm s\u1ed1 là D = $R\\setminus\\bigg\\{\\dfrac{2}{b}\\bigg\\}$<\/span>.<\/p>Ta có $\\displaystyle\\lim_{x\\to\\frac{2}{b}^+}y=\\displaystyle\\lim_{x\\to\\frac{2}{b}^+}\\dfrac{ax+1}{bx-2}=\\infty$<\/span> ⇒ $x=\\dfrac{2}{b}$<\/span> là ti\u1ec7m c\u1eadn \u0111\u1ee9ng. Khi \u0111ó theo \u0111\u1ec1 bài ta có $\\dfrac{2}{b}=1$<\/span> ⇒ b = 2.<\/p>M\u1eb7t khác, $\\displaystyle\\lim_{x\\to\\pm\\infty}y=\\displaystyle\\lim_{x\\to\\pm\\infty}\\dfrac{ax+1}{bx-2}=\\displaystyle\\lim_{x\\to\\pm\\infty}\\dfrac{a+\\dfrac{1}{x}}{b-\\dfrac{2}{x}}=\\dfrac{a}{b}$<\/span>. Suy ra y = $\\dfrac{a}{b}$<\/span> là ti\u1ec7m c\u1eadn ngang. <\/p>K\u1ebft h\u1ee3p \u0111\u1ec1 bài và b = 2 ta \u0111\u01b0\u1ee3c $\\dfrac{a}{2}=\\dfrac{1}{2}$<\/span> ⇒ a = 1.<\/p>V\u1eady a = 1 và b = 2.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-06-30 02:16:17","option_type":"txt","len":2},{"id":"5410","post_id":"7523","mon_id":"1159285","chapter_id":"1159288","question":"Có bao nhiêu giá tr\u1ecb nguyên m ∈ [–10 ; 10] sao cho \u0111\u1ed3 th\u1ecb hàm s\u1ed1 $y=\\dfrac{x-1}{2x^2+6x-m-3}$<\/span> có hai \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng?<\/p>","options":["A.<\/strong> 19","B.<\/strong> 15","C.<\/strong> 17","D.<\/strong> 18"],"correct":"3","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 C.<\/strong> 17.<\/span><\/p>Ta có \u0111\u1ed3 th\u1ecb hàm s\u1ed1 $y=\\dfrac{x-1}{2x^2+6x-m-3}$<\/span> có hai \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng khi ph\u01b0\u01a1ng trình $2x^2+6x-m-3=0$<\/span> có hai nghi\u1ec7m phân bi\u1ec7t khác 1  ⇔ $\\begin{cases}3^2-2(-m-3)>0\\\\2.1^2+6.1-m-3\\ne0\\end{cases}$<\/span> ⇔ $\\begin{cases}m > -\\dfrac{15}{2}\\\\m\\ne5\\end{cases}$<\/span>.<\/p>T\u1eeb \u0111ó ta suy ra t\u1eadp các giá tr\u1ecb nguyên c\u1ee7a m th\u1ecfa mãn là m ∈ {–7; –6; –5; –4; –3; –2; –1; 0; 1; 2; 3; 4; 6; 7; 8; 9; 10}.<\/p>V\u1eady có 17 giá tr\u1ecb nguyên c\u1ee7a m th\u1ecfa mãn.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-06-30 02:19:04","option_type":"txt","len":0},{"id":"5412","post_id":"7523","mon_id":"1159285","chapter_id":"1159288","question":"Có bao nhiêu giá tr\u1ecb nguyên d\u01b0\u01a1ng c\u1ee7a tham s\u1ed1 m \u0111\u1ec3 \u0111\u1ed3 th\u1ecb hàm s\u1ed1 $y=\\dfrac{x-1}{x^2-8x+m}$<\/span> có 3 \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn?<\/p>","options":["A.<\/strong> 14","B.<\/strong> 8","C.<\/strong> 15","D.<\/strong> 16"],"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> 14.<\/span><\/p>Ta có $\\displaystyle\\lim_{x\\to-\\infty}\\dfrac{x-1}{x^2-8x+m}=\\displaystyle\\lim_{x\\to+\\infty}\\dfrac{x-1}{x^2-8x+m}=0$<\/span> nên hàm s\u1ed1 có m\u1ed9t ti\u1ec7m c\u1eadn ngang y = 0.<\/p>Hàm s\u1ed1 có 3 \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn khi và ch\u1ec9 khi hàm s\u1ed1 có hai \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng ⇔ ph\u01b0\u01a1ng trình $x^2-8x+m=0$<\/span> có hai nghi\u1ec7m phân bi\u1ec7t khác 1 ⇔ $\\begin{cases}\\Delta^\\prime=16-m>0\\\\1^2-8.1+m\\ne0\\end{cases}$<\/span> ⇔ $\\begin{cases}m<16\\\\m\\ne7\\end{cases}$<\/span>. K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n m nguyên d\u01b0\u01a1ng ta có m ∈ {1; 2; 3; 4; 5; 6; 8; 9; 10; 11; 12; 13; 14; 15}.<\/p>V\u1eady có 14 giá tr\u1ecb c\u1ee7a m th\u1ecfa mãn \u0111\u1ec1 bài.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-06-30 02:27:06","option_type":"txt","len":0}]}