Chú ý: Để đảm bảo quyền lợi và bảo vệ tài khoản của mình
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HomeLớp 12Toán lớp 12 - Sách Kết nối tri thứcBài 2. Giá trị lớn nhất và giá trị nhỏ nhất của hàm sốBài tập trung bình
{"common":{"save":0,"post_id":"7511","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5345","post_id":"7511","mon_id":"1159285","chapter_id":"1159288","question":"<p>G\u1ecdi m, M l\u1ea7n l\u01b0\u1ee3t là giá tr\u1ecb nh\u1ecf nh\u1ea5t và giá tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a hàm s\u1ed1 <span class=\"math-tex\">$f(x)=\\dfrac{1}{2}x-\\sqrt{x+1}$<\/span> trên \u0111o\u1ea1n [0 ; 3]. Tính t\u1ed5ng S = 2m + 3M.<br \/>.<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$-\\dfrac{7}{2}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$-\\dfrac{3}{2}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$-3$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$4$<\/span>"],"correct":"1","level":"2","hint":"<p>S\u1eed d\u1ee5ng quy t\u1eafc tìm giá tr\u1ecb l\u1edbn nh\u1ea5t , nh\u1ecf nh\u1ea5t c\u1ee7a hàm s\u1ed1 trên \u0111o\u1ea1n [a ; b].<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> <span class=\"math-tex\">$-\\dfrac{7}{2}$<\/span>.<\/span><\/p><p>Ta có <span class=\"math-tex\">$f^\\prime(x)=\\dfrac{1}{2}-\\dfrac{1}{2\\sqrt{x+1}}=\\dfrac{\\sqrt{x+1}-1}{2\\sqrt{x+1}}$<\/span>.<\/p><p><span class=\"math-tex\">$f^\\prime(x)=0$<\/span> ⇔ <span class=\"math-tex\">$\\sqrt{x+1}=1$<\/span> ⇔ x = 0 ∈ [0 ; 3].<\/p><p>Khi \u0111ó f(0) = –1, f(3) = <span class=\"math-tex\">$-\\dfrac{1}{2}$<\/span> nên m = –1 và M = <span class=\"math-tex\">$-\\dfrac{1}{2}$<\/span>.<\/p><p>V\u1eady S = 2m + 3M = <span class=\"math-tex\">$-\\dfrac{7}{2}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-06-16 04:43:13","option_type":"math","len":0},{"id":"5346","post_id":"7511","mon_id":"1159285","chapter_id":"1159288","question":"<p>Tìm giá tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a hàm s\u1ed1 <span class=\"math-tex\">$f(x)=\\sin x+\\cos 2x$<\/span> trên [0 ; <span class=\"math-tex\">$\\pi$<\/span>] là<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\dfrac{9}{8}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$\\dfrac{5}{4}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$2$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$1$<\/span>"],"correct":"1","level":"2","hint":"<p>\u0110\u1eb7t \u1ea9n ph\u1ee5, s\u1eed d\u1ee5ng quy t\u1eafc tìm giá tr\u1ecb l\u1edbn nh\u1ea5t , nh\u1ecf nh\u1ea5t c\u1ee7a hàm s\u1ed1 trên \u0111o\u1ea1n [a ; b].<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> <span class=\"math-tex\">$\\dfrac{9}{8}$<\/span>.<\/span><\/p><p><span class=\"math-tex\">$f(x)=\\sin x+\\cos 2x$<\/span> <span class=\"math-tex\">$\\sin x+1-2\\sin^2x$<\/span>.<\/p><p>\u0110\u1eb7t <span class=\"math-tex\">$t=\\sin x$<\/span>, 0 <span class=\"math-tex\">$\\le$<\/span> t <span class=\"math-tex\">$\\le$<\/span> 1.<\/p><p><span class=\"math-tex\">$f(t)=-2t^2+t+1$<\/span>, <span class=\"math-tex\">$f^\\prime(t)=-4t+1$<\/span><\/p><p><span class=\"math-tex\">$f^\\prime(t)=0$<\/span> ⇔ <span class=\"math-tex\">$t=\\dfrac{1}{4}\u200b\u200b$<\/span><\/p><p><span class=\"math-tex\">$f(0) = 1;f(1) = 0;f\\bigg(\\dfrac{1}{4}\\bigg)=\\dfrac{9}{8}$<\/span><\/p><p>V\u1eady <span class=\"math-tex\">$\\displaystyle\\max_{[0;1]}f(x)=\\dfrac{9}{8}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-06-16 04:50:08","option_type":"math","len":0},{"id":"5348","post_id":"7511","mon_id":"1159285","chapter_id":"1159288","question":"<p>Giá tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a hàm s\u1ed1 <span class=\"math-tex\">$y=2\\cos x-\\dfrac{4}{3}(\\cos x)^3$<\/span> trên [0 ; <span class=\"math-tex\">$\\pi$<\/span>].<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$\\dfrac{2}{3}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$\\dfrac{10}{3}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$\\dfrac{2\\sqrt{2}}{3}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$0$<\/span>"],"correct":"3","level":"2","hint":"<p>\u0110\u1eb7t \u1ea9n ph\u1ee5, s\u1eed d\u1ee5ng quy t\u1eafc tìm giá tr\u1ecb l\u1edbn nh\u1ea5t, nh\u1ecf nh\u1ea5t c\u1ee7a hàm s\u1ed1 trên \u0111o\u1ea1n [a ; b].<\/p>","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>C.<\/strong> <span class=\"math-tex\">$\\dfrac{2\\sqrt{2}}{3}$<\/span>.<\/span><\/p><p>\u0110\u1eb7t <span class=\"math-tex\">$t=\\cos x$<\/span>, t ∈ [–1 ; 1], ta \u0111\u01b0\u1ee3c <span class=\"math-tex\">$y=2t-\\dfrac{4}{3}t^3$<\/span>.<\/p><p><span class=\"math-tex\">$y^\\prime=2-4t^2$<\/span><\/p><p><span class=\"math-tex\">$y^\\prime=0$<\/span> ⇔ <span class=\"math-tex\">$x=-\\dfrac{1}{\\sqrt{2}}$<\/span> ∈ [–1 ; 1] ho\u1eb7c <span class=\"math-tex\">$x=\\dfrac{1}{\\sqrt{2}}$<\/span> ∈ [–1 ; 1].<\/p><p><span class=\"math-tex\">$y(-1)=-\\dfrac{2}{3}$<\/span>; <span class=\"math-tex\">$y\\bigg(-\\dfrac{1}{\\sqrt{2}}\\bigg)=-\\dfrac{2\\sqrt{2}}{3}$<\/span>; <span class=\"math-tex\">$y\\bigg(\\dfrac{1}{\\sqrt{2}}\\bigg)=\\dfrac{2\\sqrt{2}}{3}$<\/span>; <span class=\"math-tex\">$y(1)=\\dfrac{2}{3}$<\/span>.<\/p><p>V\u1eady <span class=\"math-tex\">$\\displaystyle\\max_{[0;\\pi]}y=\\dfrac{2\\sqrt{2}}{3}$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-06-16 04:59:32","option_type":"math","len":0}]}