Chú ý: Để đảm bảo quyền lợi và bảo vệ tài khoản của mình
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{"common":{"save":0,"post_id":"7451","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5993","post_id":"7451","mon_id":"1158924","chapter_id":"1159187","question":"<p>Cho hàm s\u1ed1 <span class=\"math-tex\">$y=-\\sqrt{3}\\cos x+\\sin x-x^2+2030x+2040$<\/span>. S\u1ed1 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng trình <span class=\"math-tex\">$y^{\\prime\\prime}=0$<\/span> trên \u0111o\u1ea1n [0; 4<span class=\"math-tex\">$\\pi$<\/span>] là<\/p>","options":["<strong>A.<\/strong> 1","<strong>B.<\/strong> 2","<strong>C.<\/strong> 0","<strong>D.<\/strong> 3"],"correct":"2","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>B.<\/strong> 2.<\/span><\/p><p><span class=\"math-tex\">$y=-\\sqrt{3}\\cos x+\\sin x-x^2+2030x+2040$<\/span><\/p><p><span class=\"math-tex\">$y^\\prime=\\sqrt{3}\\sin x+\\cos x-2x+2030$<\/span><\/p><p><span class=\"math-tex\">$y^{\\prime\\prime}=\\sqrt{3}\\cos x-\\sin x-2$<\/span><\/p><p><span class=\"math-tex\">$y^{\\prime\\prime}=0$<\/span> ⇔ <span class=\"math-tex\">$\\sqrt{3}\\cos x-\\sin x-2=0$<\/span><\/p><p>⇔ <span class=\"math-tex\">$\\cos x.\\cos\\bigg(\\dfrac{\\pi}{6}\\bigg)-\\sin x.\\sin\\bigg(\\dfrac{\\pi}{6}\\bigg)=1$<\/span><\/p><p>⇔ <span class=\"math-tex\">$\\cos\\bigg(x+\\dfrac{\\pi}{6}\\bigg)=\\cos 0$<\/span><\/p><p>⇔ <span class=\"math-tex\">$x+\\dfrac{\\pi}{6}=k2\\pi$<\/span>, k ∈ Z<\/p><p>⇔ <span class=\"math-tex\">$x=-\\dfrac{\\pi}{6}+k2\\pi$<\/span>, k ∈ Z.<\/p><p>V\u1edbi x ∈ [0; 4<span class=\"math-tex\">$\\pi$<\/span>] thì <span class=\"math-tex\">$0\\le -\\dfrac{\\pi}{6}+k2\\pi \\le 4\\pi$<\/span>, k ∈ Z<\/p><p>⇔ <span class=\"math-tex\">$\\dfrac{1}{12}\\le k \\le\\dfrac{25}{12}$<\/span>, k ∈ Z<\/p><p>⇔ k ∈ {1; 2}<\/p><p>V\u1eady có 2 nghi\u1ec7m tho\u1ea3 mãn \u0111\u1ec1 bài.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-05-30 05:42:43","option_type":"txt","len":0},{"id":"5996","post_id":"7451","mon_id":"1158924","chapter_id":"1159187","question":"<p>Cho hàm s\u1ed1 <span class=\"math-tex\">$y=\\dfrac{1}{3}x^3-(2m+1)x^2-mx-4$<\/span>. Tìm t\u1ea5t c\u1ea3 các giá tr\u1ecb nguyên c\u1ee7a tham s\u1ed1 m \u0111\u1ec3 <span class=\"math-tex\">$y^\\prime \\ge 0$<\/span>, v\u1edbi m\u1ecdi x ∈ R.<\/p>","options":["<strong>A.<\/strong> m ∈ ∅","<strong>B.<\/strong> m ∈ <span class=\"math-tex\">$\\bigg[-1;-\\dfrac{1}{4}\\bigg]$<\/span>","<strong>C.<\/strong> m = –1","<strong>D.<\/strong> m = 0"],"correct":"3","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>C.<\/strong> m = –1.<\/span><\/p><p><span class=\"math-tex\">$y=\\dfrac{1}{3}x^3-(2m+1)x^2-mx-4$<\/span><\/p><p><span class=\"math-tex\">$y^\\prime=x^2-2(2m+1)x-m$<\/span><\/p><p>Vì a = 1 > 0 nên \u0111\u1ec3 <span class=\"math-tex\">$y^\\prime \\ge 0$<\/span> thì <span class=\"math-tex\">$\\Delta^\\prime=(2m+1)^2+m$<\/span> <span class=\"math-tex\">$\\le$<\/span> 0<\/p><p>⇔ <span class=\"math-tex\">$4m^2+5m+1 \\le0$<\/span> <span class=\"math-tex\">$-1 \\le m \\le -\\dfrac{1}{4}$<\/span><\/p><p>Mà m ∈ Z nên m = –1.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-05-30 05:55:09","option_type":"math","len":0},{"id":"5999","post_id":"7451","mon_id":"1158924","chapter_id":"1159187","question":"<p>Cho hàm s\u1ed1 <span class=\"math-tex\">$y=\\dfrac{1}{2}\\sin^22x-\\cos^2x$<\/span>. Ph\u01b0\u01a1ng trình <span class=\"math-tex\">$y^\\prime=0$<\/span> có bao nhiêu nghi\u1ec7m thu\u1ed9c kho\u1ea3ng (0; 2<span class=\"math-tex\">$\\pi$<\/span>)?<\/p>","options":["<strong>A.<\/strong> 6 nghi\u1ec7m","<strong>B.<\/strong> 7 nghi\u1ec7m","<strong>C.<\/strong> 8 nghi\u1ec7m","<strong>D.<\/strong> 9 nghi\u1ec7m"],"correct":"2","level":"3","hint":"","answer":"<p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>B.<\/strong> 7 nghi\u1ec7m.<\/span><\/p><p><span class=\"math-tex\">$y=\\dfrac{1}{2}\\sin^22x-\\cos^2x$<\/span><\/p><p><span class=\"math-tex\">$y^\\prime=2\\cos2x\\sin2x+2\\sin x\\cos x$<\/span><\/p><p><span class=\"math-tex\">$=2\\cos2x\\sin2x+\\sin2x$<\/span><\/p><p><span class=\"math-tex\">$=\\sin2x(2\\cos2x+1)$<\/span><\/p><p><span class=\"math-tex\">$y^\\prime=0$<\/span> ⇔ <span class=\"math-tex\">$\\sin2x(2\\cos2x+1)=0$<\/span><\/p><p>⇔ <span class=\"math-tex\">$\\sin2x=0$<\/span> ho\u1eb7c <span class=\"math-tex\">$2\\cos2x+1=0$<\/span><\/p><p>N\u1ebfu <span class=\"math-tex\">$\\sin2x=0$<\/span> thì <span class=\"math-tex\">$x=k\\pi$<\/span>, k ∈ Z<\/p><p>⇔ <span class=\"math-tex\">$x=k\\dfrac{\\pi}{2}$<\/span>, k ∈ Z<\/p><p>Mà x ∈ (0 ; 2<span class=\"math-tex\">$\\pi$<\/span>) nên <span class=\"math-tex\">$x=\\dfrac{\\pi}{2}$<\/span> ho\u1eb7c <span class=\"math-tex\">$x=\\pi$<\/span> ho\u1eb7c <span class=\"math-tex\">$x=\\dfrac{3\\pi}{2}$<\/span> (1).<\/p><p>N\u1ebfu <span class=\"math-tex\">$2\\cos2x+1=0$<\/span> thì <span class=\"math-tex\">$\\cos2x=-\\dfrac{1}{2}$<\/span><\/p><p>⇔ <span class=\"math-tex\">$2x=\\pm\\dfrac{2\\pi}{3}+k2\\pi$<\/span>, k ∈ Z<\/p><p>⇔ <span class=\"math-tex\">$x=\\pm\\dfrac{\\pi}{3}+k\\pi$<\/span>, k ∈ Z<\/p><p>Mà x ∈ (0 ; 2<span class=\"math-tex\">$\\pi$<\/span>) nên <span class=\"math-tex\">$x=\\dfrac{\\pi}{3}$<\/span> ho\u1eb7c <span class=\"math-tex\">$x=\\dfrac{4\\pi}{3}$<\/span> ho\u1eb7c <span class=\"math-tex\">$x=\\dfrac{2\\pi}{3}$<\/span> ho\u1eb7c <span class=\"math-tex\">$x=\\dfrac{5\\pi}{3}$<\/span> (2).<\/p><p>K\u1ebft h\u1ee3p (1) và (2), ta th\u1ea5y ph\u01b0\u01a1ng trình <span class=\"math-tex\">$y^\\prime=0$<\/span> có 7 nghi\u1ec7m trên kho\u1ea3ng (0 ; 2<span class=\"math-tex\">$\\pi$<\/span>).<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-05-31 01:48:47","option_type":"txt","len":1}]}