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9 months ago

Bộ đề thi thử THPT QG 2018 Các môn TOÁN - LÍ - HÓA Các trường THPT Cả nước CÓ HƯỚNG DẪN GIẢI (Lần 9) [DC14042018]

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https://twitter.com/daykemquynhon plus.google.com/+DạyKèmQuyNhơn www.facebook.com/daykem.quynhon http://daykemquynhon.blogspot.com http://daykemquynhon.ucoz.com Nơi bồi dưỡng kiến thức Toán - Lý - Hóa cho học sinh cấp 2+3 / Diễn Đàn Toán - Lý - Hóa 1000B Trần Hưng Đạo Tp.Quy Nhơn Tỉnh Bình Định ⎡ ⎢∆ = 0 ⎢ ⎡m = 3 ⎢ 1 m − < − ≤ 0 ⇔ ⎢ 0 ≤ m < 3 ⇔ m ≥ 0 . ⎢ 2 6 ⎢ ⎢ m 3 m 1 ⎣⎢ > ⎢− < − ≤ 0 ⎣ 6 2 Vậy m ≥ 0 thỏa mãn yêu cầu bài toán. Cách 3 ′ ≥ ∀ ∈ +∞ . Hàm số đồng biến trên khoảng [ 0;+∞ ) khi và chỉ khi y 0, x [ 0; ) ( ) [ ) 2 ⇔ 12 + 2 + 3 + ≥ 0, ∀ ∈ 0; +∞ x m x m x 2 ( 2 1) 12 6 , [ 0; ) ⇔ m x + ≥ − x − x ∀x ∈ +∞ [ ) ⇔ m ≥ −6 x, ∀x ∈ 0; +∞ ⇔ m ≥ Max − 6x = 0 . ( ) [ 0; ) x∈ +∞ Vậy m ≥ 0 thỏa mãn yêu cầu bài toán. Câu 14: Đáp án C * y′ = a cos x + bsin x − m . * Hàm số đồng biến trên R khi và chỉ khi ( ) y′ ≥ 0, ∀x ∈ R ⇔ asin x + b cos x ≥ m, ∀x ∈ R ⇔ m ≤ min f x . f x = a x + b x . với ( ) sin cos * Áp dụng bất đẳng thức Bunhiacopxki ta có: ( ) ( ) 2 2 2 2 2 2 ≤ + ⇔ − + ≤ ≤ + . f x a b a b f x a b 2 2 Vậy m ≤ − a + b . Câu 15: Đáp án B f ′ x = 3x − 18x + 24. * ( ) 2 ⎡x = 2 ′ = 0 ⇔ ⎢ . ⎣x = 4 * f ( x) x∈R * Lập bảng biến thiên và suy ra ( x ; y ) ( 4;20 ),( x ; y ) ( 2;24) Suy ra x1 y2 − x2 y1 = 56 . Câu 16: Đáp án A Tập xác định: D = R = = . 1 1 2 2 DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN MỌI YÊU CẦU GỬI VỀ HỘP MAIL DAYKEMQUYNHONBUSINESS@GMAIL.COM HOTLINE : +84905779594 (MOBILE/ZALO) https://daykemquynhonofficial.wordpress.com/blog/ DẠY KÈM QUY NHƠN OFFICIAL ST> : Đ/C 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN Trang 15 Đóng góp PDF bởi GV. Nguyễn Thanh Tú www.facebook.com/daykemquynhonofficial www.facebook.com/boiduonghoahocquynhonofficial

https://twitter.com/daykemquynhon plus.google.com/+DạyKèmQuyNhơn www.facebook.com/daykem.quynhon http://daykemquynhon.blogspot.com http://daykemquynhon.ucoz.com Nơi bồi dưỡng kiến thức Toán - Lý - Hóa cho học sinh cấp 2+3 / Diễn Đàn Toán - Lý - Hóa 1000B Trần Hưng Đạo Tp.Quy Nhơn Tỉnh Bình Định ′ = − = − . Đạo hàm y 4x 3 4mx 4x ( x 2 m) ⎡x = 0 y′ = 0 ⇔ ⎢ . 2 ⎣x = m Hàm số có 3 cực trị ⇔ phương trình y′ = 0 có 3 nghiệm phân biệt và y’ đổi dấu khi x đi qua các nghiệm đó ⇔ m > 0 . Cách 1 Giả sử 3 điểm cực trị của đồ thị hàm số là: 2 2 ( 0; 1 ), ( ; 1 ), ( ; 1) A m − B − m − m + m − C m − m + m − . Vì A∈ Oy và B, C đối xứng nhau qua Oy nên ∆ ABC cân tại A. Theo đề: 1 ∆ = − − = = = + = . 2 2 4 S ABC yB yA xC xB m m; AB AC m m, BC 2 m 4 ( + ) AB. AC. BC m m 2 m 3 R = = 1 ⇔ = 1 ⇔ m − 2m + 1 = 0 . 2 4S 4m m ∆ABC ⎡m = 1 2 ⎡m − 1 = 0 ⇔ ( m − 1)( m + m − 1) = 0 ⇔ ⎢ ⎢ ⇔ 2 − 1± 5 . m m 1 0 ⎢ ⎣ + − = m = ⎢⎣ 2 So với điều kiện m > 0 ta suy ra Cách 2 ⎡m = 1 ⎢ 5 −1. ⎢ m = ⎢⎣ 2 Vì A∈ Oy và B, C đối xứng nhau qua Oy nên tâm đường tròn ngoại tiếp I của ∆ ABC thuộc Oy. Giả sử ( 0; ) I t . Theo giả thiết ta có ⎧1− 2m − t = 1 ⎪ IA = IC = 1 ⇔ ⎨ . 2 ⎪ m + ( m − 2m + 1− t) 2 = 1 ⎩ Phương trình thứ nhất của hệ tương đương với: ⎡1 − 2m − t = 1 ⎡t = −2m ⎢ ⇔ 1 2m t 1 ⎢ . ⎣ − − = − ⎣t = 2 − 2m Trường hợp 1: Nếu t = − 2m thì DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN ( ) 2 + − 2 + 1+ 2 = 1 ⇔ + 2 + = 0 . Loại vì m > 0 . 2 4 2 m m m m m m m Trường hợp 2: Nếu t = 2 − 2m thì Trang 16 MỌI YÊU CẦU GỬI VỀ HỘP MAIL DAYKEMQUYNHONBUSINESS@GMAIL.COM HOTLINE : +84905779594 (MOBILE/ZALO) https://daykemquynhonofficial.wordpress.com/blog/ DẠY KÈM QUY NHƠN OFFICIAL ST> : Đ/C 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN Đóng góp PDF bởi GV. Nguyễn Thanh Tú www.facebook.com/daykemquynhonofficial www.facebook.com/boiduonghoahocquynhonofficial

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