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8 months ago

Bộ đề thi thử THPT QG 2018 Các môn TOÁN - LÍ - HÓA Các trường THPT Cả nước CÓ HƯỚNG DẪN GIẢI (Lần 8) [DC11042018]

https://app.box.com/s/nhxs7vnw9na94twmj7hc0inbdjzwv940

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 π ( π ) ( sin ( π )) ( π ) ( sin ) π ( sin ) ∫ ∫ ∫ I = − − t f − t dt = − t f t dt = f t dt − I π Đén đây ta suy ra được kết quả ở (D) Câu 32: Đáp án B x 0 0 Đặt F ( x) = ∫ f ( t) dt . Ta cần chứng minh ( )( − ) = ( ) Ta có F '( x) f ( x) 0 = . Khi đó a ∫ ∫ π f x a x dx F x dx 0 0 a a a a ( )( − ) = ( ) − ( ) = ( ) − '( ) ∫ ∫ ∫ ∫ f x a x dx a f x dx xf x dx aF a xF x dx 0 0 0 0 Sử dụng công thức tích phân từng phần, ta có '( ) = ( ) − ( ) Thay vào ta thu được kết quả ở B Câu 33: Đáp án A a ∫ ∫ 0 0 2 2 Ta có t = ∫ dx = − ln 20 − 3 v + C với C là hằng số 20 − 3v 3 Vào thời điểm t = 0 thì vật có vận tốc bằng 0. Suy ra 2 2 0 = − ln 20 + C ⇔ C = ln 20 3 3 2 2 Khi đó t = − ln 20 − 3v + ln 20 3 3 3 ⇔ ln 20 − 3v = ln 20 − t 2 ⇔ 20 − 3v = 20. 3 t 2 e − 3 ⎡ ⎡ 20 20 − t 2 v = − e 20 − 3v = 20. e ⎢ 3 3 ⇔ ⎢ ⎢ ⎢ ⇔ ⎢ 20 20 ⎢⎣ 20 − 3v = − 20. e ⎢v = + e ⎣ 3 3 3 − t 2 3 3 − t − t 2 2 a xF x dx aF a F x dx Để ý rằng phương trình thứ hai không thể đạt v = 0 tại t = 0 cho nên ta chỉ nhận phương trình thứ nhất là 3 20 20 t 2 v = − e − 3 3 DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN Câu 34: Đáp án A Phương trình hoành độ giao điểm của hai đường cong = ⇔ = 0hoặc x = 1 2 x x x a 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 21 Đóng góp PDF bởi GV. Nguyễn Thanh Tú www.facebook.com/daykemquynhonofficial www.facebook.com/boiduonghoahocquynhonofficial

https://twitter.com/daykemquynhon plus.google.com/+DạyKèmQuyNhơn 1 1 2 2 Diện tích hình phẳng cần tìm là ( ) ∫ ∫ S = x − x dx = x − x dx = 0 0 1 3 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 Do đó 3S ( 3S − 2) 2018 = 1 Câu 35: Đáp án D Phương trình hoành độ giao điểm của ( C ) và ( ) 3 3 x − 3x + 2 = 2x + 2 ⇔ 2x = x − 3x P là Giải phương trình này, ta thu được hai nghiệm là x = 0; x = 2 2 2 ( ) 3 Thể tích vật thể cần tìm là π ( 2 ) ( 3 ) Suy ra a = 4; b = 35; c = 0; d = 0 Kiểm tra từng mệnh đề, nhận thấy D sai vì Câu 36: Đáp án A Ta có 2 z 3 m 3mi = − + 2 z là số thuần ảo ⇔ 3 − m = 0 ⇔ m = ± 3 Câu 37: Đáp án C 2 4π V = ∫ x − x − x dx = 35 0 b + d 35 + 0 = = 7 a + c + 1 4 + 0 + 1 Ta có ABCD là hình bình hành ⇔ CD = BA ⇔ d − c = a − b ⇔ d = a + c − b ⇔ d = 8 + m − 3 i ABCD là hình chữ nhật ( ) ( ) ( ) ⇔ AC = BD ⇔ c − a = d − b ⇔ 3 + m + 2 i = 9 + m − 4 i ( ) ( ) 2 2 2 2 ⇔ 3 + m + 2 = 9 + m − 4 ⇔ m = 7 Câu 38: Đáp án B 2 2 2 Ta có ( )( )( ) ( ) ( ) z z z + z z = z . z . z + z z = 1+ 3i 2 − i 3 + 4i + 2 + i 3− 4i = 20 + 35i 1 2 3 2 3 1 2 3 2 3 Câu 39: Đáp án A Gọi z = a + bi với a, b ∈ R và a > 0 2 2 ⎧ ⎪a + b = 5 Theo giả thiết ta có ⎨ 2 2 ⎪⎩ ( a − 2) + ( b + 3) = 16 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 22 Đóng góp PDF bởi GV. Nguyễn Thanh Tú www.facebook.com/daykemquynhonofficial www.facebook.com/boiduonghoahocquynhonofficial

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