Đề kiểm tra 1 tiết Đại số Giải tích 11 chương 1 năm 2017-2018 có đáp án - Trường THPT B Bình Lục

SỞ GD&ĐT HÀ NAM<br /> TRƯỜNG THPT B BÌNH LỤC<br /> <br /> ĐỀ KIỂM TRA 1 TIẾT<br /> ĐẠI SỐ & GIẢI TÍCH 11<br /> NĂM HỌC: 2017-2018<br /> (Thời gian 45 phút)<br /> Đề 01<br /> <br /> I. Trắc nghiệm<br /> Câu 1. Hàm số y  3cosx  1 có giá trị lớn nhất là:<br /> A. 2<br /> B. -4<br /> C. 3<br /> Câu 2. Phương trình nào sau đây vô nghiệm?<br /> A. 2cos2 x  cos x  1  0<br /> <br /> B.<br /> <br /> D. 0<br /> <br /> 3 cosx  sin x  3<br /> <br /> C. 2cosx  1  0<br /> D. 3sin x  cosx  2<br /> Câu 3. Giá trị của m để phương trình: cosx  m  0 vô nghiệm là:<br /> A. m 1<br /> <br /> B. m  1<br /> <br /> C. 1  m  1<br /> <br />  m  1<br /> D. <br /> m  1<br /> <br /> 1<br /> có bao nhiêu nghiệm thỏa: 0  x  <br /> 2<br /> A. 1<br /> B. 3<br /> C. 2<br /> D. 4.<br /> Câu 5. Phương trình nào sau đây là phương trình bậc hai với một hàm số lượng giác<br /> Câu 4. Phương trình: sin2x  <br /> <br /> A. sin(x <br /> <br /> <br /> <br /> B. sin2 2x  3sin2x  2  0<br /> <br /> ) 1 0<br /> <br /> 3<br /> C. x  3x-6=0<br /> D. sin2 x  3cosx  3  0<br /> Câu 6. Hàm số y  cot 2x xác định khi:<br /> 2<br /> <br /> B. x  k , k  Z<br /> <br /> A.x  R<br /> <br /> C. x    k , k  Z<br /> <br /> <br /> <br /> D. x  k , k  Z<br /> 2<br /> <br /> II. Tự luận<br /> 3<br /> .<br /> 2<br /> <br /> Câu 2 (1 điểm): 2 cos( x  )  1  0 .<br /> 4<br /> <br /> Câu 1 (1 điểm): sin x =<br /> <br /> Câu 3 (2 điểm): 3cos2x – 2cos x – 5 = 0.<br /> Câu 4 (1,5 điểm): sin2x  cosx<br /> <br /> 3sin2x – cos2x  2<br /> <br /> Câu 5 (1,5 điểm):<br /> <br /> Bài làm<br /> Trắc nghiệm<br /> Câu<br /> Đáp án<br /> <br /> 1<br /> <br /> 2<br /> <br /> 3<br /> <br /> 4<br /> <br /> 5<br /> <br /> 6<br /> <br /> ..............................................................................................................................................................................<br /> ..............................................................................................................................................................................<br /> ..............................................................................................................................................................................<br /> 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..............................................................................................................................................................................<br /> ..............................................................................................................................................................................<br /> ..............................................................................................................................................................................<br /> ..............................................................................................................................................................................<br /> <br /> ĐÁP ÁN ĐỀ SỐ 01,08,15,22,29,36<br /> Trắc nghiệm<br /> Câu<br /> Đáp án<br /> Câu<br /> Câu 1<br /> <br /> Câu 2<br /> <br /> Câu 3<br /> <br /> 1<br /> A<br /> <br /> 2<br /> B<br /> <br /> 3<br /> D<br /> <br /> 4<br /> C<br /> <br /> Nội dung<br /> <br /> <br /> 3<br />  sin x  sin<br /> 2<br /> 3<br /> <br /> <br />  x  3  2k<br />  <br /> <br />  x     2k<br /> 3<br /> <br /> <br /> <br />  x  3  2 k<br />  <br /> ; k  Z.<br /> 2<br /> x <br />  2 k<br /> 3<br /> <br /> sin x =<br /> <br /> 0,25<br /> <br /> 0,25<br /> <br /> <br /> <br />  x  3  2 k<br /> KL: Vậy phương trình có hai họ nghiệm <br /> 2<br /> x <br />  2 k<br /> 3<br /> <br /> <br /> <br /> 1<br /> 2 cos( x  )  1  0  cos( x  ) <br /> 4<br /> 4<br /> 2<br />  <br /> <br />  x  4  3  2 k<br />  <br />  x       2 k<br /> <br /> 4<br /> 3<br /> 7<br /> <br />  x  12  2k<br />  <br /> ;k  Z<br />  x     2 k<br /> <br /> 12<br /> 7<br /> <br />  x  12  2k<br /> KL: Vậy phương trình có hai họ nghiệm <br />  x     2 k<br /> <br /> 12<br /> <br /> 3cos2x – 2cos x – 5 = 0<br />  Đặt t = cos x; ĐK t ≤ 1.<br /> <br />  t = -1  cos x = -1  x = (2k + 1)  ; k  Z.<br />  KL: PT có nghiệm x = (2k + 1)  ; k  Z.<br /> <br /> <br /> <br /> 2 cos (3x) =<br /> <br /> 0,25<br /> 0,25<br /> <br /> 0,25<br /> <br /> 0,25<br /> <br /> ;k  Z<br /> <br /> 0,25<br /> <br /> 0.5<br /> <br /> 0.5<br /> 0.5<br /> <br /> 3<br /> <br /> 2<br /> 2<br /> 3<br /> sin(3x) –<br /> cos(3x) <br /> 2<br /> 2<br /> 2<br /> <br /> 2<br />  2<br /> ; sin  <br /> ta có: cos  <br /> 2<br /> 2<br /> <br /> ; k  Z.<br /> <br /> 0.5<br />  t  1<br /> 5<br /> t  (loai)<br />  3<br /> <br /> 2 sin(3x) –<br /> <br /> 6<br /> D<br /> Điểm<br /> 0,25<br /> <br />  PT trở thành 3t2 - 2t - 5 = 0  <br /> <br /> Câu 4<br /> <br /> 5<br /> B<br /> <br /> 0.25<br /> 0.25<br /> <br /> <br /> <br /> <br /> <br /> 3<br /> Phương trình trở thành: cos sin(3x) – sin cos(3x) <br /> 4<br /> 4<br /> 2<br /> <br /> 3<br /> <br /> <br />  sin(3x  ) <br />  sin(3x  )  sin<br /> 4<br /> 3<br /> 4<br /> 2<br />  <br /> 7<br /> <br /> <br />  3 x  4  3  2 k<br />  3 x  12  2 k<br />  <br />  <br /> <br /> <br /> 11<br /> 3 x      2 k<br /> 3 x <br />  2k<br /> 4<br /> 3<br /> <br /> 12<br /> <br /> 7<br /> <br /> <br />  x  36  k 3<br />  <br /> ;k  Z.<br /> 11<br /> <br /> x <br /> k<br /> 36<br /> 3<br /> <br /> 7<br /> <br /> <br />  x  36  k 3<br />  KL: PT có nghiệm <br /> ; k  Z.<br /> 11<br /> <br /> x <br /> k<br /> 36<br /> 3<br /> <br /> <br /> Câu 5<br /> <br /> 0.25<br /> 0.25<br /> 0.25<br /> 0.25<br /> 0.25<br /> <br /> 0.25<br /> <br /> sin2x  cosx<br />  <br />  sin2x  sin  -x <br /> 2 <br /> <br /> <br />  2x  2 -x  k2<br /> <br />  <br /> <br /> 2x<br /> <br /> <br /> <br />  -x   k2<br /> <br /> 2 <br /> <br /> <br />  k2<br /> x<br /> <br /> <br /> <br /> 6<br /> 3 k Z<br /> <br />  <br />  x    k2<br /> <br /> 2<br /> <br /> <br />  k2<br /> x  6  3<br /> KL: Vậy phương trình có 2 học nghiệm: <br /> k  Z<br />  x    k2<br /> <br /> 2<br /> <br /> 0.5<br /> <br /> 0.5<br /> <br /> 0.5<br /> <br />