2024 DSE 数学延伸部分-M2 真题 答案 详解
2024-05-01序号 | 文件列表 | 说明 | ||
---|---|---|---|---|
1 | 2024-数学延伸部分-M2-paper-eng.pdf | 28 页 | 3.64MB | 真题(英文) |
2 | 2024-数学延伸部分-M2-paper-zh.pdf | 28 页 | 4.03MB | 真题(中文) |
3 | 2024-数学延伸部分-M2-answer-eng.pdf | 23 页 | 658.98KB | 答案(英文) |
真题(英文)
Level 3 Exemplar 2
2024-DSE MATH EP M2
HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY
HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION 2024
MATHEMATICS Extended Part Module 2 (Algebra and Calculus) Question-Answer Book
8:30 am – 11:00 am (2½ hours)
This paper must be answered in English
INSTRUCTIONS
(1) After the announcement of the start of the examination, you should first write your Candidate Number in the space provided on Page 1 and stick barcode labels in the spaces provided on Pages 1, 3, 5, 7, 9, 11 and 13.
(2) This paper consists of TWO sections, A and B.
(3) Attempt ALL questions in this paper. Write your answers in the spaces provided in this Question-Answer Book. Do not write in the margins. Answers written in the margins will not be marked.
(4) Graph paper and supplementary answer sheets will be supplied on request. Write your Candidate Number, mark the question number box and stick a barcode label on each sheet, and fasten them with string INSIDE this book.
(5) Unless otherwise specified, all working must be clearly shown.
(6) Unless otherwise specified, numerical answers must be exact.
(7) The diagrams in this paper are not necessarily drawn to scale.
(8) No extra time will be given to candidates for sticking on the barcode labels or filling in the question number boxes after the ‘Time is up’ announcement.
© Hong Kong Examinations and Assessment Authority All Rights Reserved 2024
2024-DSE-MATH-EP(M2)-1
1
Please stick the barcode label here.
Candidate Number
* A 0 3 2 E 0 0 1 *
真题(中文)
答案(英文)
2024 HKDSE Mathematics Extended Part (Module 2) Examination
Suggested Solutions by Jacky Chan
1a O, A and B are collinear.
b
[ |\overrightarrow{OA}| = |3\mathbf{i} + 4\mathbf{j}| ]
[ = \sqrt{3^2 + 4^2} ]
[ = 5 ]
[ |\overrightarrow{OB}| = |\overrightarrow{OA}| + |\overrightarrow{AB}| ]
20 = 5 + (|\overrightarrow{AB}|)
[ |\overrightarrow{AB}| = 15 ]
Since ( \overrightarrow{OA} // \overrightarrow{AB} ),
[ \overrightarrow{AB} = \frac{15}{5}\overrightarrow{OA} ]
= 3(3\mathbf{i} + 4\mathbf{j})
= 9\mathbf{i} + 12\mathbf{j}
P.1 (23)
© Jacky Chan Yui Him (2024)
All rights reserved.