CEDAR 2025 EMATH PRELIM P2 MS

CEDAR GIRLS’ SECONDARY SCHOOL PRELIMINARY EXAMINATION 2025 SECONDARY FOUR CANDIDATE NAME MARK SCHEME CLASS INDEX NUMBER MATHEMATICS 4052/02 Paper 2 22 August 2025 2 hours 15 minutes Candidates answer on the Question Paper. READ THESE INSTRUCTIONS FIRST Write your class, index number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all the questions. The number of marks is given in brackets [ ] at the end of each question or part question. If working is needed for any question it must be shown with the answer. Omission of essential working will result in loss of marks. The total of the marks for this paper is 90. The use of an approved scientific calculator is expected, where appropriate. If the degree of accuracy is not specified in the question and if the answer is not exact, give the answer to three significant figures. Gives answers in degrees to one decimal place. For  , use either your calculator value or 3.142. For Examiner’s Use 90 This document consists of 19 printed pages and 1 blank page. [Turn over]

2 Cedar Girls’ Secondary School 4052/02/S4/Prelim/2025 Mathematical Formulae Compound interest Total amount = n rP      + 1001 Mensuration Curved surface area of a cone = rl Surface area of a sphere = 2 4 r Volume of a cone = hr 2 3 1 Volume of a sphere = 3 3 4 r Area of triangle ABC = Cab sin2 1 Arc length = r , where  is in radians Sector area = 2 2 1 r , where  is in radians Trigonometry C c B b A a sinsinsin == Abccba cos2222 −+= Statistics Mean = f fx   Standard deviation = 22        −  f fx f fx

3 Cedar Girls’ Secondary School 4052/02/S4/Prelim/2025 [Turn over Answer all the questions. 1 (a) Solve the inequality 211 4 0 2 yy −+   . 2 8 2 1 0yy+  −  2 8 2 1yy+  − and 2 1 0y− 1 2y− 1 2y 1 2y − Answer ………………………………… [2] (b) Solve ( ) 41 1 3 5xx− − − = . ( ) 2 2 41 1 35 49 1355 4 9 10 0 xx xx xx − − = − + = − − = 2( 9) ( 9) 4(4)( 10) 2(4)x − −  − − −= 0.816x=− or 3.07x= (3 s.f.) Answer x = ………. and x = ………. [3]

4 Cedar Girls’ Secondary School 4052/02/S4/Prelim/2025 (c) (i) Given that 56243 2 6p p p −= , find the value of p . 5 5 63 2 6p p p −= 5666pp −= 56pp=− 1p= Answer p = ……………...…………… [2] (ii) Simplify 3 4 2 5 8 16 p pqq −   and leave your answer in positive index notation. = 3 8 2 5 416 q pqp   = 12 65 1 64