S4 A-Math SJI P2 2024 (with Answer) EXAM555

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2024 (YEAR 4) CANDIDATE NAME CLASS INDEX NUMBER ADDITIONAL MATHEMATICS Paper 2 Candidates answer on the Question Paper. 4049/02 20 August 2024 2 hours 15 minutes (0805 - 1020) 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 in the space provided. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions. 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 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. Give answers in degrees to one decimal place. For , use either your calculator value or 3.142, unless the question requires the answer in terms of . The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 90. This document consists of 19 printed pages and 1 blank page.

2 [Turn over 1. ALGEBRA Quadratic Equation For the equation 02  cbxax , a acbbx 2 42  Binomial expansion   nrrnnnnn bbar nbanbanaba                  ......21 221 , where n is a positive integer and ! )1)...(1( )!(! ! r rnnn rnr n r n       2. TRIGONOMETRY Identities 1cossin 22  AA AA 22 tan1sec  cosec 2 A = 1 + cot 2 A BABABA sincoscossin)sin(  BABABA sinsincoscos)cos( ∓ BA BABA tantan1 tantan)tan( ∓  AAA cossin22sin AAAAA 2222 sin211cos2sincos2cos  A AA 2tan1 tan22tan  Formulae for ABC C c B b A a sinsinsin  Abccba cos2222  Cab sin2 1

3 [Turn over 1 Solve the inequality 282 2 4xx  and represent your solution on a number line. [6] 1 282 2 4xx      2 2 82 28 0 42 0 2 or 4 xx xx xx xx               2 2 22 4 22 4 0 64 0 46 xx xx xx x            Combining both inequalities: 42 o r 4 6xx     –4 –2 0 4 6

4 [Turn over 2 The expression 3cos2 6sin2  is defined for 0  . (a) Using cos 2R  , where 0R and 0 2  , solve the equation 3cos2 6sin2 5  . [6] 2(a) cos 2 3cos 2 + 6sin 2R      = cos 2 cos sin 2 sinRR    cos 3 sin 6 R R     22 236 45 R R    = 35 or 6.7082 (6.71) 16tan3     = 1.1071 (1.11)  35 c o s2 1 . 1 1 or  6.71cos 2 1.11  3cos2 6sin2 5    35 c o s2 1 . 1 1 5   5cos 2 1.11 35    1