CHIJ STC 2023 Prelim 4E5N A Math P2 solutions

CANDIDATE NAME CLASS INDEX NUMBER ADDITIONAL MATHEMATICS 4049/2 Paper 2 25 Aug 2023 2 hours 15 minutes Candidates answer on the Question Paper as well as on the graph paper provided. READ THESE INSTRUCTIONS FIRST Write your name, class and index number in the spaces at the top of this page. Write in dark blue or black pen. You may use a pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an approved scientific calculator is expected, where appropriate. 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 17 printed pages. CHIJ ST. THERESA’S CONVENT PRELIMINARY EXAMINATION 2023 SECONDARY 4 EXPRESS / 5 NORMAL (ACADEMIC)

CHIJ ST. THERESA’S CONVENT SECONDARY 4 EXPRESS / 5 NORMAL (ACADEMIC) 2023 PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS PAPER 2 2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x = a acbb 2 42 −− Binomial expansion (a + b)n = an + ban n 1 1 −       + 22 2 ban n−       + …+ rrn bar n −       + … + bn, where n is a positive integer and !)!( ! rrn n r n −=      ( 1)...( 1) ! n n n r r − − += . 2. TRIGONOMETRY Identities sin2A + cos2A = 1 sec2A = 1 + tan2A cosec2A = 1 + cot2A sin(A ± B) = sinAcosB ± cosAsinB cos(A ± B) = cosAcosB sinAsinB tan(A ± B) = tan tan 1 tan tan AB AB  sin2A = 2sinAcosA cos2A = cos2A − sin2A = 2cos2A − 1 = 1 − 2sin2A tan2A = 2 2 tan 1 tan A A− Formulae for ABC C c B b A a sinsinsin == a2 = b2 + c2 − 2bc cos A  = 1 sin2 ab C

CHIJ ST. THERESA’S CONVENT SECONDARY 4 EXPRESS / 5 NORMAL (ACADEMIC) 2023 PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS PAPER 2 3 1 Given that f( x) = 326 7 2x x x+ − − , show that x + 1 is a factor of f( x) and hence factorise f( x) completely. [3] 2 The equation of a circle is 22( 3) ( 4) 26xy− + + = . Determine if the origin O lies inside or outside the circle. [3] f(–1) = 326( 1) 7( 1) ( 1) 2− + − − − − = –6 + 7 + 1 – 2 = 0 By the Factor Theorem, x + 1 is a factor of f(x) (shown) By inspection, f(x) = 2( 1)(6 2)x x x+ + − = ( 1)(2 1)(3 2)x x x+ − + The coordinates of the centre of the circle are (3, –4). Distance of centre from O is 2234+ = 5 units Radius of circle is 26 units.