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. Since 5 < 26 , the origin lies inside the circle.

CHIJ ST. THERESA’S CONVENT SECONDARY 4 EXPRESS / 5 NORMAL (ACADEMIC) 2023 PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS PAPER 2 4 3 Solve the equation 32xxee −+= , giving your answer(s) correct to 3 significant figures. [5] 32xxee −+=  132x xe e += Let u = xe . Then 132u u+=  23 2 1uu +=  23 2 1 0uu + − =  (3u – 1)(u + 1) = 0  u = 1 3 or u = –1  xe = 1 3 or xe = –1 There is no solution for xe = –1 since xe > 0 for all real values of x.  the only solution is x = ( )1ln 3 = –1.10 (3 s.f.)

CHIJ ST. THERESA’S CONVENT SECONDARY 4 EXPRESS / 5 NORMAL (ACADEMIC) 2023 PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS PAPER 2 5 4 Given that y = 3 16 53 x− , find (i) the value(s) of x for which d d y x = 1, [4] (ii) the value of 1 0 dyx , giving your answer correct to 3 significant figures. [4] 1 0 dyx = 11 3 0 16 (5 3 ) dxx − − = ( ) 12 3 0 (5 3 )16 2 ( 3)3 x −  − = 12 3 0 8 (5 3 )x− −  = 22 338 (5 3) (5 0)− − − −  = 10.7 d d y x = 1 3d16 (5 3 )d xx − − = 4 3116 (5 3 ) ( 3)3 x −− − − = 4 316(5 3 )x − − d d y x = 1  4 316(5 3 )x − − = 1  4 3(5 3 )x− = 16  1 3(5 3 )x− = 2  5 – 3x = 8  x = –1 or x = 13 3

CHIJ ST. THERESA’S CONVENT SECONDARY 4 EXPRESS / 5 NORMAL (ACADEMIC) 2023 PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS PAPER 2 6 5(a) (i) Using the substitution z = 1 x+ , write down all the terms in the expansion of 6[1 (1 )] x++ , leaving each term in the form (1 ) pkx + where k and p are integers. [2] (ii) Hence show that the remainder when 6(2 ) x+ is divided by 1 x+ is 1. [1] (b) Given that the coefficient of 2x in the expansion of 5(1 2 )(1 )x px+− is 20, find the two possible values of the constant p. [5] 5(1 2 )(1 )x px+− = 25(1 2 ) 1 5( ) ( ) ... 2x px px + + − + − +   = 22(1 2 ) 1 5 10 ...x px p x+ − + + The term in 2x is 22(1)(10 ) (2 )( 5 )p x x px+− = 2 2 210 10p x px− The coefficient of 2x is 210 10pp − = 20 (given) 2 2pp −− = 0 (p + 1)(p – 2) = 0 p = –1 or p = 2 From (i), 6(2 ) x+ = 2 3 4 51 (1 ) 6 15(1 ) 20(1 ) 15(1 ) 6(1 ) (1 )x x x x x x++ + + + + + + + + + + so the remainder when 6(2 ) x+ is divided by 1 x+ is 1 (shown) 6[1 (1 )] x++ = 6[1 ] z+ = 2 3 4 5 66 6 6 6 6 61 1 2 3 4 5 6z z z z z z           + + + + + +                       = 2 3 4 5 61 6 15 20 15 6z z z z z z+ + + + + + = 2 3 4 5 61 6(1 ) 15(1 ) 20(1 ) 15(1 ) 6(1 ) (1 )x x x x x x+ + + + + + + + + + + +

CHIJ ST. THERESA’S CONVENT SECONDARY 4 EXPRESS / 5 NORMAL (ACADEMIC) 2023 PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS PAPER 2 7 6 At time t seconds, the velocity, v m/s, of a metal ball falling through a very thick liquid is given by v = /38(1 e ) t−− . (i) State the initial velocity of the ball. [1] The depth of the ball from the surface of the liquid at time t seconds is s metres. The ball is initially on the surface of the liquid. (ii) Find an expression for the depth of the ball in terms of t. [4] (iii) Show that the ball is accelerating throughout its motion. [3] (iv) Deduce the speed of the ball for large values of t. [1] When t is large, /3e t−  0 so v  8(1 – 0) = 8. For large values of t, the ball moves at an almost constant speed of 8 m/s. When t = 0, v = 08(1 e )− = 0 m/s, the initial velocity of the ball Displacement, s = dvt = ( )/38d1e t t−− = /3e88 1/ 3 t tC −−+ − = /38 24e ttC −++ When t = 0,