2022 CGS AMath Prelims P1 Ans

Name: Worked Solutions Register No.: Class: CRESCENT GIRLS’ SCHOOL SECONDARY FOUR 2022 PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS 4049/01 Paper 1 23 August 2022 Candidates answer on the Question Paper. No Additional Materials are required. 2 hours 15 min READ THESE INSTRUCTIONS FIRST Write your name, register number and class in the spaces at the top of the page. 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. Write your answers in the answer spaces provided. 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. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. The total of the marks for this paper is 90. For Examiners’ Use Question 1 2 3 4 5 6 7 8 9 10 11 12 13 Marks Table of Penalties Question Number Presentation −1 Accuracy / Units −1 Parent’s/Guardian’s Signature This document consists of 21 printed pages. 90

Crescent Girls’ School 2022 Sec 4 A Math Prelims P1 2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 02 =++ cbxax , a acbbx 2 42 −−= Binomial expansion 1 2 2( ) ... ... 12 n n n n n r r n n n na b a a b a b a b b r − − −     + = + + + + + +           , where n is a positive integer and ! ( )...( 1) !( )! ! n n n n r n r r r n r r  − − +== − 2. TRIGONOMETRY Identities 1cossin 22 =+ AA AA 22 tan1sec += 22cosec 1 cotAA=+ BABABA sincoscossin)sin( = cos( ) cos cos sin sinA B A B A B= tan tantan( ) 1 tan tan ABAB AB = sin 2 2sin cosA A A= AAAAA 2222 sin211cos2sincos2cos −=−=−= A AA 2tan1 tan22tan −= Formulae for ABC C c B b A a sinsinsin == Abccba cos2222 −+=  = 1 sin2 bc A

Crescent Girls’ School [Turn Over 3 1 A trapezium of area ( ) 212 7 1 cm+ has parallel sides of length ( )2 7 m1 c− and ( )7 m6 c+ . Without using a calculator, obtain an expression for the height, h, of the trapezium in the form 7ab + , where a and b are integers. [4] Solution: ( ) ( ) 1 7 1 7 6 7 12 3 7 5 7 2 72 3 7 5 7 2 3 7 5 3 7 5 3 7 5 504 120 7 6 7 10 63 25 494 114 7 38 13 3 2 12 24 24 7 24 h h h − + + = + + = + += + +−= +− − + −= − −= =− [M1] [M1] – rationalize denoiminator [M1] – correct expansion [A1] 2 Integrate 2sin 1x x+ with respect to x. [3] Solution: 2 1 1 cos 2 1 d d 2 sin 2 2 1s sin 1 ln2 in 2 ln24 xx x xxx xxx xx x c c = = − + + −++ −+  +  =  [M1] [A1] – Integrate 1 cos 2 2 x− [A1] – Integrate 1 x –1 mark if c is not written

Crescent Girls’ School 2022 Sec 4 A Math Prelims P1 4 3 Solve the simultaneous equations. 1 11 2 5 89 2 5 11 xy xy + −− += −= [4] Solution: 1 11 5 89 2 --- (1) 2 5 11 --- (2) yx xy + −− =− −= Sub (1) into (2): ( ) 1 1 12 5 89 2 11 2 89 2 112 5 5 9 10 3 2 144 2 5 2 5 2 xx xx x x x − − +   − − = − + = = = = Sub 5x= into (1), 65 89 2 25 2 y y =− = = [M1] [M1] [A1] [A1]

Crescent Girls’ School [Turn Over 5 4 Express 2 2 7 23 6 ( 2)( 4) xx xx −+ −− in partial fractions. [5] Solution: 22 22 7 23 6 7 23 6 ( 2)( 4) ( 2) ( 2) x x x x x x x x − + − += − − − + Let 2 22 7 23 6 22( 2) ( 2) ( 2) Bx x A xxx x x C−+ = + +−+− + − 227 23 6 ( 2)( 2) ( 2) ( 2)x x A x x B x C x− + = − + + + + − Let 2, 12 4 3 xB B = − = =− Let 2, 80 16 5 xC C =− = = Let 0, 6 4 6 20 2 xA A = =− − + = 2 22 7 23 6 2 22( 2) ( 2) ( 35 2) xx xxx x x −+ = − +−+− + − [B1] – Factorise denominator completely [B1] [M2] – solve for A, B, C using substitution or comparing coefficients. 2 marks if all correct, 1 mark if 2 correct. [A1]

Crescent Girls’ School 2022 Sec 4 A Math Prelims P1 6 5 The gradient function of a curve is 2 4 (1 2 ) p x + − . It is given that the gradient of the curve at the point (1, –3) is 7. (a) Find the value of p. [2] Solution: 2 4 7 (1 2) 3 p p += − = [M1] [A1] (b) Find the coordinates of the point where the curve meets the y-axis. [4] Solution: 2 1 4 3 d (1 2 ) 4(1 2 ) 3( 1)( 2) 2 312 yx x x xc xcx − =+ − −= + +−− = + +−  Sub (1, –3) into y, 233 1 4 c c − = + +− =− 2 3412yx x= + −− Since the curve meets the y-axis, x = 0. y = –2. The coordinates are (0, –2). [M1] [M1] [M1] [A1]

Crescent Girls’ School [Turn Over 7 6 (i) Express 32 2 4log 16 n n − in the form an b+ , where a and b are integers. [3] Solution: ( ) 3 2 6 4 22 4 24 2 42log log 16 2 2 1 2 1 2 log 2 nn nn n n −− − =   = =− [M1] [M1] [A1] (ii) Solve the equation 2 4 2log ( 4) lo 1)g (7 4xx+ =+ + . [4] Solution: 2 2 2 2 2 2 2 2 2 2 4 2 2 2 2 2 2 log ( 4) log (7 4 log (7 4log ( 4) log 4 log (7 4log ( 4) 2 2log ( 4) log (7 4 log 1) 2 ) 1 2 ) 1 2 ( 4) log (7 4 ( 4) 174 ( 6 )1 )1 ) lo 4) 274 8 1