2026 HGV P2 MS

HILLGROVE SECONDARY SCHOOL PRELIMINARY EXAMINATION 2026 SECONDARY 4G3/5G2 CANDIDATE NAME MARK SCHEME ( ) CLASS - CENTRE NUMBER S INDEX NUMBER ADDITIONAL MATHEMATICS Paper 2 Candidates answer on the Question Paper. No Additional Materials are required. 4049/02 31 Aug 2026 2 hours 15 minutes 10.05 AM to 12.20 PM READ THESE INSTRUCTIONS FIRST Write your class, index number and name in the spaces at the top of this 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. DO NOT WRITE ON ANY BARCODES. 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. 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 number of marks for this paper is 90. For Examiner’s Use TOTAL 90 Parent’s/ Guardian’s Signature: ____________________ Setter: Mr Ng Choon Chooi This document consists of 17 printed pages.

Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 2 0ax bx c+ + = , 2 4 2 b b acx a−  −= Binomial expansion ( ) 1 2 2 ... ... ,12n n n n n r r nn n na b a a b a b a b b r− − −     + = + + + + + +           where n is a positive integer and ( ) ! ( 1)...( 1) ! ! ! n n n n n r r r n r r  − − +== − 2. TRIGONOMETRY Identities 22sin cos 1AA+= 22sec 1 tanAA=+ 22cosec 1 cotAA=+ sin( ) sin cos cos sinA B A B A B =  cos( ) cos cos sin sinA B A B A B=  tan tantan( ) 1 tan tanABAB AB=  sin 2 2sin cosA A A= 2 2 2 2cos 2 cos sin 2cos 1 1 2sinA A A A A= − = − = − 2 2 tantan 2 1 tan AA A= − Formulae for ABC sin sin sin a b c A B C== Abccba cos 2222 −+= 1 sin2 ab C=

HGV 4G3/5G2 A MATH PRELIM EXAM P2 2026 3 1. (a) Express 23 24 58xx−+ in the form 2()a x b c++ where a, b and c are constants. [3] 23 24 58xx−+ 2 583[ 8 ] 3xx= − + [M1] 2 2 2 583[ 8 4 4 ] 3xx= − + − + [M1] 2 103[( 4) ] 3x= − + 23( 4) 10x= − + [A1] (b) Hence explain why there is no intersection between the graph 23 24 58y x x= − + and the line 9y= . [1] 2 The minimum point of 3 24 58 is (4,10)y x x= − + or Minimum value of y = 10 So there no intersection between the curve and the line 9.y= [B1]

HGV 4G3/5G2 A MATH PRELIM EXAM P2 2026 4 2. Solve the equation ( ) ( ) 14log1log5log2 3 82 2 4 +−=−− xxx . [6] ( ) ( ) 14log1log5log2 3 82 2 4 +−=−− xxx ( ) ( ) 2 4 2 82log 5 log 1 3log 4 1x x x− − = − + [M1] ( ) ( )2 22 2 22 log 4log 52( ) log 1 3( ) 1log 4 log 8 xx x −− − = + [M1] ( ) ( )2 22 2 log 4log 52( ) log 1 3( ) 123 xx x −− − = + ( ) ( ) 2 2 2 2 2log 5 log 1 log 4 log 2x x x− − = − + ( ) ( ) 2 22 5log log 2 41 x xx =−− [M1] ( ) ( ) 25 241 x xx =−− [M1] ( )( ) 25 2 4 1x x x= − − 23 10 8 0xx+ − = ( )( )3 2 4 0xx− + = [M1] 2 or 43xx= =− [A1]

HGV 4G3/5G2 A MATH PRELIM EXAM P2 2026 5 3. In the binomial expansion of 8 2 kx x + , where k is a positive constant, the coefficient of 2x is 28. (a) Show that 1 4k = . [4] 88General term (2 ) ( ) rr kxr x−= [M1] 888 2 r r r rx k xr − − −= 8 8 28 2 r r rkxr −−= [M1] Let 8 2 2, 3rr− = = 8 3 38 2 283 k− = [M1] 31792 28k = 3 1 64k = 1 4k = [A1] (b) Hence, find the term in x in the expansion of 8 162 kxx xx  −+     . [4] Let 8 2 0, 4rr− = = 8 4 48 1 352 ( )4 48− = [M1] 8 2 162 1 356 ...... 28 ...... 8 kxx xx xx x   −+       = − + + +     [M1] 235 1Term in 6 28 8x x x x=  −  [M1] 7 4 x=− [A1]

HGV 4G3/5G2 A MATH PRELIM EXAM P2 2026 6 4. The diagram shows a quadrilateral ABCD in which the point B is (0, 3) and the point D is (7, 0). The equation of the line AD is 5 2 14yx=− and C lies on the line yx= . The line CD is parallel to the y-axis. Given that A lies on the perpendicular bisector of BD, (a) find the coordinates of A and of C, [8] 7 0 0 3 1 1Mid-point of ( , ) (3 ,1 )2 2 2 2BD ++== [B1] 0 3 3Gradient of 7 0 7BD −= =−− [M1] 7Gradient of perpendicular bisector 3= [M1] Equation of perpendicular bisector is 1 7 1( 1 ) ( 3 )2 3 2yx− = − 72 633yx=− [M1] 72Subst. 6 into 5 2 14,33y x y x= − = − 725( 6 ) 2 1433xx− = − [M1] 29 58 33x= y x O A D C B(0,3)

HGV 4G3/5G2 A MATH PRELIM EXAM P2 2026 7 2, 2xy= =− Coordinates of (2, 2)A=− [A1] Subst. 7 into ,x y x== [M1] 7y= Coordinates of (7,7)C= [A1] (b) find the area of the quadrilateral ABCD. [2] 2 7 7 0 21Area of 2 0 7 3 22ABCD= −− [M1] 39= [A1]

HGV 4G3/5G2 A MATH PRELIM EXAM P2 2026 8 5. (a) Express 2 28 ( 4)( 1) x xx + +− in partial fractions. [6] 22 28Let ( 4)( 1) ( 4) ( 1) Cx Ax Bx x x x++ =++ − + − [M1] 22 8 ( )( 1) ( 4)x Ax B x C x+ = + − + + [M1] Subst. 1, 10 (5), 2x C C= = = [B1] Subst. 0, 8 ( 1) (4), 0x B C B= = − + = [B1] Subst. 1, 6 ( )( 2) (5), 2x A B C A=− = − + − + =− [B1] 22 2 8 2 ( 4)( 1) ( 4) ( 1) 2xx x x x x +− =++ −