AISS Prelim EMath Paper 2 Marking Scheme

1 2025 Sec 4E Math Prelim Paper 2 Marking Scheme Qn Working Mark Awarded Sub- total Remarks 1 (a) 5 2 1 3xx−  −  + 5 2 1x−  − and 2 1 3xx−  + 42 x− 4x 2 x− 24 x−   Primes that satisfy the inequality: 2, 3 M2 A1 3 M1 for 2 x− M1 for 4x (b) 22 330 11 25 22 xz y xy z  = 22 3 30 22 11 25 x y y z xz = 3 4 12 5 xy z B3 2 4 terms to simplify: 12 5 , x, y3, z4. B1 for one correct term in the correct position (numerator or denominator). B2 for two or three correct terms. B3 for all four correct terms. (c) 3 12 4 1681 p q −   = 3 16 4 12 81q p   = 12 9 27q p B2 2 B1 for numerator, B1 for denominator. (d) 4212 x xx +=−− 2( 1) 4 12 xx xx +− =−− 3 2 4 12 x xx − =−− (3 2)(2 ) 4( 1)x x x− − = − 6x – 3x2 – 4 + 2x = 4x – 4 –3x2 +4x = 0 x(–3x + 4) = 0 x = 0 or –3x + 4 = 0 x = 4 3 M1 M1 A1 3 M1 for correct quadratic equation. M1 for correct factorisation or correct substitution into quadratic formula A1 for both correct answers.

2 Qn Working Mark Awarded Sub- total Remarks 2 (a)(i) ( –4, 9) B1 1 (a)(ii) riseGradient = run 5 = 3 5 = 3 − − y = mx + c y = 5 3 xc−+ When x = –1 and y = 4, 54 ( 1)3 c=− − + 7 3c= 57 33yx=− + M1 A1 2 M1 for correct gradient. (b) D(3, y) Length of CD = 3 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 22 2 2 2 2 22 2 3 ( 2) 5 13 3 ( 2) 5 13 5 5 13 5 13 5 5 144 5 12 5 12 y y y y y y y − − + − = − − + − = + − = − = − −= − = = y = –7 or y = 17 D(3, –7) D(3, 17) M1 M1 A1 3 M1 for identifying that the x-coordinate of D is 3. M1 for correct formula for length of line segment. A1 for both correct answers.

3 Qn Working Mark Awarded Sub- total Remarks 3 (a) Cosine Rule 2 2 2 2 2 2 2 2 2 2 2 2 1 62 37 50 2(37)(50)cos 62 37 50 2(37)(50)cos 62 37 50 cos2(37)(50) 62 37 50cos 2(37)(50) 89.61286341 89.6 (to 1 d.p.) ABC ABC ABC ABC ABC ABC − = + −  − − =−  −− =−  −−=  −  =   =  M1 M1 A1 3 M1 for correct substitution of values into Cosine Rule. M1 for rearranging and making ABC the subject of the equation. (b) Method 1: Interior angles 180o – 124o = 56o (interior angles, parallel lines) 360o – 56o = 304o (angles at a point) Method 2: Alternate angles 180o + 124o = 304o