ACSI 2023 Y3EXP FYE AMath P1 Solutions

2023 Yr 3 EXP End Year Exam P1 Mark Scheme Markers’ Comments 1 Making y the subject the linear equation: 355 3xy y x+=⇒=− ---- (1) 2225x xy y+−= --------------(2) Sub Eqn (1) into Eqn (2), ( ) ( ) ( ) 22 22 2 2 2 2 53 53 5 2 5 3 25 30 9 5 10 35 30 0 2 7 60 ( 2)( 2 3) 0 xx x x x xx xx xy xx xx + − −− = +− − − + = − + −= − + −= −−+= 2 or 1.5x∴= 1y∴= − or 0.5y= • Quite well attempted. • Some students forgot to find the other variable to complete the solutions 2 58 1 24 2 ABm −= =−− 18 ( 4)2YX−= − OR ( )184 2 c= + 6c∴= 1 62YX∴= + 21 62 x xy = + 2 2 2 12 2 2 12 2 12 xx y y xx xy x += ⇒= + ∴= + • Some students input the given points as (x, y) instead of (X, Y): ( ) 241 482 15 2 c c = + ∴= − 3 2 22 12 111 x xx + = +−− ( )( ) 2 11 11 AB xx x x = ++− +− ( ) ( )21 1Ax Bx= −+ + Let 1: 1xB= = Let 1: 1xA= −= − 2 2 1 1111 11 x x xx +∴ = −+− +− • Many students did not express use long division to express given algebraic fraction in proper form 4(i) ( )00100 2 ln100 6.64ln2 tPP t = = = ∴7 hours

4(ii) 5 3 3 3 3 3 3 log 3log 9 1 log 9log 3 1 log 6log 1 log xx x x x x −= −= −= Let 3log x be a. ( )( ) 26 1 60 3 20 3 or 2 a aaa aa a −= ⇒ −−= − += = − 323 or 3 127 or 9 x x −= ∴= 6(i) 2 2 2 cot sec 1 1 tantan 53 135 89 3 25 227 75 θθ θθ + = ++ = ++  = + = 6(ii) 3sin 34 θ =− 6(iii) ( )cos 180 cos 5 34 θθ°− =− = 0P

6(iv) ( ) ( ) ( ) cosec cos cos sec sin sin 5 534 3 3 34 θ θθ θ θθ − = = −− − = =− 7(a) Comparing coefficient of 3x : 5B= Sub 1:x= 33A=− Sub 0x= : 33 ( 3) 11CC−= − ∴= 7(b) ( ) ( ) ( )( ) ( ) 2() 2 3 2 21 2 fx x x Qx R x x Q x ax b = +− + = − + ++ 1 4 (1)2 2 7 (2) ab ab + = −−− − + =− −−− Taking (2) – (1): 2.5 11 4.4aa− = − ∴= 1.8b⇒= 8(a) Given 2 2 22 0.56 xx xx ++ ≥−+ Since ( ) 22 2 2 1 10xx x+ += + +> for all real values of x 2 5 60xx⇒ − +> ( )( )2 30xx− −> 23 x∴<< 8(b) y xk= + -------------(1) 2xy y= + -----------(2) Sub (1) into (2) : ( ) 2x xk xk+ =++ ( ) 2 1 20x k xk+ − −−= ( ) ( ) 2 1 20x k xk+− −+= ( ) ( ) 22 2 2 2 4 1 4(1)( 2) 2 14 8 29 18 b ac k k kk k kk k − = − − −− = − ++ + =++ =++ Since ( ) 2 1 8 0 k+ +> for all real values of k, ∴2 intersections

9(i) 21 or 0.02100 50p π π= = 9(ii) ( )sin 100 1 100 2 0.005 t t t π ππ = = = 9(iii) 9(iv) ( ) ( ) 1 230sin 100 115 1sin 100 2 1sin 26 5100 , 66 11 ,600 120 t t t t π π πα πππ − = = = = = =   10(i) Area of triangle ABC 2 22 7 21 3 411 32 47.5 units −−= − = 10(ii) 7 4 5 AB AD m m =− =− Since AB ADmm ≠ , A, B and E are not collinear. 10(b) Let the foot of the perpendicular from A to BC be E. 11 ( 4) 372 BCm −−= = − Equation of line : ( 4) 3( 2) BC y x−− = − 3 10yx∴= − ---------- (1)

1 3 AEm =− 1Equation of line : 3 ( 2) 3AE x−= − + 17 33yx∴= − + ---------- (2) (1) = (2): 173 10 33xx−= −+ 9 30 7 10 37 3.7 and 1.1 xx x xy − = −+ = ∴= = ( )3.7,1.1E∴