SAJC 9758 2025 Prelim P2 Solutions

2025 JC2 H2 Maths Prelim Paper 2 Solutions Page 1 of 18 No Solution 1(i) Method 1: Algebraic Method 23 4 2xx−  + 2 2 2 2 2 2 2 22 2 3 4 2 9( 4) ( 2) 9( 2) ( 2) ( 2) ( 2) 9( 2) 1 0 ( 2) (3 5)(3 7) 0 xx xx x x x xx x x x −  + −  + − +  + + − −  + − −  + + ─ + 572 33− 57 or 233 xx  =− Method 2: Graphical Method Using GC to find x-coordinates of the points of intersection between | 2|yx=+ and 23 | 4 |yx=− : 572 or or 33xx=− =

2025 JC2 H2 Maths Prelim Paper 2 Solutions Page 2 of 18 From the graph, 57 or 233 xx  =− (ii) 2 2 2 2 2 3 1 4 12 1 4 132 113 4 2 x x xx x xx xx − + − + −  + Replace x with 1 x , 5 1 7 1 or 233 xx  =− From the graph, 3 3 1 or7 5 2xx  =− 2(a) (i) Since x = is a root, f ( ) 0 = 64 0p qr + + = 64 64 ) ) ((( )f 0 qp r qrp     −− = + − + = + + = x  =− is also a root. 2 (a) (ii) 5x= and x = are roots 33 75   1 2−

2025 JC2 H2 Maths Prelim Paper 2 Solutions Page 3 of 18 5x =− and x =− are also roots. Since coefficients of f are all real, *x = and *x =− are also roots. Therefore, the remaining roots are –5, an,* d*,  −− . (b) Method 1: Since 1 i+ is a root, ( )( ) 3 2 2(3 ) (2 6i) 6 1 i Bz a z z z zCz+ − − + − = − ++− Comparing coefficients, Constant: 6 ( 1 i) 3 3i ---(1) C C − = − − =− 2 : 3 1 i (2) 4i BaB az =− = − − −−−−− −+ : 2 6i 3 3i ( 1 i) (1+i) 5 3i 5 3i = 1i =4 i zB B B B − − = − + − − =+ + + − 4 i 4 (4 i) i 2iaB= − + = − − + = ( )( ) 32 2 (3 ) (2 6i (4 i) ) 6 0 1 i 0 3 3i z a z z z z z + − + − + − − + − = − − = 1iz=+ or 2 (4 i) 3 3i 0zz + − + − = ( )( ) or 1 3 1 i 0 i3 z z z z+ + − = = =− +− OR 2 2 4(1)(3 (4 i) 3 3i 0 (4 i) (4 i) 2 4 i (2 i) 2 3i 13 ) or i z z z z z z −− + − + − = − −  −= − +  += =− =− + Therefore, the other roots are and3 1i− −+ . Method 2: Since 1i+ is a root,

2025 JC2 H2 Maths Prelim Paper 2 Solutions Page 4 of 18 32(1 i) (3 )(1 i) (2 6i)(1 i) 6 0 2 2i (3 )(2i) ( 4 8i) 6 0 (3 )2i 4 6i 4 6i3 2i 3 3 2i 2i a a a a a a + + − + − + + − = − + + − − − + − = − = + +−= − = − = ( )( ) 3 2 2(3 2i) (2 6i) 6 31 3iiz z z z z Bz + ++ − − + − = − − − Comparing coefficients, 2 : 3 2i 1 i 4i zB B − = − − =− ( )( ) 32 2 (3 ) (2 6i (4 i) ) 6 0 1 i 0 3 3i z a z z z z z + − + − + − − + − = − − = 1iz=+ or 2 (4 i) 3 3i 0zz + − + − = ( )( ) or 1 3 1 i 0 i3 z z z z+ + − = = =− +− OR 2 2 4(1)(3 (4 i) 3 3i 0 (4 i) (4 i) 2 4 i (2 i) 2 3i 13 ) or i z z z z z z −− + − + − = − −  −= − +  += =− =− + Therefore, the other roots are and3 1i− −+ . 3(a) BC = 2 E is the midpoint of BC. 1BE EC = = By Pythagoras’ theorem, 222 1 3AE= − = In AED , 3tan 4 x ED  += 3 tan 4 ED x=  +

2025 JC2