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 H2 Maths Prelim Paper 2 Solutions Page 5 of 18 BD BE ED = + 31 tan 4 x=+  + (shown) (b) 1 2 22 2 31 tan 4 31 tan tan4 1 tan tan4 31 1 tan 1 tan Since tan , 1 3(1 )(1 ) 1 3(1 )(1 ) 1 3(1 ) 1 3 2 3 2 3 BD x x x x x xx BD x x x x x x x x x xx    − =+  + =+ + − =+ + −   + − + = + − − + = + − + − + = + − + where 1 3, 2 3, 2 3a b c= + =− = 4(i) Note that from 123 19 16 rruu+ =+ 1 19 16 23 23 rruu+ = + From GC, 10 3.6417u = (ii) As 1, and rrr u l u l +→ → → : 23 19 16 4 16 4 ll l l =+ = = (iii) When ru exceeds 99.9% of l : 99.9 100 99.9 (4)100 3.996 r r r ul u u   

2025 JC2 H2 Maths Prelim Paper 2 Solutions Page 6 of 18 1 19 16 23 23 rruu+ = + By GC, r ru 33 3.9956 < 3.996 34 3.9963 > 3.996 35 3.9970 > 3.996  Smallest 34r = (iv) From ( ) 1 n nr r S u c = =− where 1 1942 23 n ru − =−  , ( ) ( ) ( ) ( ) 1 1 1 11 21 1942 23 19 4 2 23 19 19 19 4 2 1 23 23 23 1911 23 4 2 191 23 23 19 4 1 2 23 kn n k knn kk n n n Sc c cn cn cn − = − == −  = − −  = − −      = − − + + + +         − = − −  −  = − − −     Hence, 4Ac=− and 23 2B=− (v) ( ) 23 1941 2 23 n nS c n  = − − −   As n→ , 19 023 n  → and ( )4 cn− → if 4c  . Hence if nS diverges if 4c 

2025 JC2 H2 Maths Prelim Paper 2 Solutions Page 7 of 18 5(i) 2 2 2 2 22 22 22 22 22 22 22 πtan cot , tan cot , 0 2 dd sec cos sec cosdd d d d d d d sec cos sec cos 11 cos sin 11 cos sin sin cos sin cos sin cos cos 2 (Shown) xy xy ec ec y y x x ec ec                   = − = +   = + = − = −= + − = + −= + =− =− Alternative method ( ) 2 2 2 2 22 22 2 2 2 2 2 2 2 2 2 2 2 2 2 πtan cot , tan cot , 0 2 dd sec cos sec cosdd d sec cos d sec cos cossec 1 sec cossec 1 sec 1 cot 1 cot 1 cot cos 1 coscos sin xy xy ec ec y ec x ec ec ec ec ec                      = − = +   = + = − −= + −= − − = + −= =− =− ( ) 2 22 cos cos sin cos 2 (Shown)    =− − =−

2025 JC2 H2 Maths Prelim Paper 2 Solutions Page 8 of 18 (ii) 22 2 At , Equation of : (tan cot ) cos 2 ( (tan cot )) (cos 2 ) tan cot cos 2 tan cos 2 cot (cos 2 ) tan (1 cos 2 ) cot (1 cos 2 ) (cos 2 ) tan (2cos ) cot (2sin ) sin co(cos 2 ) (2cos )cos p L y p p p x p p y p x p p p p p p y p x p p p p y p x p p p p py p x p p  = − + =− − − + = + + − + = + + − + = + + = + 2s (2sin )sin (cos 2 ) 2sin cos 2sin cos (cos 2 ) 2sin 2 (shown) p pp y p x p p p p y p x p + = + += (iii) dpGiven 0.2dt = When 0y= , ( )cos 2 2sin 2 2 tan 2 Note: Since 0 0 0 24 4 2 tan 2 0 p x p xp pp p    = =          ( )2tan 2 ,0Qp When 0x= , 2sin 2yp= Note: Since 0 0 0 24 4 2 sin 2 0 pp p            ( )0,2sin 2Rp ( )( )1 2sin 2 2 taAr n 2 2 tan 2 sin 2ea f , 2o O ppQ pR pA == ( ) ( ) 2 2 d 2 tan 2 (2cos 2 ) (sin 2 )2sec 2d 4 sin 2 tan 2 sec 2 At , 6 d 3 5 4 sin tan sec 4 2 3 4 3 10 3d 3 3 3 2 2 d d d 10 3 0.2 2 3 units /sdt d d A p p p pp p p p p A p A A p pt     =+ =+ =         = + = + = =                   = =  = 

2025 JC2 H2 Maths Prelim Paper 2 Solutions Page 9 of 18 6(a) (i) Case 1: Two married couples are in the same taxi Number of ways = 2 Case 2: Two married couples are in different taxis Number of ways = 2 1C × 4 2C × 2 2C = 12 Total number of ways = 2 + 12 = 14 (ii) The two married couples must sit at the back in different taxis. Number of ways to sit the two married couples = 2 × (2 × 2) × (2 × 2) = 32 Number of ways to sit the 4 singles = 4! or 4 2C 2! 2! = 24 Required number of ways = 2 × (2 × 2) × (2 × 2) × 4! = 768 7 (a) If A and B are independent, P( ) P( )P( )A B A B= Since BA , P( ) P( )A B B= and since AS , P( ) 1A  So P( ) P( ) P( )P( )A B B A B =  . Hence A and B are not independent. OR A and B are not independent since the occurrence of B implies the occurrence of A. (or the non-occurrence of A implies the non-occurrence of B) OR From the Venn Diagram, B A S . Hence ( | ) 1 ( )P A B P A= as seen from the Venn Diagram. OR From the Venn Diagram, B A S . Hence ( | ) ( )P B A P B as the reduced sample space AS as seen from the Venn Diagram. ( | ) 1P A B = 7(b)( i) ( 2) ('2'appears once and 0' appears twice or '1' appears twice and '0' appears once) 1 1 1 3! 1 1 1 3!= 2 6 6 2! 3 3 6 2! 7 72 PX P = =    +    = (ii) ( 6) ('2'appears 3 times) 111 222 1 8 PX P = = =   =

2025 JC2 H2 Maths Prelim Paper 2 Solutions Page 10 of 18 ( 5) ('2'appears twice and '1' appears once) 1 1 1 3!= 2 2 3 2! 1 4 PX P = =    = ( 3) ('2','1','0' all appear once or '1' appe ar 3 times) 1 1 1 1 1 1= 3!2 3 6 3 3 3 11 54 PX P = =    +   = Or Complement Method 𝑃(𝑋 = 3) = 1 − 1 216 − 1 36 − 7 72 − 7 24 − 1 4 − 1 8 =