ANDSS AM2024WA2 (ANSWER)

3E A Math WA2 2024 Marking Scheme 1 2( 3) 4k x x k+ − + Coefficient of 2x must be 0 , 30k+ 3k − M1 and Discriminant must be 0 , ( ) ( )( ) 2 4 4 3 0 kk− − +  M1 24 12 16 0kk− − +  24 12 16 0kk+ −  2 3 4 0kk+ −  ( )( )4 1 0kk+ −  M1 4k − (rej 3k − ) or 1k  A1 – marks will not be awarded without rejection k 4− 1

2(a) ( ) 3f 9 10x x x= − − ( ) ( ) ( ) 3 f 2 2 9 2 10 0− = − − − − = M1 By factor theorem, ( )2x+ is a factor. ( ) ( )( ) 2f 2 5x x x Ax= + + − ( ) ( ) ( ) 32 2 2 5 10f x A x A xx = + + + − − By comparing coefficients of 2x (or alternatively coefficients of x), 20A+= 2A=− Or 259A− =− 2A=− --- Alternatively, ( ) ( ) ( ) 2 3 32 2 2 25 2 9 10 2 2 9 2 4 5 10 5 10 0 xx x x x xx xx xx x x −− + − − −+ −− − − − −− − − − M1 – if student expand ( )( ) 22x Ax Bx C+ + + the method mark will only be given for getting linear equation solving for unknown B in one of the forms shown in the marking scheme. For long division method, the method mark will be awarded if the student does the first 2 rounds of division correctly. ( ) ( )( ) 2f 2 2 5x x x x= + − − A1

2(b) ( )f0 x = ( )( ) 22 2 5 0x x x+ − − = 20x+= or 2 2 5 0xx− − = 20x+= or ( ) ( ) ( )( ) ( ) 2 2 2 4 1 5 21x − −  − − −= M1 20x+= or 2 24 2x = 2x=− or 1.45x=− (3s.f.) or 3.45x= (3s.f.) A1 (if student leaves in the simplest surd form 16 , these are accepted too) 3 Base 3 75 2 2 108 4 += − M1 – correct formula applied using area of triangle Base 3 75 108 2 += − Base 3 5 3 6 3 2 += − M1 – simplifying surds to simplest form Base 3 5 3 6 3 2 6 3 2 6 3 2 ++= −+ M1 - rationalising Base 28 3 96 104 += Base 24 7 3 26 += cm A1 4(a) 3 11 2xx−= 23 11 4xx−= M1 24 11 3 0xx+ − = ( )( )4 1 3 0xx− + = M1 – alternatively, quadratic formula will be accepted too 1 4x= or 3x=− (rej 3 11 0x− , thus 20x ) A1 – marks will not be awarded if rejection with the correct reasons is not provided

4(b) 15 12 5 5 ab+ = + ( )( )15 12 5 5 5 a b a b+ = + + M1 – squaring both sides of equation 2215 12 5 2 5 5a ab b+ = + + 2215 12 3 5 2 5a b ab+ = + + 2 12ab= 6a b= --- (1) 2215 5 ab=+ --- (2) M1 – correctly identified simultaneous equations by comparing LHS and RHS Subs (1) into (2), 2 2615 5 bb =+ 2 2 3615 5 bb=+ M1 – substitution A1 5 y x k=− --- (1) 22 39xy+= --- (2) Subs (1) into (2), ( ) 22 39x x k+ − = M1 2 2 23 6 3 9 0x x kx k+ − + − = 224 6 3 9 0x kx k− + − = For the line to be tangent to the curve, Discriminant 0= , ( ) ( )( ) 2 26 4 4 3 9 0kk− − − = M1 2236 48 144 0kk− + = 212 144 0k −= 2 12 0k −= ( )( )12 12 0kk+ − = M1 – simplify quadratic equation 12 23 k = = A1 – 3.46 (3s.f.) will be accepted too but surds not in the simplest form will not be accepted to be consistent with what we apply for topic on surds

6(a) By Factor theorem, 2f0 3  = 32 2 2 23 5 03 3 3 ab     + − + =           4 22 99ab+= --- (1) M1 – applying factor theorem By Remainder theorem, ( )f 2 40− =− ( ) ( ) ( ) 32 3 2 2 5 2 40ab− + − − − + =− 4 26ab+ =− --- (2) M1 – applying remainder theorem ( ) ( )2 1 :− 32 256 99a=− M1 – elimination or substitution method to solve simultaneous equation 8a=− 6b= A1 A1 6(b) ( ) 32f 3 8 5 6x x x x= − − + By Remainder Theorem, ( ) ( ) ( ) ( ) 32 f 4 3 4 8 4 5 4 6− = − − − − − + ( )f 24 94− =− B1 – if student uses long division, it will be accepted too.