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