ANDSS 4E2024EM Prelim P1 MS

Sec 4 Express Mathematics 2024 Prelims Marking Scheme 1 ( )( )5 2 5 2a a b a b−+ ( ) ( )( ) 22 52a a b=− ( ) 2225 4a a b=− M1 Correct expansion 3225 4a ab=− A1 2 23 3 90xx ++= 23 3 3 90xx+  = M1 ( )3 1 9 90x += 39x = 233x = 2x= A1 3 2 22 2 4 3 6 26 x xy x y x xy y + − − +− ( ) ( ) 22 2 2 3 2 26 x x y x y x xy y + − += +− M1 Factorise by grouping for numerator ( )( ) 22 2 2 3 26 x y x x xy y +−= +− ( )( ) ( )( ) 2 2 3 2 3 2 x y x x y x y +−= −+ M1 Factorise denominator 23 23 x xy −= − A1 4 (a) LCM of A and B is 237mq + and at the same time the LCM of A and B is 33 5 7 . 1m= B1 5q= B1 (b) 337A= and 3 5 7B=   HCF of A and B 37= 21= B1

5 The size/height of the bar graph. B1 Stating that the graph does not start from zero is accepted too. It can be misleading because for example, the size/height of the graph in 2019 is twice that of the graph in 2010 but this does not represent their actual temperatures. B1 A relevant example has to be given. 6 (a) 2 9 3 627 y x − −   29 3 2236 333 y x −     − −  −       = 6 243 y x − −= M1 Multiply 2 3 − to each index 2 46 3 xy= A1 46 9 xy is also accepted (b) 2 25 5 125 y x− = ( ) 2 23 5 5 5 y x− = 2 63 5 5 5 y x− = 26355 xy−+ = M1 Any other equivalent form of 55mn= will be accepted too 3455xy− = Comparing index, 34xy−= 4 3 yx += A1

7 (a) 229 24 16x xy y++ ( ) 2 34xy=+ B1 (b) Let 4xa= , ( ) 8 8 4 2144 9 24 16a a a y y− + + ( ) 2 2 2144 9 24 16x x xy y= − + + ( ) ( ) 22 12 3 4x x y= − + M1 Attempt to factorise an expression in the form 22ab− , after making use of part (a) answer ( )( )12 3 4 12 3 4x x y x x y= + + − − ( )( )15 4 9 4x y x y= + − ( )( ) 4415 4 9 4a y a y= + − A1

8 (a) 2 kq r= , where k is the proportionality constant. q and r are the initial intensity and distance respectively. When the distance is reduced by 40%, the new distance is 3 5 r . New intensity 2 3 5 k r =  M1 29 25 k r= 2 25 9 k r= 25 9 q= A1 (b) Percentage difference New Intensity Initial Intensity 100%Initial Intensity −= 25 9 100% qq q − = 16 100%9= 178%= (3 s.f.) B1 7177 %9 will be accepted too.

9 B1 Both shape and y- intercept needs to be drawn and indicated correctly 10 SC RD= (given) No marks for this step SCD RDA = (alternate angles) M1 CD DA= (sides of a rhombus are equal) M1 By SAS congruency test, triangle SCD is congruent to triangle RDA. A1 11 2 10 2xx−− ( ) 2 5 25 2x= − − − ( ) 2 5 27x= − − M1 Minimum point is ( )5, 27− A1 y x ( )0,1

12 (a) ( )n  33= 9= B1 (b) P ( ) ( ) ( ) ( ) 1,0 , 0, 1 , 0,0 , 0,1= − − B1 (c) Q ( ) ( ) 2,1 , 1,1= − − B1

13 The diagram below shows a triangle ABC. Note that marks will not be awarded if relevant arcs are not drawn for parts (a) and (b) respectively. (a) B1 (b) B1

14 (a) ( )n2W = B1 (b)        , 0 , 0 , 0, 0 B1 (c) 'AB / ( )''AB / ( ) 'A B A / ( ) 'A B B B1 15 513 32 xxx −+−   53 3 xx −− 3 9 5xx−− 4 14x 13 2x M1 51 32 xx−+  10 2 3 3xx−  + 75 x 215 x M1 211352 x A1

16 o90BAC= (right angle in a semicircle) M1 cos ABABC BC= 8 17 34 AB= M1 Only awarded with the relevant values substituted in this step. 16AB= cm AC 2234 16=− 30= cos cosACD ACB =−  M1 cos AC BCACD=− 4cs 30 3o ACD − = 7cs 15 1o ACD − = A1 17 (a) 2 7 87T =− 7 57T = B1 (b) ( ) 2 1nT n n= + − B1 (c) 1nnTT+ − ( ) ( ) ( ) 22 2 1 1n n n n = + − + − + −  M1 22 4 4 1 2 1n n n n n n = + + − − − + + − 22n=+ ( )21n=+ M1 I agree. Since the difference 1nnTT+ − is a multiple of 2, hence the difference will always be an even integer. A1

18 (a) 15 3 5= ( )2 5 2,1Q= − +  ( )8,1Q= B1 (b) Gradient of PR ( ) 41 13 2 −= −− 1 5= M1 Let the equation of the line PR be y mx c=+ 14 135 c=  + 7 5c= Equation of line PR: 17 55yx=+ A1 (c) ( )13,0W = B1 (d) Length of line segment QR ( ) ( ) 22 13 8 4 1= − + − 34= M1 Perimeter of PQRS 2 34 2 10=  +  M1 31.7= units (3 s.f.) A1 19 320.3 10 365 7409500= 67.41 10= (3 s.f.) B1