MFSS AMath Prelim P1 Solutions for Students

Sec 4 Express A Math Prelim Paper 1 2024 Marking Scheme 1 Given that 334(5 ) 20xx+− = , evaluate 10x without using a calculator. [4] 33 3 3 3 3 4(5 ) 20 204(5 )(5 ) 20 205 (20 ) 4(5 ) 100 16 10 4 xx x x xx x x +− = = = = =

2 [Turn over 2 Solve the equations. (a) [4] (b) 510log 5 3 logy y+= [4] a 22 22 2 22 2 2 2 1 2 log ( 4) 2log 1 2log log ( 4) 1 log log ( 4) 1 log 1 4 24 2 8 0 ( 4)( 2) 0 4, 2 (rej) xx xx xx x x x x xx xx xx + = − − + = − + = =+ =+ − − = − + = = =− b 5 5 5 5 5 5 5 2 2 55 52 10log 5 3 log log 510 3 loglog 10 3 loglog Let log 10 3 10 3 3 10 0 ( 5)( 2) 0 5, 2 log 5 log 2 55 13125 25 y y yy yy yu uu uu uu uu uu yy yy yy − +=  +=  += = += += − − = − + = = =− = =− == == 22log ( 4) 2log 1xx+ = −

3 [Turn over 3 The variables x and y are related by 28 21 xy x += + . When values of (1 )xy − are plotted against y, a straight line is obtained. The straight line intersects the vertical and horizontal axes at A and B respectively. (i) Find the coordinates of A and of B. [4] (ii) State the value of tan . [1] i 28 21 2 2 8 2 2 8 2 (1 ) 8 1(1 ) 4 2 (0, 4) sub (1 ) 0, 104 2 8 (8,0) xy x xy y x x xy y x y y x y y A xy y y B += + + = + − = − − = − − = − =− −= =− = = ii 1tan 2 =

4 [Turn over 4 (i) Factorise completely 322 3 5 6x x x− − + . [4] (ii) Hence, solve 322 3 5 6 0y y ye e e− − + = . [3] i 32 2 32 32 2 2 ( ) 2 3 5 6 (1) 2 3 5 6 0 ( 1) is a factor 26 1 2 3 5 6 (2 2 ) 5 6 ( ) 6 6 ( 6 6) f x x x x f x xx x x x x xx xx xx x x = − − + = − − + = − −− − − − + −− − − + − − + −+ − − + 0 3 2 22 3 5 6 ( 1)(2 6) ( 1)(2 3)( 2) x x x x x x x x x − − + = − − − = − + − ii 322 3 5 6 0 31, ( ), 2 2 0, ln 2 y y y y y y e e e e e rej e yy − − + = = =− = ==

5 [Turn over 5 (i) Prove that   tan1)cos(sinsec sec2 2 −=+ − . [4] (ii) Hence solve the equation 2 24 2sec sec 2sec (sin cos )     − =−+ for 02  . [5] i 2 2 2 2 2 secLHS = sec (sin cos ) 2 (1 tan ) 1 (sin cos )cos 1 tan sin cos cos cos 1 tan tan 1 (1 tan )(1 tan ) tan 1 1 tan RHS               − + −+= + −= + −= + +−= + =− = OR 2 2 2 2 2 2 2 2 22 2 secLHS = sec (sin cos ) 112 [ (sin cos )]cos cos 2cos 1 cos cos sin cos 2cos 1 cos (sin cos ) 2cos (sin cos ) cos (sin cos ) cos sin cos (sin cos ) (cos sin )(cos sin )                            − + = −  + −= + −= + −+= + −= + +−= cos (sin cos ) (cos sin )(cos sin ) cos (sin cos ) cos sin cos sin1 cos 1 tan RHS                + +−= + −= =− =− =

6 [Turn over ii 2 2 2 2 2 4 2sec sec 2sec (sin cos ) 2(1 tan ) sec 2 2 2 tan 1 tan 2 tan 2 tan 3 0 (tan 3)(tan 1) 0 tan 3, tan 1 1.2490 4 1.2490, 2 1.2490 , 44 51.8925,5.0341 , 44 1.89,5.03 (3sf)                