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)                  − =−+ − = − − = + − + − = + − = =− = == = − − = + == =

7 [Turn over 6 The first two non-zero terms in the expansion of ( )( ) 6 11 axbx ++ in ascending powers of x are 1 and 221 4 x− . Find the value of each of the constants a and b, where ab . [7] ( ) ( ) 6 6 5 4 2 22 6 22 2 2 2 2 2 2 661 1 (1 )( ) (1 )( ) ...12 1 6 15 ... (1 ) 1 (1 )(1 6 15 ...) 1 6 15 6 ... 60 6 (1) 2115 6 (2) 4 Sub (1) into (2): 2115 6 ( 6 ) 4 1 ax ax ax ax a x bx ax bx ax a x ax bx a x abx ab ba a ab a a a a    + = + + +       = + + + + + = + + + + = + + + + + += =− −−− + =− −−− + − =− = 4 11, (rej)22 3, 3 (rej) aa bb =− = = =−

8 [Turn over 7 The diagram shows a circle passing through the points A, B, C and D. AC is a diameter of the circle. The line EA is a tangent to the circle and it intersects the straight line EDC at E. (i) Show that angle AED = angle DAC. [2] (ii) Show that 2AD CD DE= . [4] i Let 180 90 90 ( in semicircle or sum of s in a ) 90 (90 ) (tangent radius) = AED EAD DAC AED      =  = − − = −     = − − =⊥  ii 2 ( s in alternate segment) 90 ( in semicircle) (sum of s in a ) similar to (AAA) EAD ACD ADE CDA DEA DAC DEA DAC DE EA DA DA AC DC DE DA DA DC AD CD DE  =   = =    =    == = = OR (from i) 90 ( in semicircle) (sum of s in a ) similar to (AAA) DEA DAC ADE CDA EAD ACD DEA DAC  =  = =    =   

9 [Turn over 8 A vessel in the shape of an inverted right pyramid has a square base of side 12 cm and a height of 30 cm. Water is leaking from the vessel at a constant rate of 5 cm3/s. (i) Show that the volume of water in the vessel, V cm3, is given by 34 75 hV = , where h is the depth of the water. [2] (ii) Find the rate of change of the depth of water when the water is 6 cm deep. [3] i Let x be the length of the side of the water surface. 2 3 12 30 2 5 12( ) ( )35 4 75 xh hx hVh hV = = = = ii 3 2 2 4 75 4 25 4 (6) 525 125 cm/s144 hV dV hdh dV dV dh dh dt dt dh dt dh dt = = = =−  =−

10 [Turn over 9 f(x) is such that 1f '( ) sin cos 44x x x=− . Given that f (2 ) 1 = , show that 16f ''( ) f ( ) sin 4x x a x b+ = + , where a and b are constants. [6] 11( ) 4cos sin 444 21Sub f(2 ) 1, 4cos sin 8 144 1 11( ) 4cos sin 4 144 11''( ) cos 4sin 444 16 ''( ) ( ) 1 1 1 116( cos 4sin 4 ) 4cos sin 4 14 4 4 4 1 1 1 14cos 64sin 4 4cos sin 44 4 4 4 363 sin 44 f x x x c c c f x x x f x x x f x f x x x x x x x x x x  =− − + = − − + = = =− − + =+ + = + − − + = + − − + =+ 1