ACSI 2020 Y3EXP FYE AMath P1 Solutions

Solutions – End of Year Exam 2020 Year 3 Express Additional Mathematics Paper 1 1 Solve the following pair of simultaneous equations 2243 1x xy y+ += 1xy+= [4] Solutions: 1 2243 1x xy y+ += -- (1) 1xy+= -- (2) From (2): 1xy= − -- (3) Sub (3) into (1): 224(1 ) 3(1 )( ) 1y yy y− +− += 2 2248 4 3 3 1yy yyy−+ +− += 22 5 30yy− += ( 1)(2 3) 0yy− −= 10y−= or 2 30y−= 1y = or 3 2y = 0x= or 1 2x=−

2 Given that 12tan 5θ = and that θ is acute, find the exact value of (i) cos( ),θ− [1] (ii) cos(90 ), θ°− [1] (iii) tan(180 ). θ°− [1] Solutions: 2 (i) cos( ) cosθθ−= 5 13= (ii) cos(90 ) sinθθ°− = 12 13= (iii) tan(180 ) tanθθ°− =− 12 5=− θ 12 5 13

3 (a) The graph of log ( 1)ay kx= − passes through the points with coordinates (1, 0) and (5, 2). (i) Determine the value of each of the constants a and .k [4] (ii) Write down the range of values of x such that y is defined. [1] (b) Sketch the graph of 4logyx= . [2] Solutions: 3 (a) (i) Sub (1, 0) into log ( 1)ay kx= − 0 log ( 1)a k= − 01ka−= 2k = Sub (5,2) into log ( 1)ay kx= − 2 log (2 5 1)a= ×− 2 log 9a= 29 a= 3a∴= (rej 3a=− ) (a) (ii) log (2 1)ayx= − 2 10x−> 1 2x> (b) x y 1 4logyx=

4 It is given that 32( ) 4 16 21 9fx x x x=− +− . (a) Find the quotient when ()fx is divided by 2 1x + . [2] (b) Prove that 1x− is a factor of ()fx . [1] (c) Hence, factorise ()fx completely. [3] (d) Express () x fx in partial fractions. [5] Solutions: 4 (a) Quotient = 4 16x− (b) ( ) ( ) ( ) ( ) 32 1 4 1 16 1 21 1 9f =− +− 0= By factor theorem, 1x− is a factor of ()fx . (c) 32( ) 4 16 21 9fx x x x=− +− 2( 1)( )x Ax Bx C=− ++ By synthetic division, 2( ) ( 1)(4 12 9)fx x x x= − −+ 2( 1)(2 3)xx= −− (d) 2( ) ( 1)(2 3) xx fx x x= −− 21 23 ( 23 ) AB C xx x= ++−− − ( ) ( )( ) ( ) 2 23 1 23 1x A x Bx x Cx∴= − + − − + − Sub 1x= , 1A∴= Compare coefficients of 2x : 324 16 21 9xxx− +−2 1x + 4 16x− 3(4x− 4)x+ 216 17 9xx− +− 2( 16 x−− 16)− 17 7x+ 4 16− 21 9− 1 4 4 12− 12− 9 9 0

0 4(1) 2 B= + 2B∴= − Compare constants: 0 9(1) 3( 2) C= +−− 3C∴= 22 12 3 (1 ) ( 23 ) 1 23 ( 23 ) x xx x x x∴ = −+−− − − −

5 The equation of a graph is 2sin 2 1yx= + for 0 x π≤≤ . (i) State the period and amplitude of .y [2] (ii) Solve 0y = for 0, x π≤≤ giving your answer in exact form. [3] (iii) Sketch the graph of 2sin 2 1yx= + for 0 x π≤≤ . [3] (iv) By drawing a suitable straight line on the same axis in (iii), find the number of solutions to the equation 2sin 2 1x= . [3] Solutions: 5 (i) period 2 2 π= π= amplitude 2= (ii) 0y = 2sin2 1 0x+= 1sin2 2x=− 1 1sin 2α − =  6 π= 0 x π≤≤ 02 2 x π≤≤ 2 ,2 66x ππππ= +− 7 11,66 ππ= 7 11,12 12x ππ= ✓ ✓

(iii) (iv) 2sin 2 1x= 2sin2 1 1 1x+=+ 2y∴= 2∴ solutions 2y = 1 3 x y 0 2sin 2 1yx= + 1− π 2 π 4 π 3 4 π 1 3 x y 0 2sin 2 1yx= + 1− π 2 π 4 π 3 4 π

6 (i) Given that 3lg( ) 2 2lg lgxy x y= +− , express x in terms of .y [4] (ii) Solve the equation 4 16log ( 2) 4 log ( 1) 1xx+− −= . [4] Solutions: 6 (i) 3lg( ) 2 2lg lgxy x y= +− 3lg( ) 2lg lg 2xy x y− += 32lg( ) lg lg 2xy x y− += 33 2lg 2xy yx  ×= 4lg( ) 2xy = 42 10xy = 4 100x y= OR 3lg 2 2lg lgxy x y= +− 3 22lg( ) lg10 lg lgxy x y= +− 2 3 100lg( ) lg xxy y =  2 3 100() xxy y= 2 33 100xxy y= 4 100x y= OR 3lg( ) 2 2lg lgxy x y= +− 32lg( ) 2 lg lgxy x y= +− 33 2lg( ) 2 lg lgxy x y= +− 33 2lg lg 2 lg lgxy xy+= +− 32 3lg lg 2 lg lgx x yy− = −− 3lg 2 (lg lg )x yy= −+ 4lg 2 lgxy= − 42 lg10 yx −= or 2 4lg10 yx −= (ii) 4 16log ( 2) 4 log ( 1) 1xx+− −= 4 4 16log ( 2) log ( 1) 1xx+− − =

4 4 4 4 log ( 1)log ( 2) 1log 16 xx −+− = 4 4 4log ( 1)log ( 2) 12 xx −+− = 44log ( 2) 2log ( 1) 1xx+− −= 4 2 2log 1( 1) x x + =− 2 2 4( 1) x x + =− 22 4( 1)xx+= − 224 8 4x xx+= − + 24 9 20xx− += (4 1)( 2) 0xx− −= 4 10x−= or 20x−= 1 4x= (N.A.) 2x=

7 ()fx is a cubic polynomial such that ( ) ( 1)( )( 3 ),fx x x m x m= +− − where m is an integer. It is given that ()fx has a remainder of 10 when divided by ( 1).x− (i) Find the value of .m [3] (ii) With the value of m found in (i), write down the expression for ()fx in descending powers of x . [2] (iii) Hence, solve the equation 32( 1) 7( 1) 4 16 0y yy+− +++= . [2] Solutions: 7 (i) ( ) ( 1)( )( 3 )fx x x m x m= +− − By remainder theorem, (1) (1 1)(1 )(1 3 ) 10f mm= +− −= 2(1 )(1 3 ) 10mm−−= 23 4 40mm− −= ( 2)(3 2) 0mm− += 20m−= or 3 20m+= 2m= 2 3m=− (N.A.) (ii) ( ) ( 1)( 2)( 6)fx x x x= +−− 2( 2)( 6)xx x= −− − 32 7 4 12xxx=− ++ (iii) Sub 1xy= + into 32 7 4 12 0xxx− ++= 32( 1) 7( 1) 4( 1) 12 0yyy+ − + + ++ = 32( 1) 7( 1) 4 4 12 0y yy+ − + + ++ = 32( 1) 7( 1) 4 16 0y yy+− +++= 11y+= − or 12y+= or 16y+= 2y =− 1y = 5y =