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] Sol