ACSI 2022 Y3EXP FYE AMath P2 Solutions

FINAL EXAMINATION 2022 YEAR THREE EXPRESS ADDITIONAL MATHEMATICS PAPER 2 Solutions with Markers’ Comments 1 Given the function ( ) 322 12f x x ax bx= + +− is divisible by ( )3x− and leaves a reminder of 14− when divided by( )21x+ . Find the value of a and of b. [5] Solutions Given ( ) 322 12f x x ax bx= + +− . Factor Thm: ( )30f = Remainer Thm: ( )1 2 14f −= − ( ) ( ) ( ) 32 2 3 3 3 12 0ab+ + −= ( ) ( ) ( ) 32111 2222 12 14ab− +− +−−= − 54 9 3 12 0ab++−= 1 12 14442 ab−+−− = − 3 14ab+= − -------(1) 27ab= − -------(1) Sub (2) into (1): ( )3 2 7 14bb− += − 6 21 14bb− += − ( ) 1 21 7 5 b a ∴= ∴= −= − ( ) 322 5 12fx x x x∴ = − +− Markers’ Comments: • Some students cannot differentiate between factor theorem and remainder theorem, and just equated the two equations they got after subbing in the roots of the divisor. • Students are reminded to be careful and to check their workings. Quite a few students subbed 𝑥𝑥 = 1 2 instead of 𝑥𝑥 = − 1 2

2 2 In the diagram, the circle with centre C passes through the point ( )3, 4A − and touches the line 35yx−= at the point ( )1, 2B . Find (a) the equation of the line BC, [3] (b) the coordinates of C, [5] (c) the equation of the circle. [2] Solutions (a) Given tangent: 15 33yx= + 3BCm⇒= − Equation of the line BC: ( )23 1yx−= − − 35yx∴= − + --------------(1) (b) Midpoint of AB = ( )3 14 2, 1, 322 −+ + = − Gradient of AB, 42 1 231 2 ABmm ⊥ −= = −⇒=−− Equation of the perpendicular bisector of AB : ( )32 1yx−= + 25yx∴= + -----(2) (1) = (2): 3 52 5xx− += + 0, 5xy∴= = ( )0, 5C∴ (c) Radius of circle = ( ) ( ) 22 1 0 2 5 10− +− = units Or ( ) ( ) 22 3 0 4 5 10= −− + − = units Equation of circle: ( ) 22 5 10xy+− = or 22 10 15 0xy y+− += Markers’ Comments: • A lot of students were stuck at part b as they could not apply the property that perpendicular bisector of a circle cuts through the centre • Students are reminded that they cannot assume properties of the diagram without being able to verify it, ie AC parallel to tangent Perpendicular bisector of chord AB passes through the center of circle

3 3 Given that ( ) ( ) 324 16 13 3x x x x mgx+ + +≡ − , where ( )gx is a polynomial and m is an integer. (a) Find the value of m. [2] (b) Find ( )gx . [2] (c) Hence, or otherwise solve for 324 16 13 3 0xxx+ + += . [1] (d) Explain why the equation 6 424 16 13 3 0x xx+ + += has no real solutions. [2] Solutions (a) Let ( ) 324 16 13 3fx x x x=+ ++ . Let 3x=− : ( ) ( ) ( ) ( ) 32 4 3 16 3 13 3 3 0fx = − + − + −+=