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 = − + − + −+= ( )3x⇒+ is a factor 3m∴= − (b) By synthetic/long division: ( ) ( )( ) 234 4 1fx x x x=+ ++ ( )( ) 2 32 1xx= ++ ( ) ( ) 2 21gx x∴= + (c) For ( ) ( )( ) 2 0 32 1 0fx x x=⇒ + += 13, 2x∴= − − (d) In 6424 16 13 3 0xxx+ + += replacing x with 2x : 2 13, 2x⇒ = −− ⇒ These equations have no real solutions Markers’ Comments: • This question was not done well. Students could not tell that (𝑥𝑥 − 𝑚𝑚) is a factor of 4𝑥𝑥3 + 16𝑥𝑥2 + 13𝑥𝑥 + 3 and they tried to find the value of m by subbing in values of m, ignoring 𝑔𝑔(𝑥𝑥). Eg, Sub 𝑥𝑥 = 0, 3 = (0 − 𝑚𝑚)𝑔𝑔(0) 𝑚𝑚 = −3 • Some students did not show explicitly how to get the value of m – they should use factor theorem to verify that (𝑥𝑥 + 3) is a factor • For part (d), it seems that many students are conditioned to think that 𝑏𝑏2 − 4𝑎𝑎𝑎𝑎 can be used in any situation to show that there are no real solutions

4 4 (a) Find the values of x and y which satisfy the equations, 93xy= , ( )2 1 72 2xy+= . [5] (b) Solve the equation ( )4241 log log 8 log 3aa++= + . [4] Solutions 93xy= 233xy⇒= 2xy= ------------------(1) ( )2 1 72 2xy+= ----------(2) Sub (1) into (2) : Let 2 be :x a ( ) 22 1 72 2xx+= 21 72aa+= ( ) 272 2 2 1 0xx − −= 272 1 0aa−−= ( )( )92 182 1 0xx×+ ×−= ( )( )9 18 1 0aa+ −= (NA) 3 12 2 or 9 x −∴= − (NA) 11 or 89a∴= − 3, 6xy∴= − = − 312 =2 8 x −∴= 3, 6xy∴= − = − 4(b) ( )4241 log log 8 log 3aa++= + ( )441 log 3 log 3aa+ += + ( )44log 3 log 4aa+− = 4 3log 4a a + = 43 4a a +⇒= 434aa+= 443aa−= ( ) 4413a −= 4 31 4 1 85a∴= = − Markers’ Comments: • For part (a), quite a few students forced their way to simplify equation 2 using laws of indices. For eg, 72(2𝑦𝑦) = 144𝑦𝑦 Markers’ Comments: • For part (b), some students are not able to see that 𝑙𝑙𝑙𝑙𝑔𝑔28 can be expressed as a constant and used ‘change of base’ law which could make the working more prone to mistakes. • 𝑙𝑙𝑙𝑙𝑔𝑔4(𝑎𝑎 + 3) ≠ 𝑙𝑙𝑙𝑙𝑔𝑔4𝑎𝑎 + 𝑙𝑙𝑙𝑙𝑔𝑔43

5 5 (a) Given that ( ) 4 log 32 0log x y y x += , express y in terms of x. [4] (b) (i) Simplify ( )( ) ( )( ) 2 22 2 9 25 . 15 5 3 nn nn n ++ [4] (ii) Hence find the value of x if ( )( ) ( )( ) 2 22 215 5 3 31 9 25 nn n nn x ++ = − . [2] Solutions (a) ( ) 4 log 32 0log x y y x += ( ) ( ) 4 log log 32xx yy =− ( ) 5 log 32x y =− log 2x y=− 2 2 1yx x −∴= = (b) (i) ( )( ) ( )( ) 2 22 2 9 25 15 5 3 nn nn n ++ ( )( ) ( )( ) 22 2 2 22 35 15 5 5 3 nn nn n= +× ( ) 2 22 15 15 25 15 n nn= + ( ) 2 2 15 15 1 25 n n= + 1 26= (ii) 26 3 1x= − 9x∴= Markers’ Comments: • Students are generally inadept in using the Log Laws & changing the log base - Power Law: ( ) ( ) 4 log 4 logxx yy ≠ - Quotient Law: ( ) 4 log 32 0log x y y x += (erroneous) ( ) 4 log log 32 0xy yx − += - Correct + efficient way of changing the base: 1 loglog x y yx = Markers’ Comments: • Many students did not cancel out the common term from every term ie ( )( ) ( )( ) 22 22 2 22 35 1 ,15 2515 5 5 3 nn nnn n = ++× • Students not able to recognize: 2 2215 3 5n nn= ×

6 6 (a) Prove that cos 1 tan1 sin cos θ θθθ−=− . [3] (b) Hence solve the equation cos 1 3cot1 sin cos θ θθθ−=− for 0 360θ°≤ ≤ ° . [3] Solutions (a) LHS: cos 1 1 sin cos θ θθ−− cos 1 1 sin cos θ θθ−− ( ) ( )( ) 2cos 1 sin 1 sin cos θθ θθ −−= − cos 1 sin 1 1 sin 1 sin cos θθ θ θθ +=×−−+ ( )( ) 21 sin 1 sin 1 sin cos θθ θθ − −+= − ( ) 2 cos 1 sin 1 1 sin cos θθ θθ += −− ( )( ) 2sin sin 1 sin cos θθ θθ −= − ( ) 2 cos 1 sin 1 cos cos θθ θθ +== − ( ) ( )( ) sin 1 sin 1 sin cos θθ θθ −= − 1 sin 1 cos cos θ θθ +== − sin tan RHScos θ θθ= = = 1 sin 1 sin tan RHScos cos θθ θθθ +−= = = (b) tan 3cotθθ= 2tan 3 tan 3θθ= ⇒= ± 1tan 3 60α −= = ° 60 ,120 , 240 ,300θ∴= ° ° ° ° Markers’ Comments: • Many students ‘forced’ their answers in this proof question. Students are strongly reminded of the importance of integrity in obtaining their answers. • Common errors to note: - cos 1 sinθθ≠− - sec 1 tanθθ≠− - 13cot 3tanθ θ≠ • Quite a few students started the proof with both sides of the given equation. Students are reminded to start on one side of the identity (usually the more complex side), and use logical step