ACSI 2023 Y3EXP FYE AMath P2 Solutions

1 FINAL EXAMINATION 2023 YEAR THREE EXPRESS ADDITIONAL MATHEMATICS PAPER 2 SOLUTIONS 1 It is given that a and b are the roots of the quadratic equation 2 2 10xx− −= and that ab> . Show that 3 22a b = −− . [5] ( ) ( ) ( )( ) ( ) 2 2 2 41 1 21x −− ± − − −= 28 2 ±= 222 122 ±= = ± 1 2 and 1 2ab∴=+ =− 12 12 1 21 2 a b ++= × −+ 122 2 322 12 1 ++ += = −− 322= −− 2 A circle with centre O passes through the points ( )1, 7P − and ( )0,8Q . (i) State the relationship between the perpendicular bisector of PQ and the point O. [1] (ii) Find the coordinates of O, given that the line 22yx= − passes through the centre of the circle.[5] (iii) Hence find the equation of the circle. [2] (i) The perpendicular bisector of the chord PQ passes through O, the centre of the circle. (ii) ( ) 87 101 PQm −= =−− 1m⊥⇒= − Midpoint of PQ 10 78,22 −+ += 1 15,22 = − Equation of perpendicular bisector of PQ : 15 1 22yx  − = − −−   MARKER’S COMMENTS: Students should infer from the question that exact answers are required. Therefore, they should not be relying on non-exact roots from the calculator. COMMON MISTAKES: 1) The perpendicular bisector is the radius. No, the perpendicular bisector is a continuous line. 2) A line drawn through point O is perpendicular to the perpendicular bisector. No, there are so many ways to draw this line. COMMON MISTAKE: Finding the gradient of the line perpendicular to 22yx= − .

2 7yx∴= −+ ------------(1) 22yx= − -------------(2) Sub (1) into (2): 72 2xx−+= − ( )3 ; 2 3 2 4xy∴= = −= ( )3, 4O∴ (iii) Radius of circle ( ) ( ) 22 3 0 4 8 5 units= − +− = Equation of circle: ( ) ( ) 22 3 4 25xy− +− = 3. The polynomial ( )fx is given by ( ) 329 30 23 4fx x x x=− −− (a) Factorise ( )fx completely. [4] (b) Hence, prove that the equation 9 23 4 30xx x x−= + has only one real root. Find the solution. [3] (a) By Trial and Error Let 4:x= ( ) ( ) ( ) ( ) 32 9 4 30 4 23 4 4 0fx = − − −= ( )4 is a factorx⇒− . 9 30 23 4 4 36 24 4 9 6 10 −−− ( ) ( )( ) 249 6 1fx x x x=− ++ ( )( ) 2 43 1xx= −+ (b) For ( ) 0fx = , 14, 3x= − 9 23 4 30xx x x−= + 329 30 23 4 0xxx⇒ − − −= Replace “ x ” with “ x ” (NA) 14 or 3x⇒= − 16 one real rootx∴= ⇒ MARKER’S COMMENTS: • Students should remember to write all their coordinates in brackets. • Some students used an alternative method by equating the distance from P and Q to the centre O. This is not recommended as it is highly tedious and likely to lead to careless mistakes. COMMON MISTAKE: Using the midpoint of PQ to find the radius. This point is not even on the circle. MARKER’S COMMENTS: Factor theorem working was not clearly shown by many students, who likely just used their calculator to get the relevant roots.

3 __________________________________________________________________________________________ 4. (a) Solve the equation 12 23xx+− += . [4] (b) The equation of a curve is 224y x ax b=−+ where a and b are constants. Explain why 0y> if 22ba> . [3] (a) 1223xx+− += 122 3 2 x x⇒ ×+ = Let 2xa= : 123a a+= 22 3 10aa− += ( )( )21 10aa− −= 1 or 12a∴= 1022o r 22xx −⇒= = 1 or 0x∴= − (b) 2 4b ac− ( ) ( )( ) 2 4 42ab= −− 216 8ab= − ( ) 282 ab= − If ( ) 282 0ab−< 220ab−< 22ba⇒> 22 2 When 2 , 4 0. And since 2 0, 0 when 2 b a b ac a y ba ∴ > −< => ∴> > __________________________________________________________________________________________ COMMON MISTAKES: • Instead of squaring x to get x, many students applied square root again. • Many students rejected by claiming that you cannot square root a negative number. In this context, the rejection reason should be that a surd cannot be negative ( in this case, the surd is a square root; the square root of a non-negative real number is a non-negative real number, while the square root of a negative real number is a complex number). There was also no need to waste time substituting all the roots back into the original equation to check. MARKER’S COMMENTS: • Students need to improve on their indices laws. • There was no need to use logarithm in this question. • The factorisation step for solving quadratic equations must be clearly shown. COMMON MISTAKES: • Misconception that 2 4b ac− > 0. • Forgot to include the condition a > 0. • Substituting before finding discriminant. •

4 5. (a) Find an expression for x, in terms of e, for which ( )ln 3 ln 3xx−= + . [3] (b) Solve the equation 5 25log log 4xx+= . [3] (a) ( )ln 3 ln 3xx−= + ( ) 3ln 3 ln lnxxe−= + ( ) ( ) 3ln 3 lnx xe−= 33 x xe−= 33 xe x= + ( ) 331 xe= + ( ) 3 3 1 x e ∴= + (b) 5 25log log 4xx+= 2 25 25 3 25 log log 4 log 4 xx x += = 34 25x = ( ) 1 4 325x= 73.1x∴= COMMON MISTAKES: • • • • • Expressing e in terms of x. MARKER’S COMMENTS: • When using shortcuts to the change-of-base law, many students get confused and make careless mistakes • Students are required to simplify their answers to 3s.f. as instructed on the cover page and not leave final answer as .

5 6. (ai) Prove the trigonometric identity: ( ) 2 2 2 cosec 2cot cosec cos sin AA A AA + = + [4] (aii) Hence solve the equation: ( ) 22cosec 2cot 4 cos sinA A AA+= + for 0 360A≤≤ ° . [4] (ai) LHS: ( ) 2 2 cosec 2cot cos sin AA AA + + ( ) 2 2 1+ cot 2cot cos sin AA AA += + ( ) ( ) 2 2 1+ cot cos sin A AA = + ( ) 2 2 cos1+ sin cos sin A A AA = + ( ) 2 2 sin cos sin cos sin AA A AA += + ( ) ( ) ( ) 2 22 sin cos 1 sin cos sin AA A AA += × + ( ) 2 1 sin A = 2 (proven)cosec = RHS A= Alternatively, LHS: ( ) 2 2 cosec 2cot cos sin AA AA + + 2 (proven)cosec = RHS A= (aii) ( ) 2 2 2 cosec 2cot 4 cos sin cosec 4 AA AA A + = + = 2 1 4sin A = 2 1sin 4 1sin 2 A A = =± 1 1sin 302α −= = ° 30 ,150 ,210 ,330A∴=° ° ° ° MARKER’S COMMENTS: • All working must be clearly shown in a proving question. Quite a few students skipped from: COMMON MISTAKE: • Did not apply identities correctly when expanding:  MARKER’S COMMENTS: • Students must clearly write the degree symbol as otherwise the angle will be treated as radians. COMMON MISTAKE: • Forgot to take .

6 6b. Given that 1 sin 2cos 11 2sin cos xx xx ++ =++ and x is acute, find the exact value of cos x. [4] 6b. 1 sin 2cos 11 2sin cos xx xx ++ =++ 1 sin 2cos 1 2sin cosx x xx++ = + + cos sinxx= sin 1 tan 1cos x xx = ⇒= 1tan 1 45 45x α −= = ° ∴= ° 1cos 2 x∴= _________________________________________________________________________________________ 7. (ai) Write down the first four terms in the expansion of ( ) 6 14 x− . [2] (aii) Hence find he coefficient of 3x in the expansion of ( )( ) 623 14xx+− . [2] (ai) ( ) ( ) ( ) ( ) 6 123 66 6 12 31 4 1 4