NYGH 2015-S3EOY-IM2 ans

2015 Sec 3 IM2 EOY Mark Scheme 1a (6 – x)2 > x 36 – 12x + x2 > x x2 – 13x + 36 > 0 (x – 9)(x – 4) > 0 ∴ x < 4 or x > 9 1b α + β = 3, αβ = 5 βα 12 − + αβ 12 − βαβα 1122 −−+= ( )       +−−+= αβ βααββα 22   −−= 5 3109 5 8−=    −      − αββα 11 22 ( )      +−−= αββααβ 12 ( ) ( )      ++−= αββααβ 12   +−= 5 1325 5 111= New equation: 05 111 5 82 =++ xx 5x2 + 8x + 111 = 0 2a e2x – 7(ex) – 2 = 0 (ex)2 – 7(ex) – 2 = 0 2 )2)(1(4497 −−±=xe 2 577+=xe or 2 577− (NA) ∴ 2 577ln +=x or 1.98 2b yy 3 3 3 log29log log =− yy 33 log2log2 1 =− 2log2 1 3 −=y 4log3 −=y ∴ y = 81 1

3a 2f(x – 5) = 2x2 – 16x + 32 f(x – 5) = x2 – 8x + 16 = (x – 4)2 = (x + 1 – 5)2 ∴ f(x) = (x + 1)2 OR Reverse 2nd transformation: y = (2x2 – 16x + 32) ÷ 2 = x2 – 8x + 16 Reverse 1st transformation: y = (x + 5)2 – 8(x + 5) + 16 = x2 + 2x + 1 = (x + 1)2 3bi 3bii 1 e 1 3 += − xx x – 1 = ex– 3 ln (x – 1) = x – 3 Draw the line y = x – 3 ∴ number of solutions is 2 4 27x ÷ 3y = 9  33x ÷3y = 32  3x – y = 2 ----(1) 22x × 41–y = 64  22x × 22–2y = 26  x – y = 2 ----(2) (or 2x – 2y = 4) (1) – (2): 2;0 −== yx 5i 84000 5ii 84000e3k = 42000 e3k = ½ 3k = ln ½ ∴ 2 1ln3 1=k (or –0.231) 5iii 84000 ekt = 16800 x=1

ekt = 0.2 kt = ln 0.2 ∴ t = 6.97 (3sf) 6a (x + 4)2 + (x + p)2 = 8 2x2 + (8 + 2p)x + (8 + p2) = 0 Eqn has real roots  Discriminant ≥ 0 (8 + 2p)2 – 4(2)(8 + p2) ≥ 0 –4p2 + 32p ≥ 0 p2 – 8p ≤ 0 OR –p2 + 8p ≥ 0 p(p – 8) ≤ 0 p(8 – p) ≥ 0 ∴ 0 ≤ p ≤ 8 b x2 – 4x + c = x2 – 4x + 4 + c – 4 = (x – 2)2 + (c – 4) c – 4 = 3 ∴ c = 7 7a 7b 8i Max = 0, Min = – 2 8ii 4π y = g(x) 2 –1 (2, –2) y x 0 (3, 1) (1.5, 2) (1, –1) 3 (–0.5, 1) y = g(2x) – 1 y = h–1(x) y = h(x) 2 1 (2, –1) (–2, 3) y x 0 (–1, 2) (3, –2) y = x 1 2

8iii 1 8iv 12sin −=   x 2 3 2 π=x ∴ x = 3π 8v 9ai ii iii tan θ – sin θ sin θ 9b tan x + 2 (1 + tan2 x) – 5 = 0 2 tan2 x + tan x – 3 = 0 (tan x – 1)(2 tan x + 3) = 0 tan x = 1 or tan x = 2 3− Ref. ∠ = 4 π or 0.98279 4 5,4 ππ=∴ x , 2.15 or 5.30 10a LHS = A A A sin cos sin 1 − = A A sin cos1−

= A A A A cos1 cos1 sin cos1 + +×− = ( )AA A cos1sin cos1 2 + − = ( )AA A cos1sin sin2 + = A A cos1 sin + = RHS (shown) 10bi A in 2nd quadrant, hyp = 5 5 4cos −=A 10bii B in 3rd quadrant, opp = 12 5 12tan =B 10biii BABecA cos 1 sin 1seccos ×=   −  = 5 13 3 5 3 14or 3 13 −−= 11i xx =+12 3 2x2 + x = 3 2x2 + x – 3 = 0 (2x + 3)(x – 1) = 0 ∴ x = 2 3− or 1 11ii f2(x) = f( 12 3 +x ) = 112 32 3 +   +x = 12 126 3 + ++ x x = ( ) 72 123 + + x x or 72 36 + + x x f2 ( ) 72 123: + + x xx  where 2 1,2 7 −≠−≠ xx 11iii g(x) = x2 – 3x + 1

= 4 912 3 2 −+   −x = 4 5 2 3 2 −   −x Turning point is    − 4 11,2 11 The range of g is g(x) ≥ 4 11− 11iv Max. k = 2 11 Let y = g(x) Let y = g(x) Let g –1(x) = y 4 5 2 3 2 −   −= xy y = x2 – 3x + 1 x = g(y) 4 5 2 3 2 +=   − yx x2 – 3x + 1 – y = 0 4 5 2 3 2 −   −= yx 4 5 2 3 +±=− yx 2 )1(493 yx −−±= 4 5 2 3 2 +=   − xy 2 3 4 5 ++±= yx 2 453 yx +±= 4 5 2 3 +±=− xy g –1(x) 2 3 4 5 ++±= x g –1(x) 2 453 x+±= 2 3 4 5 ++±= xy Since g –1(x) ≤ 2 11 , g –1: x 2 3 4 5 ++− x 4 5, −≥x or 2 453 x+− Bonus qn ababab ab abab logloglog 1 log 1 −=− )1(log)1(log +−+= ba ab ba ab loglog −= ab ba log 1 log 1 −= 2 log 1 log 1       −= ab ba            −      += abab baba log 1 log 14log 1 log 1 2 4293−= = 17