ACSI 2018 Promo Paper 2 ans

HL P2 2018 Section A QN Solution Remarks 1a. [Sketch of graph is optional here.] Max sunlight = 14.4h Min sunlight = 9.65h 1b. Intersection points: x = 80.2, x=262.9 No. of days = (80.23 1) (365 262.9)− + − days = 181 days OR No. of days = 364 – (263-80) = 181 days 2a Point of intersection is 11 8 24,,5 5 5 −−

2b No solution. The planes are non-parallel and do not intersect. 3a 4 3 , 3 4z i i=− + − − Use GDC to solve. 3b 4 3 , 3 4z i i=− + − − are represented by (-3, -4) and (-4, 3). 31 ( 3 4 ) 4 3z iz i i i= = − − = − 42 ( 4 3 ) 3 4z iz i i i=− =− − + = + Common mistake: 31zz = 4a cosOA = sinBA = 4b From triangle COA cos2OA OC = cos2 cosOC = cos 3cos2OC  == [Use GDC to solve.] OR 3 sin2CA = By Pythagoras Theorem, 𝑂𝐶2 = 𝑂𝐴2 + 𝐶𝐴2 22cos 3sin 2 3+= [Use GDC to solve.]

0.523599 30.0rad == Hour hand 2 60.0 = Minute hand On the clock the time is 2:05. 5a By Sine rule, sin(180 ) sin40 86 A−  = 8180 arcsin sin40 58.98 59.06A − =  = =  180 59.0 121A= − =  Common mistake: Not realising ACB is obtuse. 5b 59BCD= By Cosine rule, 2 2 26 6 2 6 cos59.0CD CD= + −     2 12cos59 0 ( 12cos59 ) 0 12cos59 6.18 CD CD CD CD CD −  = −  = = = 6a For , – 3 ≤ x ≤ 3 and for 2arcsin , – 3 ≤ x ≤ 3 Hence the domain is – 3 ≤ x ≤ 3 Or 29 xx −      3 x

Hence the domain is – 3 ≤ x ≤ 3 6b [Chain Rule and formula to differentiate arcsinf(x) ] = = = Recall: 𝑑 𝑑𝑥 𝑠𝑖𝑛−1𝑓(𝑥) = 1 √1−𝑓(𝑥)2 × 𝑓′(𝑥) 7a 3 3 11 1 x x += + [Sketch of graph is optional here.] ( 1.26, 1)P −− or 3( 2, 1)P −− 91 3 2 )9( )9(d d 2 2 1 2 2 2 1 2 xx xxx y − + − −−= 2 1 22 1 2 2 2 1 2 )9( 2 )9( )9( xx xx − + − −− 2 1 2 22 )9( 29 x xx − +−− 2 2 9 211 x x − −

7b [Purpose of graph is to show how to find the value of the gradient at a point graphically. Sketch of graph is optional here.] From GDC, ( )3'( 1.2599) 1 4.762dfx dx− = + = Angle between tangent of f(x) to the horizontal, 1 1 11 tan tan '( 1.2599) tan 4.762 78.14y fx − − −= = − = =  3 1'( 1.2599) 4.763 1 dg dx x − = =−  + Angle between tangent of g(x) to the horizontal, 1 1 12 tan tan '( 1.2599) tan ( 4.763)y gx − − −= = − = − 180 78.14 101.86= − =  Angle between the two tangents = 21 101.86 78.14 23.7− = − =  OR From GDC, ( )3'( 1.2599) 1 4.762dfx dx− = + = and 3 1'( 1.2599) 4.763 1 dg dx x − = =−  + Direction vector of the tangent of f(x) at P = (∆𝑥 ∆𝑦) = ( 1 4.762) and Direction vector of the tangent of g(x) at P = (∆𝑥 ∆𝑦) = ( 1 −4.762) . This is a difficult question and difficult to solve without the help of the GDC. Recall: Gradient = ∆𝑦 ∆𝑥 Tip: Use GDC to solve for the gradient at a point Ie. '( 1.2599) 4.762f −=

Angle between the two tan