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 tangents = cos−1 ( ( 1 4.762)∙( 1 −4.762) |( 1 4.762)||( 1 −4.762)| ) = 23.7° OR Note: Let 1 be the angle between the tangent of f(x) and the vertical. And let 2 be the angle between the tangent of g(x) and the vertical. From GDC, ( )3 1 4.762d xdx += [A1] 3 1 4.763 1 d dx x  =− + [A1] Required angle = 12 11arctan arctan arctan arctan ( ) ( )ddf x g xdx dx + = + [M1] 11arctan arctan4.762 4.763=+ − [M1] 11.86 11.86 23.7= + =  [A1] 8a Asymptotes at: 2 2 0 2 1.41x x x− =  =   =  [Since denominator = 0 at vertical asymptote] 4y = [Since | 4𝑥2+8𝑥 𝑥2−2 | = |4 + 8𝑥+8 𝑥2−2| → 4 as 𝑥 → ∞.] A straight forward question that is done well. From the graph, Intersections at: x = -7.16, -1.66, -0.838, 0.517 (3 s.f) 15 7.16, 1.66 0.838, 0.517 15, 1.41x x x x−  − −  −    Most didn’t remember to exclude the vertical asymptote.

Section B QN Solution Remarks 9a ( ) arctan( 3 ) sin( ) arctan3 sin f x x x x x x x − =− − − + − =− + − ( )arctan3 sin () x x x fx =− − + =− OR From graph, f(x) is symmetrical about the origin. Note: A function that is not even does not imply that it must be odd. Recall: Odd function: f(x) = -f(-x) Even function: f(x) = f(-x) 9b The inflexion points are when f’’(x) = 0. OR By differentiation, 22 54''( ) sin (1 9 ) xf x x x =− + + Equate ''( ) 0fx = , then 22 54sin (1 9 ) xx x = + . Recall: Inflexion points for functions of degree ≤ 3 are at f”(x) = 0 OR at the turning points of f’(x). Tip: Use GDC to graph f”(x) by using the 𝑑2 𝑑𝑥2 ( ) template. Then find the zeros to solve f”(x) = 0.

x-coordinates of inflexion points: x = -0.8719, 0, 0.8719 Coordinates: (-0.872, -0.432), (0,0), (0.872, 0.432) 9c OR 22 3'( ) 1 cos (1 9 ) f x x x = + − + Since '( 0.872) 0, '(0) 0, '(0.872) 0f f f−    , they are all non-stationary points of inflexion. Recall: Non-stationary point, x occurs when the gradient f’(x) ≠ 0. Hence non-stationary inflexion point, x occurs when f’(x) ≠ 0 and f”(x) = 0.

9d f(x) is concave up when f’’(x) > 0. Hence, 0.8722 x−  − or 0 0.872x . Recall: The inflexion points mark a change in curvature. f”(x) > 0, concave upwards f”(x) < 0, concave downwards 10a 22112 2 sin222 2 sin2 S r r   =−  =− Question is well done. 10b ( ) 212 sin2 22Sr   = − = = 2 sin2 2 −= θ=1.15 rad Use the GDC to solve. 10c Note: Since both circles are equal and the diagram is symmetrical, the problem can be simplified by just considering one half of the figure. Ie. Consider triangle ABD expanding as X approaches A. x

Since AC decreases at a rate of 2 cm/min, 1dAX dt =− From triangle ABX, cos 1 adj AX hyp == Differentiating with respect to time cos 1 d d AX dt dt  = sin d dAX dt dt −= ( )11 1sin sin d dt   −=  − = OR cos cos 1 adj AX AXhyp= =  = sindAX d  =− 11 sin sin dAX d dt dAXdt d    −= = = − 10d ( )2 sin2S =− ( ) ( )2 2cos2 2 1 cos2dS d  = − = − ( ) 12 1 cos2 sin dS dS d dt d dt  =  = −  ( ) 12 1 cos2 sinr  = −  21 1 2sin2 4sinsin   −+== Done well. Recall: cos 2𝛼 = 1 − 2𝑠𝑖𝑛2𝛼 11a At the x-z plane, y = 0 12 00 1 tx tt tz +       − =  =      −+    Coordinates of the point: 1 0 1   − Done well. 11b 21 12 11 k        −       −    Hence the lines are not parallel. Remember to check if the lines are parallel. Note: There is no need to find the point of intersection of the question doesn’t specifically asks for it.