ACSI 2019 Promo Paper 2 Ans

1. B A C Using cosine rule 2 2 23 6 2(6) cos18AB AB= + −  2 12cos18 27 0AB AB−  + = AB = 3.35 or 8.06. Surprising numbers of students could not find the additional answer or did not attempt to do so. 2a z = *z = 1.41 or 2 from GDC arg(z)= - arg(z*) = 2.88 or 011 or 16512  from GDC Quite a number of students used long tedious way to find the modulus and argument when this could be found more simply using the GDC. Several students misinterpreted the question and found the wrong argument. (b) arg(-z) = arg(-1)+ arg(z) = 2.88 6.02 6.02 2+ = = − = -0.263 or 0 or 1512 −− 3. (a) x=1, y=0, z= -1 Generally ok. When using GDC, the workings must be shown or else marks deducted. (b) Based on the result of RREF above, the system has no solutions.

4(a) k = 10. Generally ok. (b) Max H=227m when cos( ) 15 t  =− Min H = 1m when cos( ) 15 t  = (c) Duration = 6.12 - 3.88 = 2.24 min 5(a) 19 5 11628C = Good. (b) 5 14 14 5005CC= [1 adult from 5 to choose and 4 remaining children to choose from 14] 6(a) If cos and sinP R y Q R y== 2 2 2 2 2 2cos and sinP R y Q R y = = 2 2 2 2 2 2 (cos sin )P Q R y y R+ = + = Or a suitable diagram to show proof Generally well done. Its possible to show diagrams as part of proof. (b) sin tancos Q R y yP R y== arctan Qy P =  Some used inverse trigonometric functions and could not find the proof or steps are unclear.

(c) cos sin cos cos sin sin cos( ) P x Q y R x y R x y R x y + =+ =− Generally good. (d) 3cos 4sin 5cos( 0.927)x x x+ = − This implies 5cos( 0.927) 5x−= ( 0.927) 0x−= = 00.927 or 53.13x Not well done. Students could not link the previous parts to solve this part. Generally weak in trigonometry 7(a) Largest possible domain is 4x  Alternative answer: ] ,4]− . Some give the wrong domain x < 4. (b) Some did not plot the correct inverse function or the correct intersections. Some did not label all the intercepts and stationary point. (c) Draw y = f ’ (x) 0.863x  Alternative answer: ] ,0.863[− Depending on previous part, usually if graph is correct, they could find the intersections. Some give the wrong domain or wrong inequality sign. 8(a) 2DC  = 2DC = [using similar triangles] arcsinAOB = Mostly ok.

Area = 211(2 )(2 ) (1) arcsin22    − =  − 12 arcsin2 sq units [SHOWN] (b) If