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 2 2 2 2 arcsin 4 & 2 2 1[ 1 2 1 ]         =  = = = + =  = Therefore the area is −1 8 sq units [Shown] Some could not figure out the angle as 45 degrees and could not show the relevant step in the proof. 9(a) Using the graph drawn in GDC, the graph y axis is the line of symmetry, ie, symmetrical about the y axis hence it is an even function. Or ( ) cos( ) arcsin 5 xf x x −− = − − cos arcsin ( ) 5 xx f x= − = Some students did not show the graph or the algebraic working to show that it is even. (b) Quite a number of students are not able draw the correct graphs. Either completely wrong or only the right side is correct. Some will do their own differentiation first but they are not correctly done. (c) the x coordinates are -4.39, -1.59, 1.59 and 4.39 Some give

coordinates but out of those who give the coordinates, some gave wrong y values. (d) 4.39 1.59 & 1.59 4.39xx−  −   Some gave wrong range as they use f’(x) not f”(x). 10(a) If &OA OB OC⊥ therefore 1 1 1 2 2 0 & 1 0 1 1 4 aa b −               • = • =              −−        2 4 0 2aa+ − =  = 1 4 0 3bb− + − =  = Alternative 1 2 8 1 2 1 4 2 5 3 & 2 4 5 1 b b k a k b a b −−                = + =  =  = =              −−        Many did cross product and some made mistakes. They did not realise they could just use OA instead. With the mistakes error carry into other parts. (b) the eqn of the plane 1 :2 1 d   =− r 11 1 1 2 12 11 d        ==      −    Therefore 1 : 2 12 2 12 1 x y z   =  + − =− r Generally method is correct but accuracy not there. Some students gave wrong form, vector eqn form. (c) Eqn of line AC: 1 2 1 1 1 2 1 2 , 2 1 1 4 1 1 5 t t t −               = + −  = + −               − + −        rr 1 4 2 1 5 12 1t t t t+ + − + − =  =− Pt of intersection is P(0, 3, -6). Generally students get the idea but they are not presenting the working accurately. Did not state . Some students gave wrong form, position vector form. t

(d) 1 0 1 2 3 1 1 6 5 PA            = − = −          −−      A Distance, d cosPA = P AP= n n since cosPA PA =nn = 11 12 51 6        −      −    = 6 units Or OA= 6 units Distance of the Plane from the origin = 12 6 units Need to check if A and the plane is the same side as O. 11 2 2 6 11         =      −−    [ie: an =d] Since d=6 same side as the eqn of plane, then same side. Hence distance is 12 66 6 −= Common Method: Line passing through A perpendicular to Plane 1 2, 1 tt  =− r Substitute into plane: 2 12 2 t t t t + − = = Point of intersection F(2, 4, -2) 2 1 1 4 2 2 2 1 1 AF            = − =          − − −      1 4 1 6AF = + + = A lot of students assume PA is the distance. Those who use the alternative distance method, did not check if the point and plane are on the same side of O.

(e) x = 17 2 & 4y = 2z + 7 Many students did not give the correct form, they might have given but final gave vector form which reflect clearly that they do not know their definition well. 11(a) k = - 1 Ok except some either gave k=1 or x=-1. (b) Asymptotes x= - 1 and y = 0. Many students could not give complete information. Either not able to give correct x or y- intercepts or y = 0, x=-1. (c) 23 14 2( 1) xxx ex −− =+ 23 14 2( 1 ) xx x x e− − = + 23 14 2( 1 ) 2( 1 ) 2( 1 ) xx x x x e x− − + + = + + + 23 12 2( 1 ) 2( 1 ) xx x x e x+ − = + + + 23 12 2( 1)( 1) x xx xe +− =++ Many students are not able to manipulate the algebra and hence could not get the straight line. Some did not simplify but the GDC will still draw the y=2. But

Draw y = 2 x = - 1.98 students will be penalized. (d) (i) - 2.17 < x < -1 or x > 1.84 (ii) - 2.17 < tan θ < -1 or tan θ > 1.84 2< θ < 2.36 or 1.07< θ < 1.57 Alternative answer: 2< θ < 3 4  or 1.07< θ < 2  (i)Ok if they draw the graph correctly. (ii) not many draw the graph hence a number of the students did no get the inequality right, the upper bound was not given.