RVHS 23 Prelim H2MA P2 Std Soln

River Valley High School 1 2023 H2MA Prelim Paper 2 1 Solution [5] Complex Numbers P2 Q1 (i) 12 12 22 Distance between and 4 2 2(4)(2) cos - Using Cosine Rule33 28 zz zz          Many students converted the z=r form of the complex number to the cartesian form z=x+iy first, before plotting out the points on the Argand Diagram. This probably shows that the student is not conversant with the z=r form. Students must be familiar with all forms of complex numbers given by a question. The r from the z=r form must not be seen as just a value obtained by applying a formula. Students ought to know that it measures the “distance” of complex number z from the Origin. Similarly, the represents the angle measured from the Real axis. Majority of students do not know how to handle complex numbers ( suggesting that students lack preparation for linking complex numbers on argand diagrams to simple Vectors addition/subtraction concept. (ii) 1 12 2 zz az a z   Same thing, many students solved for a by converting everything to cartesian form, not knowing that they could have just made a EM BE D Eq uati on. DS MT 4 EM BE D Eq uati on. DS MT 4 E M B E D E q u at io n. D S M T 4 E M B E D E q u at io n. D S M T 4 E M B E D E q u a ti o n . D S M T 4 E M B E D E q u a ti o n . D S M T 4

River Valley High School 2 i 3 ii33 i 3 2i 3 4 2 2 2 eae e e         Either: 1z is the scaling of 2z by factor of 2 and rotating 2z 2 3  anti-clockwise about the Origin, Or: 2z is the scaling of 1z by factor of 1 2 and rotating 1z 2 3  clockwise about the Origin, the subject immediately and working with z=r form. Some students do not understand the phrase “geometrically… relationship…”, leaving this part blank. For those who understand, tried answering using phrases such as “reflection”, “symmetry”, “half of the complex number”, etc. Note that students should use words such as suggested by the solution.

River Valley High School 3 2 Solution [6] P2 Sequence (i) When 0n , 00 10 4 1 2 3 3 3u u Q              00 4 4 1 233 3 3 3 Q             81 333Q   and 11 21 4 1 1 2 3 3 3 3uu              2 4 1 1 2 4 2 3 3 3 3 3 3u              Many students made arithmetic mistakes while solving for Q and . No evidence to show that students do not understand how to start this part of the question. (ii) 1 4 1 1 2 3 3 3 3 nn nnuu               1 1 1 2 1 2 2443 3 3 3 3 3 n n n n nnuu                           1 52 33 n nnuu   Thus 1nnuu   when 52 33 n   23 35 n   3ln 5 2ln 3 n   0n can be 3ln 5 2ln 3       or celling of 3ln 5 2ln 3   . ___________________________________________ For the first half of the question, most students do not know what the question is asking for. The question can be rephrased as “...show that there can be a maximum value (indicated by the inequality sign), in which | | is bounded to. Subsequently, many students went on to solve for the second half of the question by using GC and listing out integer values which the inequality is satisfied.

River Valley High School 4 When 0.001  , 3ln 5 2ln 3 n   18.29 0n can be 19.

River Valley High School 5 3 Solution [7] Transformation of graphs (i) Graph of  fyx for 14 x   Some students still do not know that for each piece of graph, the coordinate of end-points must be clearly indicated. Otherwise, most students scored full credit for this part of the question. (ii) Graph of 1f1 2yx  , for 14 x   The end-points are at ( 1, 1) and (4, 0.5) Note: When 4x  1 1 3 1f 1 f 2 1 (2) 12 2 2 2yx         Many students show understanding for the transformation required for this question (scaling in x and translation in y). However, many students can be observed to use inappropriate approaches (such as translating and scaling each piece of graph), when they could have actually just transformed their original graph and cut off t he graph at end-points. (b) Scale parallel to y-axis by factor 3. Translate 2 units in the negative x-direction. Reflect about the y-axis. OR Reflect about the y-axis. Translate 2 units in the positive x-direction. Many students have difficulty handling the index (that is, the index 3) in the logarithm, suggesting transformation steps such as scaling by , etc. Some students used transformation phrases such as “flipping” when they mean reflection, etc.

River Valley High School 6 4 Solution [10] P2 3D Vectors (a) 1 01 : 0 1 , 30 l                     r Let 0 0 3 OB   . Given that 1 0 4 OA   0 1 1 0 0 0 3 4 1 AB OB OA                                  Let 1 1 1 1 0 1 1 1 0 1                              n 1 0 1 1 0 1 1 3 1                              r 1 1 :r 1 3 1     1:3x y z    Generally well done. Some mistook the dot product form or the parametric form as the cartesian equation. Some included point P in their calculation of the equation leading to a wrong answer while some used Vector OA in the cross product to find normal which leads to the w rong answer. (ii) Q is the foot of perpendicular of P on 1l 01 0 1 for some 30 OQ                   0 1 3 0 1 4 3 0 5 PQ OQ OP                                  31 41 20 PQ                     PQ perpendicular to the line 1l 0PQd Many are able to do this part using the zero do t product method or the projection method. Many who used the projection method m isused the vector OP in their projection vector which gives a wrong answer. The projection should be from a point on the line to point P projected on the line. Some answers use d the idea that the line is in the plane so Q is the foot of the perpendicular from P to the

River Valley High School 7 3 1 1 4 1 1 0 2 0 0                                1 2 0   1 2  0 1 0.5 10 1 0.523 0 3 OQ                              plane which is not point Q but rathe point N found in part (iii). (iii) 2l parallel to 1 21 is perpendicular to  dn 21 =0dn 31 1 1 0 1 3 1 0 2 m m m                         _______________________________________________ h distance between 2l and 1 h distance between ( 3,4,5)P  and 1 1 :r 1 3 1     Let N be the foot of perpendicular of P on 1 . 31 : 4 1 , 51 PNl                     r --- (1) 1 1 :r 1 3 1     --- (2) To find N, sub (1) into (2): 3 1 1 4 1 1 3 5 1 1                                   Many able to find m but there are some who did not realise the fact that a line being parallel to the plane means the line is perpendicular to the normal of the plane. Many mistook point Q as the foot of the perpendicular from P to the plane. Many also mistook the length as th