2020_H2_Prelim_P2_Marker_s_Report

1 2020 ACJC H2 Math Prelim P2 Marker’s Report Marking Annotations used in scripts: K:Knowledge Gap; C:Carelessness; R:Read/Interpret question wrongly; P:Presentation issue Qn Solutions Comments 1(a) By Cosine rule; 2 2 2 2 2 2 2 2 2( )( )cos 2 cos 53 4 2 2 42 AB OA OB OA OB AOB AOB             a-b a b a b b b b (since length |b| is postive) Many Students have forgotten the Cosine Rule formula 2 2 2 2 cos   c a b ab 2 2 2 2( )( )cosAB OA OB OA OB AOB Some students did not realise that if angle AOB is used, c has to be the length AB. Many are able to start with 5 2cosa b=|a||b| but didn’t know how to use it properly. 1(b)(i) rk represents the length of projection of r onto the z-axis. OR rk represents the distance of point  ,,abc from the xy-plane. Some students did not portray r or k as vectors. 1(b)(ii) rk represents the area of a parallelogram with adjacent sides given by the vectors r and k. OR rk represents the perpendicular distance of point  ,,abc to the z-axis (or to the line with direction vector k). Some students used ‘adjacent sides OR and OK’ without defining what is point R & K. Students who used ‘perpendicular distance’ didn’t specify that it is from a point (a, b, c) towards the vector k 1(b)(iii) From k r p 0 0 10 p ab ba c                                From r p k 22 0 0 01 a b ac b a bc c a b                                          22 0 (1) and 0 (2) and 1 (shown) (3) ac bc ab    From (1) : 0 or 0 From (2) : 0 or 0 From (3) : and cannot both be 0, 0 ac bc a b c    Many students were able to perform cross product correctly to get to 0 0 10 p ab ba c                                but got the next cross product wrong eg. 22 ac bc ab      or 22 ac bc ab     For the last part getting to c = 0, many students didn’t support with clear explanation. O A B

2 OR From r p k , r is perpendicular to k. 0 0 0 0 0 1 rk a bc c                        From k r p 0 0 10 p ab ba c                                Substitute into r p k 22 22 0 0 0 0 1 00 00 1 1 ab ba ab ab                                                    Some students who were able to see r and k being perpendicular to one another, didn’t proceed to use the property of their scalar DOT product = 0 2(i) d d xk t x  3/2 dd 2 3 x x k t x kt c    When 0t , 21 3xc   When 10t , 16 2 74 10 3 3 15x k k      3/ 22 7 2 3 15 3xt  2/ 3 7 110xt   Many students have the misconception that the “rate of… inversely proportional to square root of…” is d1 d x xtk . A small number of students mis