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 misread the question without “root”, thus wrote 2 d d xk tx . Majority did not notice “x, in hundreds”, and used 100x and x = 400 to calculate for the unknown constants. Many students did not make x the subject even though the question stated “x as a function of t ”. Some expressed t in terms of x instead. Misconceptions of law of indices led to answers such as 3/ 2 7 1,10    xt 2 3 7 110xt 2/3 7and 1. 10   xt 2(ii)     2d 100 100d x m x x mx xt     Majority are able to write

3   1 dd100  x m txx Using partial fraction, 1 1 1 dd100 100  x m txx 11 d 100 d100 ln 100100 x m txx x mt cx         100 100 100 100 100 100 100 e , where e100 e 100 1 e 100 e 100 e 1e 100 1e 100           mt c mt mt mt mt mt mt x AAx x A x x A A Ax A x A H   1 d d and100  x m txx simplify the integral either using completing the square or by partial fraction. Some did the integration wrongly by writing   22 2 11 dd100 100   xxxx xx   2 11or d d100 100   xxx x and attempted to use the formula 22 11 d ln .2  axxCa x a a x A few made the careless mistake of re-writing the integral as   2 11 dd100 100  xxx x x . Many students did not realise that 100 0x , where modulus is necessary. Some students stopped at this step and could not proceed on. 100 100 100 e .1e  mt mt Ax A 2(iii) The scientist’s model. The scientist’s model suggests that in the long term, the number of bacteria increases and tends to 10000. In a petri dish, there should be a limit to the maximum population of bacteria. Many students were able to give the correct answer. There were many interesting answers given but did not fit the context of the question. Eg. The scientist since the scientist is smarter than the student. 3(a)(i) Using similar triangle, 2 2 yx a x a 2 2axy xa  Area of S 1 2 xy 2 12 2 axx xa    2 2 ax xa  (shown) Most students attempted to express y in terms of x. A number of students used similar figures, 2 11 22   Al Al , and were successful in showing. Some students attempted to use the equation of a straight line   22   yY a X a x , where x and X got cancelled out when simplifying, which is incorrect. 3(a)(ii)         2 2 2 2222 22d d x a ax ax ax x aS x x a x a      Many students did long division to simplify S, which is not necessary.

4 For maximum or minimum, d 0d S x      22 22 2 0 x a ax ax xa    23 2 0ax a x   2 2 0x x a 0x  (reject since x > 0 ) or 2 2xa           22 3 2 3 22 42 2 2 2 2 2d d x a ax a ax a x x aS x xa                222 5 3322 2 2 2 2 a x a x a ax a x a x a      When 2 2xa ,   25 32 22 d 2 2 d 2 Sa xa aa   > 0 Hence by the second derivative test, S is minimum at 2 2xa . Alternative for checking minimum: Using first derivative: x   22a  22a   22a  d d S x - ve 0 + ve When 2 2xa , 220xa , and since 0ax and   22 0xa ,     2 22 2d d ax x aS x xa    < 0 Similarly, when 2 > 2xa ,     2 22 2d d ax x aS x xa    > 0 Hence by the first derivative test, S is minimum at 2 2xa . Minimum value of     2225 3 22 22 2 4 4 2 aaax aSa x a a aa      Majority found d d S x correctly and equated to zero. However, when simplifying, there were many algebraic errors, resulting in the wrong value of x found. Almost all students did the “max. or min.” check using either 1st or 2nd derivative, but very few successful ones. Students who used 2nd derivative test, majority did not know that they have to find the value of 2 2 d d S x when 2 2xa . Most of them left the expression as       222 32 2 2 2    a x a x a ax xa and tried to explain that it is > 0. Students who did 1st derivative test, very few explained why d d S x < 0 when   22  xa etc, thus get the credit. Students who simply drew the table without explaining why d d S x < 0 or d d S x > 0, got penalized. Note: It is not encouraged to use 1st derivative test when the value of x contains unknown constants. Most of the students did not realise that they are required to find the value of S at 2 2xa , as stated in the question. 3(b) 3 1 1 2 3 1 1 1 1 1 iPiR R R R R     This question was badly done. A common mistake is

5 1 2 3 d 1 d 1 1 1 dd Pt R t R R R             12 2 2 2 12 d d d1 1 1 d d d P P R R R t t tR R R                          2 2 2 12 d12 d P P R rr tR R R          2 22 12 d 2 d P P R rr Rt RR     