2020 H2 Prelim P1 Marker s Report

1 2020 ACJC H2 Math Prelim P1 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(i) Given u = 2x – 1. Then 1 ( 1)2xu and d1 d2 x u  .       2 1 2 2 22 1 21 2 1 2 2 1 1 2 12 12 d 1 (2 1) ( 1) 1 d21 11 dd4 11 11 2 1 d sin42 111 sin42 1 sin 14 11sin (2 1) 1 (2 1)44 where is an arbitrar x x x u u u u uu uu u u u u C u uC u u C x x C C                                              y constant. The common mistakes are  Changing ‘dx’ to ‘2du’  When applying of the formula 1 f ( )[f ( )] d [f ( )] ,1 n n u u u u Cn      for the 1st integral, the negative sign or ½ in the denominator is left out.  The result of the 1st integral 2 d 1 u u u becomes 2 3/ 2(1 ) 3/ 2 u or 2ln 1 u Some students are able to integrate correctly but do not substitute u as 2x – 1 and add the arbitrary constant C in the final answer. 1(ii) 1sin (2 1) dxx  1 2 dsin (2 1) 1 d d2 d 1 (2 1) vux x u vxx x      1 1 2 sin (2 1) d 2sin (2 1) d 1 (2 1) xx xx x x x          The part is badly done. Not many students are able to see this is a question on ‘Integration by Parts’. Some students are able to choose ‘u’ as sin-1(2x – 1), but fail to differentiate it correctly.

2 1 1 2 12 11sin (2 1) sin (2 1) 1 (2 1)22 11 sin (2 1) 1 (2 1)22 where is an arbitrary constant. x x x x C x x x C C                   2(i) Note: Do indicate coordinates of points when question requires it. Many students do not indicate the coordinates of the x- intercepts and the equation of the three asymptotes as required in the question. The common mistakes are  The asymptote y = 3 still remains in the graph  The maximum turning point at x = 2.5.  The asymptote y = 1 is missing 2(ii) Note: Do indicate coordinates of points when question requires it. Generally well done. However, the parts of the curve when x  are not properly drawn. The curve should get closer and closer to the asymptote y = 1/3 or the asymptote y = 0 when x . The common mistake is that the curve is either too short, turns away from asymptote or seems going to cut the asymptote. For example, Another common mistake can be seen when x . The curve cuts the negative x-axis and approaches to the asymptote y = 1/3 instead of y = 0. 3(i)   3f e sec2 xxx    33f ' 3e sec2 2e sec2 tan 2xxx x x x Some students do not differentiate sec 2x directly. x (2, 0) (-1, 0) y y =1 y =0 x =0 y x O (2,-2) (-1,-0.5) y =1/3 Asymptotes A short curve

3 3 2 tan 2y y x   3 2tan 2yx They differentiate 1 cos 2x by quotient rule which required more working steps. 3(ii)       2df '' 3 2 tan 2 4sec 2d yx x y x x   When x = 0,  f 0 1  ,  f ' 0 3 ,  f '' 0 13 32 13e sec 2 1 3 2 x x x x    Some students differentiate   3f e sec2 (3 2tan 2 )xx x x  by using product rule instead of doing implicit differentiation, the working becomes very tedious and often ends up with mistakes. 3(iii)     3e sec 2 3 2 tan 2x xx    3d e sec 2d x xx 2d 13 1 3d2 xxx       3 13 x   OR         3e sec 2 3 2 tan 2 1 3 3 4x x x x x          3 1 3 4 1 3 3 13 x x x x          Some students only give the coefficients of the two terms, that are 3 and 13 instead of the terms 3 + 13x. 3(iv)     133 12 2 e sec 2 e cos 2 3 41 3 1 22 xx xx x xx               2 29 1 3 1 2 2 xxx       2 29 1 3 2 2 xxx     213 1 3 2xx   (verified) Many students did not read the question carefully. They verify part (iii) instead of (ii). Common mistakes are  2 cos 2 1 2 xx     22 1 1 1 1 cos 2 1 2 2 2x x x   4(i) xx a a b   or xa x a b   0xx b or 2x xab  x = 0 or 2 1 abx b  Most students know that to find the roots they need to consider ()x a x a   or ()x a x a    OR square both sides. For method 1 some students used ‘AND’ instead of ‘OR’ For method 2 they prefer to expand rather than factorise using A2-B2 = (A – B)(A + B) thus arriving at a more complicated equation. Other issues: Did not simplify the expression for one of the root example:

4 2 2 ( 1) 1 ab bx b   Some thought that b is a root . Thus when x(b – 1) = 0 they tend to conclude that x = 0 or b = 1. Some even went on further to simplify the other root 2 2 (1) 21 1 1 ab axa b    . Some were able to point out that the solution exist only under certain conditions example 1b  . 4(ii)(a) To have a negative root, 2 0 1 abx b  . Since ab > 0, 1 0 b    1b  Many gave the answer as 1b   with no OR long explanation But some gave possible values of x for the equation to have negative roots example x < ab. Some even consider discriminant < 0. 4(ii)(b) For b x a x ab   , from graph, 2 01 ab xb  . For e exxb a ab    , e exxb a ab     e exxb a ab     e exxb a ab     From above, replace x with e x . 2 e < 01 xab b  20 e 1 x ab b    2 ln 1 abx b    Not well done. Many gave a V shape curve . Some gave a graph with the correct shape but gave the x-intercept as (- a , 0) . Some gave the x- intercept as (a, 0) but marked the point on the right side of the y axis thus assuming that a > 0. Quite a number of candidates did not show both the graph and the line intersecting at (0, -ab) Not well done. Many students did not use the above graph to solve this part Some common mistakes: 2 01 ab xb  or 20 1 abx b  Not well done. Students tend to replace x with e x or b with -b . Even with the correct substitution some common mistake: 20 ln 1 abx b      or 2 ln 1 abx b    2 ln 1 abx b   

5 5(i) 3f ( ) n n n=   1 1 f ( 1) f ( ) 33 1 3 3 ( 1) ( 1) 3 3 ( 1) ( 1) 3 (2 1) ( 1) nn nn n n nn nn nn nn nn nn n nn            A familiar question and done well by many students. Some errors made were due to weak foundation in indices. Others did not show their working, jump straight fr 3rd to last step, hence no marks awarded. 5(ii)   1 1 1 31 112 3 (2 1) 2 ( 1) 1 f ( 1) f ( )2 nN n nN n N n nn n nn nn                 f (2) 1 2 f (1) f (3)   f (2) f (4) f (3) f ( 1) f ( )NN     11 1 3f ( 1) f (1) 32 2 1 N N N                   Students are able to express general term as a single sum.. Regards to evidence of using (i), some students were unable to link accurately to (i) by taking out the factor ½. Method of difference were still not done correctly by some students, even in wrongly listing the terms. And finally, many left their answers as f ( 1) f (1)N  . The final answer should be expressed in terms of N. 5(iii)   1 11 11 3 (2 1) ( 1)( 2) 3 (2 1) Replace by 1( 1) nN n nnN n n nn n nnnn          1 2 1 2 3 (2 1) 3 ( 1) 2 3 (2 1) 3 2 ( 1) nnN n nnN n n nn n nn           1 1 2 3 (2 1) 3(2 1) 3 2 ( 1) 2 2 nnN n n nn       Attempt at changing index, whether from (ii) to (iii) or vice versa are accepted if done correctly. However, not all students uses this approach. They found their answer by using method of difference (or result of) which is easier.     1 1 1 1 3 (2 1) 3 ( 1) 1 f ( 2)