2021 J2 RVHS H2 Math CT Markers Report

River Valley High School 1 2021 JC 2 H2 Maths Common Test Markers Report 1 Solution [6] Abstract Vector (a) A M C a P O b N B Let OP OM , then 1 2abOP   . Let AP AN , then   2 3AP OP OA      -a b . Next, we have OA AP OP  a + 2 3-a b  = 1 2ab  Since a and b are non-zero and non -parallel vectors, upon comparing a and b on both sides of equation, we have: 11 21 232 ( ) ()        From (2) we have 4 3 and 4 7 311 3 3 7         and so 4 7 Therefore,  2 27 abOP Many students struggled to form a vector equation in terms of λ and μ. Of those who could, a number did not correctly use λ and μ as defined in the question (i.e. they defined their own λ and μ).

River Valley High School 2 (b) A M C a P a O b N B Q Q lies on AN produced such that 21::AN NQ  . Applying Ratio Theorem, 21 33ON OQ OA 31 22 3 2 1 2 3 2 1 2 ba ba OQ ON OA     Then we have QB OB OQ = b  1 2ba  = 1 2 a . Since BC parallel QB and B is a common point, the points Q, B and C are colliear. Generally students had the right idea on how to attempt this question, but some struggled with algebraic errors.

River Valley High School 3 2 Solution [6] Complex Numbers * 4 4* * 4 (1) zw zw zw    Sub (1) into *2i 1zw  2i 4 1 ww      8i 2i 1 2i 1 8i 1 2i 1 8i 1 8i 1 2i 1 2i 1 2i 1 16 8i 2i 5 3 2i ww ww w w                   Therefore  4 3 2i * 1 2i z    Students struggled to correctly manipulate the conjugate. Students generally preferred to allow w = x + i y rather than solving for w directly. Almost all forgot to indicate that x, y needed to be real numbers. Students should be encouraged to make greater use of the GC in their working or to check their answers. Alternative Method Let iz a b and iw c d where , , ,a b c d  . From *2i 1zw ,         2i i i 1 2 i 2 i 1 2 2 i 1 a b c d a b c d b c a d             Comparing real and imaginary parts,     2 1 1 2 0 2 bc ad   From * 4zw ,     i i 4 i4 a b c d a c b d         Comparing real and imaginary parts,     43 04 ac bd   Solving all 4 equations using GC, we get 1, 2, 3, 2a b c d    Therefore, 1 2i, w 3 2iz    . This was a preferred method by students, and there were some good solutions. However, problems similar to the previous method were present: students forgot to indicate the real and imaginary parts needed to be real. Students also made algebraic errors after establishing the 4 equations – they should be encouraged to use GC to solve simultaneous equations or minimally check their answers using GC.

River Valley High School 4 3 Solution [5] Complex numbers (i) 22 2 cos isin44 1i zw w          This was generally well done. (ii)        arg arg arg 7 3 4 12 n zw n z w nn       For   n zw to be purely imaginary,   arg 2 n zw k   where k . 7 12 2nk      2 1 12 27 6 2 1 7 kn k   Therefore, the least integer value of n = 6, (when k = 3 ) Students generally had the right idea, but did not always appreciate that n needed to be an i nteger. 6 7n was a common incorrect answer. Some students were not able to get that there were infinite values for n and k, and instead incorrectly gave 7 or 12 2 2n  

River Valley High School 5 4 Solution [6] Summation Comments 33f f 1 1 1 2 ! 1 ! 3 3 3 2 3 1 ! 2 ! 2 ! 31 Hence, 32! ( ) ( ) ( ) ( ) () ( ) ( ) ( ) () .() rr rr r r r r r ar                       Most students are able to get the value of a. (i) From (i),  11 f f 12 ! 3 ( ) ( )() r rrr     Hence, 1 1 2!() n r r r   =   1 1 f f 13 ( ) ( ) n r rr     1 103 + f 2 f 1 + f 3 2 + ................... + ................... + f 1 f 2 + f f 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ff f nn nn         1 1 3 3 0 1 13 3 2 ! 2! 1 3 3 3 2 2 ! 11 2 2 ! ( ) ( ) () () () f n f n n n                Most of students are also able to demonstrate method of difference. (ii) For the sum 3 1 ! n r r r   , we replace ‘r’ by ‘k+2’. Then we have: 2 3 2 3 2 1 1 2 1 2! 1 2! 11 (using result in part (i))2 2 2 ! 11 2! ! ( ) () () n k n rk n k rk rk k k n n                  This part proves to be challenging to some students as they are not able to replace according. For some they may recognize what to replace but forget to change the starting and ending values of r.

River Valley High School 6 5 Solution [7] Function (i) Sketch of 2 1f 1 for R, 1 1()x x x x    : y 1x 1y 2 x To find 1f ()x : Let 2 11 then 1,y x  2 2 2 1 11 11 1 11 1 11 1 yx x y x y x y          Since 1,x 11 1x y   Thus, 1 1f1 1()x x     ,  1 ff 1,DR    (from sketch of f(x)) The most common mistake is fail to justify the choice of x.

River Valley High School 7 (ii) Sketch of 2g 4 5 for R, 1()x x x x x     : y 1 10( , ) 1 2 0 1 x g 1 R  , For fg,  gf 1 1,,RD       , thus fg does not exist For gf,  fg 1 1,,RD     , thus gf exists. To find range of gf, we use the mapping method:  1,  f   1, g  1 ,  Thus, gfR  1 ,  Alternatively. Graph of y gf x () Thus, gfR  1 ,  This part is not very well done especially finding the range of f & g. Some students are not sure how to check whe ther composite functions exist. Another common statement made by students is that gf gRR which is not necessary true. Students also state the range of gf without justification.

River Valley High School 8 6 Solution [7] Integration & its application (i)  sin d sin cos d sin cos sin d sin cos sin d t tt t t t t t t t e t e t e t t e t e t e t t e t e t e t t             Since sin d sin cos sin dt t t te t t e t e t e t t     2 sin d sin cos 1sin d sin cos2 t t t t t t e t t e t e t c e t t e t e t c           (ii) Many students failed to sketch the graph according to the specified domain. Many students failed to indicate the correct x - intercept. (iii)   1 1 0 d sin d sin d a