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