RVHS 23 J2 H2 MA CT Std Soln

River Valley High School 1 2023 JC2 H2 MA CT Markers Comments 1 The complex numbers z and w satisfy the equations: 22 * 2z ww w−= and 2122 2z w w+ = − . (i) Show that 2 2 2 0ww+ + = . [2] (ii) Hence find all possible pairs of values for z and w. [3] 1 Solution [6] Complex Numbers (i) 2 22 2 * 2 22 z ww w z w w −= =+ Substitute 221 22z w w=+  into 2122 2z w w+ = − ( ) ( ) 222 2 2 11 2 2 222 2 2 2 2 2 0 2 2 0 Shown w w w w ww ww  + + = − + + = + + = Majority were able to show the result either via the substitution or e limination method. A number of s tudents did not see that ww* = |w| 2, and hence una ble to continue or instead wrote w = x + iy and expanded. (ii) From GC, 1iw=−  When 1iw=− + , 2 2iw =− , 2 2w = and ( )2 2i 2 1 2i2z −+= = − When 1iw=− − , 2 2iw = , 2 2w = and ( )2 2i 2 1 2i2z += = + Thus, 1 2iz=− and 1iw=− + 1 2iz=+ and 1iw=− − w can be found via the GC. However some students went on to substitute w = x + i y and solve v ia comparing real and imaginary parts (which was too much work for 1 mark). Though the m ethod to find z was correct, many made careless mistakes in the computation.

River Valley High School 2 2 The complex numbers z, w and v are such that 13π 13π2 cos isin30 30z =+  , ( ) πarg 12w = and 4 i* zwv w= . (i) Find the modulus of v and show that ( ) 3arg π5v =− . [4] (ii) It is known that nv is a negative real number. Find the least positive integer value of n. [3] 2 Solution [6] Complex Numbers (i) ( ) 4 4 4 44 i* i* 1 2 4 units zwv w zw w zw w z = = = = = = 13π 13π2 cos isin30 30z =+  ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 4 arg 4arg arg arg i arg *i* 4arg arg arg arg 13π π π42 30 12 2 7π 3π Principal argument Shown55 zw z w ww z w i w  = + − + = + − − −    = + −       −== Most candidates can work out that 4v = Some candidates are not aware that ( ) ( )arg * argww =− ; ( ) πarg 2i = ; ( )7π 3π Principal argument55 −= since 7π 3π2π55 −−= (ii) ( ) ( ) 3 πarg arg 5 n nv n v −== , ( ) ( )arg negative real number 2 1 π, where kk= +  ( )3 π 21 π5 n k− =+ ( )5 213nk −=+ Least positive value of n = 5 when 2k =− . Many candidates are wrongly state that ( )arg negative real number kπ, where k=

River Valley High School 3 Note: Setting 3 π π5 n k− = is not correct.

River Valley High School 4 3 The position vectors of distinct points A, B and Q are given by a , b and q respectively with reference to the origin. (i) Given ( ) ( )−  − =q a b a 0 , explain why A, B and Q are collinear. [2] (ii) It is given further that (1 )= + −q a b where 01  , 1| | | | 2=ba , 3 ||6AQ= a and 60AOB =  . By finding the value of ( ) ( )−  −q a q a in terms of  and ||a , find the value of  and write down the ratio of AQ:QB. [5] 3 Solution [7] Abstract Vectors (i) ( ) ( )−  − =q a b a 0 where a , b and q are distinct ( ) ( )// − −q a b a since −q a 0 , −b a 0 ( ) ( ) for some kk AQ k AB  − = −  = q a b a AQ is parallel to AB with common point A. The points A, B and Q are collinear. Most students were able to deduce that the vector AQ and AB are parallel as their cross product is the ze ro vector. Students should be spe cific in naming the vectors properly and not st ating that ( )−qa is par allel to ( )−ba . Also, expanding the c ross product will not be useful in deriving the par allel property of the vector s as needed. (ii) Consider ( ) ( )−  −q a q a ( ) ( )(1 ) (1 )   = + − − + − −a b a a b a ( ) ( )( 1) (1 ) ( 1) (1 )   = − + − − + −a b a b 2222 2222 ( 1) 2( 1)(1 ) (1 ) ( 1) 2( 1)( 1) ( 1)         = − + − − + − = − − − − + − a a b b a a b b 222 2 2( 1) 2( 1) cos60 ( 1)  = − − − + −a a b b There we re students who were not able to quot e the result that 2 =a a a and hence deduce that it should be ( ) ( ) 2 −  − = −q a q a q a . Very few students were able to correctly expand ( ) ( )−  −q a q a to obtain 2 23( 1) 4 − a due to t he following errors:

River Valley High School 5 222 2 2 222 2 22 2 2 ( 1) ( 1) ( 1) ( 1) 11( 1) 22 3( 1) 4       = − − − + − = − − +     = − − +        −= a a b b a a b b a a a a a ( ) ( ) 2 23( 1) 4 −−  − =q a q a a 2 23( 1) 4AQ AQ −= a 2 2 2 2 2 3 3( 1)||64 3 3( 1) 36 4 1( 1) 9 11 3 24 or (rejected)33       −=  −= −=  − = = aa Thus, for (1 )= + −q a b 21 33 2 =+ += q a b abq 3 By ratio theorem, : 1: 2AQ QB = ( ) ( ) 22 (i) writing 2 instead of q aq a −  − = − + q a q a 2 q 2− q a + 2 a (ii) replacing b = 1 2 a in expression for vector q (note t hat 1 2=ba does not imply b = 1 2 a ) (iii) wrongly expanding the expression ( 1) (1 )− + −ab as ( 1) (1 )− + −ab . Note that in general, +  +a b a b • • • y

River Valley High School 6 4 The function f is defined by 2 1f : , , 1, 1 1x x x x  −− (i) Show that f is not a one-one function. [1] Another function g is defined by 2 1g : , , 1x x a x bx − , where ba . (ii) State the maximum value of a such that 1g− exists and write down the value of b. [2] For the rest of the question, take 1a=− . (iii) Find 1g ( )x− . Sketch the graphs of g and 1g− on a single clearly labelled diagram, showing the geometrical relation between the graphs. [4] A function h is defined by 2h : , 0x x x  . (iv) Explain why the composite function hg exists and define hg in a similar form. [2] 4 Solution [9] Functions (i) Since the line 1y= cuts the graph of f more than once, Therefore f is not a one-one function. Students generally had the right idea for how to solve this, but many did not use the standard explana tion and made mistakes as a result. Common mistakes: - No graph drawn - General y = k instead of a specific value - Claims that any line would cut twice, rather than a specific value. - Omitting conclusion. 1y=

River Valley High School 7 Alternatively Consider 1 2x =− and 2 2x = . Then ( ) ( ) 1 2 11f 321 x == −− , and ( ) ( ) 2 2 11f 321 x == − . Since 12xx , but ( ) ( )12ff xx = , then f is not one-one. Some students attempted the alternative method and were generally successful. (ii) 0a= 1b=− Generally well done. (iii) 2 2 2 2 1let 1 1 1 1 11 or , rejected since 0. y x yx y yx y yx y yyx x x yy = − −= =+ += ++=− =  ( ) 1 1g xx x − +=− Students generally had a good idea about how to approach this question. Common mistakes: - Forgot ± when ta king square root. - Incorrectly rejected negative root. - Forgot to explicit ly state g-1(x) - Forgot to s wap back x and y for g-1(x)

River Valley High School 8 Students were usually able to draw y = g( x). Some forgot to d raw in the asymptotes. Some students dre w the graph o n (-∞,0) instead of (-∞,-1). The graph of y = g-1(x) was not as well drawn. Asymptotes were often incorrect. A number of students drew the graph on the domain of (-∞,-1) instead of (0,∞). Labelling of the graph was not good . Students have taken to labelling y = g( x) as g. No marks were deducted, but students should be mindful to write in the equation of the graph or the label of the curve, rather than just the n ame of the function. ( ) 0, 1  − ( ) 1,0  −