RVHS 2023 JC2 H2 MA Test 3 Solutions (Student Copy)

RVHS 2023 JC2 H2 Maths Lecture Test 3 1 Name : Index Number : Class : Date : 12 May 2023 Duration : 50 min Max. No. of Marks : 30 Formulae from MF26 Vectors The point dividing AB in the ratio : has position vector   + + ab Vector product: 1 1 2 3 3 2 2 2 3 1 1 3 3 3 1 2 2 1 a b a b a b a b a b a b a b a b a b −            =  = −           −      ab [Answer all the questions on writing papers. Up to 1 mark will be deducted for poor presentation.] 1 The line l has equation ( )2 = + + + − +r i j k i k where  is a parameter. The plane π1 has equation 2 2 2x y z+ + =− and the point A is at ( )1,1, 2 . (i) Find the acute angle between the plane π1 and the line l. [2] (ii) The plane π2 contains the line l and is perpendicular to plane π1. Show that the equation of plane π2 is 2 3 2 3x y z− + = . [2] (iii) Find a vector equation of the line of intersection between π1 and π2. [2] (iv) Find the shortest distance from the point A to the plane π1. [3] 2 (a) The complex number 1iv=+ is a root of the equation 22 1 i 0z cz− + − = where c is a complex number. (i) Without the use of a calculator, find the value of c. [4] (ii) Without calculating the other root, explain why v* is not the other root of the equation 22 1 i 0z cz− + + = . [1] (b) The complex number w is given by ππ2 cos isin55w =−  . Showing your working clearly, find the modulus and argument of 2i 2i 2 3 w − . [6] River Valley High School 2023 JC2 Mathematics (9758) Lecture Test 3 (Term 2)

RVHS 2023 JC2 H2 Maths Lecture Test 3 2 End of Paper 3 A disc with sectors labelled 1, 2 and 3 with a free spinning arrow and a fair coin are used in a game booth at a bazaar. When the arrow on the disc is spun, it comes to a stop pointing at the sector labelled 1 with probability of p while it has a probability of 2 3 p pointing at the sector labelled 2. In a game, the arrow is spun, the n the coin is tossed 1, 2 or 3 times according to the sector where the arrow points on the disc. The random variable X denotes the score of the game obtained by adding the total number of heads from the resulting coin toss(es). (i) Show that ( ) 35P1 8 24Xp= = + . Construct a probability distribution table for X with the expressions of the probabilities in terms of p. [5] (ii) Given that ( ) 5E 6X = , find the value of p. [2] (iii) It is given that ( ) 5Var 9X = and that X1 and X2 are two independent observations of X. Find 12 1Var 32 XX − . [3]

RVHS 2023 JC2 H2 Maths Lecture Test 3 3 RVHS 2023 JC2 H2 Maths Lecture Test 3 Solution Question 1 [9] Vectors (Planes) (i) Let the acute angle be  . 1 1 222 1 11 20 21 11 20 21 12 1 32 13.6 sin sin 2 (1 d.p. 2 i ) sn  − − − −                  −                 = +  = = + = Some responses have mixed up the formula with others. (ii) The normal of plane π1: 1 22 2   =− r is parallel to the plane π2. 2 1 1 2 A normal vector to plane 2 0 3 2 1 2  −           =  = −                And point A is on the plane. 12 1 3 2 3 4 3 22        − = − + =          . Therefore, the equation of plane π2 is 2 3 2 3x y z− + = . (Shown) Generally most responses are able to identify the 2 vectors to cross for the normal vector for π2 and substituting the point OA for the constant. (iii) Using GC to solve the system of equations: 2 2 2 2 3 2 3 x y z x y z + + =− − + = Some responses crossed the normals of the 2 planes to get the direction of the line. However, many of these responses have difficulties getting a particular point on the line of intersection.

RVHS 2023 JC2 H2 Maths Lecture Test 3 4 100 7 21 7 xt yt zt =− =− − = for some t . Line of intersection: 10 70 21, 70 1 tt −    = − + −     r or 0 10 1 2 , 07 tt        = − +        −    r (iv) Let ( )0, 1,0N − be a point on π1. 0 1 1 1 1 2 0 2 2 AN ON OA −           = − = − − = −           −      Shortest distance from point A to plane π1 222 1 2 2 11 122 1 2 222 9 9 3 units AN  =  −       =−    ++   −    −= = Alternatively Let ln be a line passing through point A and perpendicular to π1. ln: 11 1 2 , 22         = +           r Let point N be the intersection of ln and π1, Some responses did not identify a point on the plane to get a vector from the plane to point A for the formula. Using OA for the formula gives the distance of point A to a plane containing the origin parallel to π1.

RVHS 2023 JC2 H2 Maths Lecture Test 3 5 ( ) ( ) ( ) 1 1 1 1 2 2 2 2 2 2 1 1 2 1 2 2 2 2 2 1 1 1 0 1 2 1 2 2 0 ON                 + =−                + + + + + =− =−             = − = −                Shortest distance from point A to plane π1 ( ) ( ) ( ) 222 01 11 02 1 2 2 1 2 2 3 units AN ON OA=−        = − −          − =−− = − + − + − =

RVHS 2023 JC2 H2 Maths Lecture Test 3 6 Question 2 [11] Complex Numbers (ai) Sub 1iz=+ into 22 1 i 0z cz− + − = , ( ) ( ) ( ) ( ) ( ) 2 2 1 i 1 i 1 i 0 2 1 2i 1 1 i 1 i 0 1 i 1 3i 1 3i 1 i 1 i 1 i 1 i 3i 3 2 2i c c c c + − + + − = + − − + + − = + = + +−= +− − + += =+ Most responses are able to substitute 1iz=+ to find c. Most are able to use operations to make c the subject. (aii) If v* is also a root, then all coefficients of the quadratic equation must be real. However, that is not the case. Therefore, v* is not a root. Many responses fall short in quoting the theorem although their response shows sign of knowledge. (b) ( ) 2i 2 3 4 5πarg 2i 2 3 6 −= −= ( )( ) 22 2 ii 2i 2 3 2i 2 3 12 4 1 unit ww = − − = = Note that π π π π2 cos isin 2 cos +isin5 5 5 5w      = − = − −             . Therefore ( ) πarg 5w =− and 2w = . ( ) ( ) ( ) 2i 2i 2 3 2i 2 3 π π 5π22 5 6 arg arg i 2arg ar 11π rad15 gww − =+ − = + − − =− − . Many are not successful in getting the correct argument for 2i 2 3− . Many are able to use the properties of modulus to get the correct answer. Few responses did not notice that w is not given in standard polar form so the argument is not π 5 .

RVHS 2023 JC2 H2 Maths Lecture Test 3 7 Question 3 [10] Discrete Random Variable (i) ( ) ( ) 23 231 2 1 5 1P 1 1 112 3 2 3 2 35 shown8 24 X p p p p           = = + + −                      =+ ( ) 23 1 2 1 5 1 1 11P 0 1 2 3 2 3 2 8 24X p p p p       = = + + − = +             ( ) 23 32 1 5 1 3 11P 2 1 23 2 3 2 8 24X p p p      = = + − = −             ( ) 3 35 1 1 5P 3 1 33 2 8 24X p p    = = − = −         Xx= 0 1 2 3 ( )P Xx= 1 11 8 24 p+ 35 8 24 p+ 3 11 8 24 p− 15 8 24 p− Many students are unable to cope with the complexity of the context. Some good response includes drawing a tree diagram to visualise all outcomes and calculate the probabilities based on the tree. Most responses missed out the outcome where there is no heads i.e. X = 0 (ii) E( ) P( )X x X x== ( ) ( )3 5 3 11 1 5 51 2 38 24 8 24 8 24 6 3 4 5 2 3 6 1 2 p p p p p      + + − + − =           −= = Generally, those who did not manage to get the probability distribution table up are not able to do