RI 2023 Year 5 H2 Mathematics Common Test (Questions and Solutions)

RAFFLES INSTITUTION 2023 Year 5 H2 Mathematics Common Test Questions and Solutions with comments Page 1 of 17 1 A cubic curve has two stationary points at  2, 0 and 4 500, 3 27     . Find the equation of the curve. [4] 1 Method 1 Let the cubic equation be 3 2y ax bx cx d    . Then 2d 3 2d y ax bx cx    . At  2, 0 , we have 8 4 2 0 ---- (1)a b c d    12 4 0 ---- (2)a b c   At 4 500, 3 27     , we have 3 2 4 4 4 500 ---- (3)3 3 3 27a b c d                      2 4 43 2 0 ---- (4)3 3a b c              Solving the system of linear equations (1) –(4), we have 1, 1, 8 and 12.a b c d     Therefore, the equation is 3 2 8 12.y x x x    Method 2 Since  2, 0 is a stationary point, then x = 2 is a repeated root for the cubic equation. So, we can write the equation as     2 2y x ax b   . At 4 500, 3 27     , 500 100 4 4 5 27 9 3 3 3 a b b a                 2d 2 2 2 2 3 2 2d y x ax b a x x ax b ax          Since 4 500, 3 27     is a stationary point,  4 2 4 2 2 03 3 0 a b a b a            Solving, 1, 3a b  . Thus, the cubic equation is     2 2 3y x x   There are a few solutions which did not start with a general form of a cubic equation:  3 2y ax bx cx   (missing y-intercept)      2 2y x x a   (assumed that the coeff of 3x is 1) There are also quite a number of careless mistakes and copying down the wrong values/signs. There are also some who do by hand to solve the systems of linear equation, instead of using the Plysmlt2 app in the GC.

Page 2 of 17 2 Given that the complex numbers w and z satisfy the equations  * 2 i and 2 i 2 6iw z w z      find w in the form ia b , where a and b are real. [4] 2 * 2 iw z  (1) (2 i) 2 6iw z    (2) From (1), substitute i * 2 wz  into (2) i *(2 i) 2 6i 2 2 (2 i)(i *) 4 12i 2 *(2 i) 3 10i ww w w w w                 Let i, ,w a b a b    2( i) ( i)(2 i) 3 10i 2 2 i+ i 2 2 i 3 10i ( 4 )i 3 10i a b a b a b a a b b b a b                 Compare real parts: 3b (3) Compare imaginary parts: 4 10a b  (4) Solve equations (3) and (4) : 2a   and 3b 2 3iw  

Page 3 of 17 3 With reference to the origin O, the points A and B are such that OA    a and OB    b. It is given that the point X is on OA such that : 1: 2OX XA  and the point Y is on OB such that : 3 :1OY YB  . M is the mid-point of the line segment XY. (a) Find the vector OM   in terms of a and b. [1] Point N lies on the line AB such that O, M and N are collinear. (b) Find the ratio :AN NB . [4] 3(a) 1 3OX    a and 3 4OY    b 1 1 3 2 6 8OM OX OY             a b There are a few who has conceptual error:  1 2OM XY      (b) Method 1 Since O, M and N are collinear, we can write 1 3 6 8ON k OM k k       a b , where k   Let : :1 , where 0 1AN NB       . By Ratio Theorem, (1 )ON       a b . Since a and b are non-parallel and non-zero vectors, 3 8: 8 3 1 8 9: 1 6 3 13 k k                 b a 9 4: : 9 : 413 13AN NB   Method 2 Equation of line OM is 1 3 ,6 8k k      r a b  . Equation of line AB is   ,t t   r a b a  . At point of intersection N,   1 3 for some ,6 8t k k t          a b a a b  It is important to note that in order to compare the coefficients of a and of b, a and b must be non-parallel and non-zero vectors A B O X Y M

Page 4 of 17 1 3(1 ) 6 8t t k         a b a b Since a and b are non-parallel and non-zero vectors, 3 8 8 3 1 8 9 16 3 13 k t k t t t t             9 9 13 13ON AN AB           a b a : 9 : 4AN NB 

Page 5 of 17 4 A plane p is parallel to the line L with equation 2 (2 2 ) t    r i j k + i j k , t  and passes through the points  5, 4,1A  and  3, 4, 2B  . (a) Find a cartesian equation of p. [3] (b) Find the exact distance between L and p. [3] 4(a) Vector parallel to p = 5 3 8 4 4 8 1 2 1 BA                                   Since p is parallel to the line L, a vector perpendicular to p = 8 2 10 5 8 2 6 2 3 1 1 32 16                                          Equation of p is 5 5 5 5 3 4 3 3 29 16 1 16 16                                            r r Cartesian equation of p is 5 3 16 29x y z   . It is important to check your work. Quite a few students made careless mistakes or copy the wrong value. (b) Consider  1, 2, 1C  which is a point on L. 1 5 4 2 4 6 1 1 2 AC                                  Distance between L and p 2 2 2 5 3 16 5 3 16 4 5 6 3 2 16 290 34 17 290or 145290 AC                              Some students approached the question by finding the foot of perpendicular from A to L, note that this would not give the distance between the line and the plane. A C A B

Page 6 of 17 5 Do not use a calculator in answering this question. By completing the square, or otherwise, show that 24 10 13x x  is always positive. Hence solve the inequality 2 2 9 15 3 52 x x x x     . [5] Hence solve the inequality 2 2 9 15 3 52 x x x x     . [3] 5 Method 1 2 2 2 2 2 54 10 13 4 13 2 5 54 13 4 4 5 274 4 4 x x x x x x                                 Since 2 5 4x     0 x  , 2 5 274 0 4 4x      so 24 10 13x x  is always positive. Method 2 Since the discriminant of 24 10 13x x  is      2 10 4 4 13 108 0    , and the coefficient of 2 (4)x is positive, so 24 10 13 0x x   for all x  . Most students are able to do the complete the square correctly.   2 2 2 2 2 2 2 2 2 9 15 3 5 (*)2 9 15 3 5 02 9 15 3 5 2 02 4 10 13 02 x x x x x x x x x x x x x x x x x x                         Since 24 10 13x x  is always positive, then the inequality is reduced to    2 2 0 1 2 0 x x x x       The solution is 1 2 x   . The question is mostly well done

Page 7 of 17 2 2 2 2 2 2 9 15 3 5 (#)2 9 15 3 52 9 15 3 52 x x x x x x x x x x x x               Thus, 1x  or 2x  . Note that 1x  has no solution. 2x and 2x cannot be a solution to equation (#). Thus, 2x  is reduced to 2x  So, the solution to inequality (#) is 2x or 2x . There is need to rewrite the inequality to look like inequality (*). After rewriting, notice that the inequality is 5 instead of 5 . Thus, the solution should be a complement of the original solution.

Page 8 of 17 6 (a) Use differentiation to find the x-coordinate of the stationary point of the curve 2 2 ln xy x p q       , where p and q are positive and negative constants respectively, and determine the nature of the stationary point. [4] (b) (i) The tangent to the curve lny x at the point  , lnA a a passes through the origin. Find the value of a. [2] (ii) The normal to the curve lny x at the point  , lnB b b passes through the origin. Show that B lies on the curve 2y kx , where k is a constant to be determined. [2] 6(a) 2 2 2 2 d 1ln 2 2 . d x y q py x p x p x q x x q x 