RI_9758_2024_Promo_Solutions

RAFFLES INSTITUTION H2 Mathematics (9758) 2024 Year 5 Page 1 of 23 Q1 [4] Method 1 2 d d y b cax x x= + − At 3 ,2x= d 0d y x = 24 0 (1)39a b c+ − = −−−− The gradient of D at x = 1 is equal to the gradient of the line 5.yx=− At 1,x= d 1d y x = 1 (2)a b c+ − = −−−− y-coordinate of D at 1x= is 1 5 4− =− . Substituting 1x= and 4y=− into equation of D, we get 4 (3)ac+ =− −−−− From GC, 2, 7, 6.a b c= =− =− Method 2 2 d d y b cax x x= + − At 3 ,2x= d 0d y x = 24 0 (1)39a b c+ − = −−−− Gradient of D at 1x= is 1 d d x y a b cx = = + − . y-coordinate of D at 1x= is ac+ . So, the equation of tangent to D at 1x= is ( ) ( )( 1) ( ) 2y a c a b c x y a b c x b c− + = + − −  = + − − + Comparing this line with 5yx=− , we get 1 (2)a b c+ − = −−−− 2 5 (3)bc− + =− −−−− From GC, 2, 7, 6.a b c= =− =−

2024 H2 Math Year 5 Promotion Examination: Solutions _____________________________________________________________________________________ Page 2 of 23 Q2 Solution (a) [1] The set of all possible positions of R is the line that passes through points A and B, or The set of all possible positions of R is the line that passes through point B (or A) and parallel to the vector .AB (b) [1] cos90 0a b a b = = since cos90 0= Additional Note: That the scalar product is 0 because 2 vectors are perpendicular (or the angle between them is 90 ) is a consequence of this definition. (c) [4] ( ) * 1OR = + −ab for some  . ( ) ( ) ( ) ( ) ( ) * 22 0 10 1 1 2 0 OR      − = + −  − = − − + −  = ab a b a b a b a b Since a and b are perpendicular, 0=ab 2 22= + b ab 22 * 2 2 2 2OR =+ ++ ba ab a b a b ** 22 2 2 2 2 22 : 1 : : : AR BR =− = ++ = ab a b a b a b O a A B b

2024 H2 Math Year 5 Promotion Examination: Solutions _____________________________________________________________________________________ Page 3 of 23 Q3 Solution [4] ( )( ) ( ) 2 2 6 2 3 2 1 2 1 10, 2 1 2 21 021 112 2 1 0 22 x x x x xx x x x x x + − − + − − − −   − + −      1 1 1 or 222 xx−    Additional Notes: You are strongly advised to use ( ) if you are making algebraic errors in arriving at 221 021 x x − − . For students who multiply by ( ) 2 21x− in the first step, you should always factorize first before any expansion, as shown below: ( )( ) ( )( ) ( ) ( ) ( )( ) 22 2 6 2 3 2 1 2 1 2 1 0 2 1 6 2 3 2 1 2 1 0 x x x x x x x x x x + − − − + −   − + − − + −  This will avoid unnecessary algebraic manipulation. [3] By replacing x with ,x

2024 H2 Math Year 5 Promotion Examination: Solutions _____________________________________________________________________________________ Page 4 of 23 1 1 1 or 222 1 1 1 1or or 2222 xx x x x −    − −    Addition