2026 RI H2 Math Year 6 Preliminary Exam Paper 1 (Solutions with comments)

RAFFLES INSTITUTION H2 Mathematics (9758) 2026 Year 6 Page 1 of 26 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with comments 1 A curve has parametric equations 2 1x t  , 3 2 ,y at bt ct   where 2t and a, b and c are constants. The curve has stationary points at 1t and 2t . Given that the curve passes through the point (5, 2) , find the values of a, b and c. [4] Solution Comments [4] 3 2y at bt ct   2d 3 2d y at bt ct   At stationary points, d d0 0d d y y x t   When 1t , 3 2 0 (1)a b c    When 2t , 12 4 0 (2)a b c    Since curve passes through the point 5, 2 , we have 25 1 2t t    . Since 2t , 2t . When 2t , 3 2(2) (2) (2) 2a b c   8 4 2 2 (3)a b c    Using GC, 1 5a , 3 10b , 6 5c This question was generally well-done. Mistakes were mostly due to carelessness in simplification.

2026 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with comments _____________________________________________________________________________________ Page 2 of 26 2 (a) The sequence 1 2 3, , ,u u u is defined by 1 4u and 1 1 2, for 13n nu u n    . (i) Write down the value of 6u . [1] (ii) Given that as , nn u l  , find the exact value of l using an algebraic method. [2] (b) Using the result  2 1 ( 1) 2 16 n r nr n n     , find 0 ( )( 1) n r n r r    , giving your answer in terms of n, in fully factorised form. [3] Solution Comments (a) (i) [1] From GC, 6 362 or 1.49243u  (3 s.f) A handful of students did not realise that this part can be done using GC. (a) (ii) [2] As , nn u l  , 1nu l  . From 1 1 2, 13n nu u n    , as n , we have 1 2 3 4 23 3 2 l l l l     The critical step in this part is to state that “As , nn u l  , 1nu l  ” which many failed to do so. (b) [3] 2 0 0 2 0 0 0 2 1 1 ( )( 1) ( ) ( 1) ( 1) ( 1)( ) ( 1) ( 1) ( 1) ( 1)(2 1)2 6 n n r r n n n r r r n n r r n r r nr n r r n r n r n r n n r n n nn n n n n                                          ( 1) 3( 1) 6 (2 1)6 ( 1)(5 10)6 5 ( 1)( 2)6 n n n n n n n n n n             The common mistake in this part was not realizing that there are a total of n+1 terms in 0 . n r n   There were students who did not read the question carefully, failing to leave the answer in fully factorized form.

2026 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with comments _____________________________________________________________________________________ Page 3 of 26 3 Water is poured at a rate of 30.5 m per minute into an empty container in the form of an open hemisphere with inside radius 3 m (see diagram). At time t minutes after the start, the radius of the water surface is mr and the volume of the water in the hemispherical container is  21 9 ,3 h h  where h m is the depth of the water. When the depth of the water is 1 m, (a) find the rate of change of the depth of water, [3] (b) find the rate of change of the radius of the water surface. [3] Solution Comments (a) [3]    2 2 3 2 2 1 1Given 9 33 3 d (6 )d d d dNow, d d d 1 0.56 d 0.5 1When 1, 0.0318 (3 sf)d 5 10 V h h h h V h hh h h V t V t h h hh t                           The depth of water is increasing at a rate of 0.0318 m per minute. You need to simplify and give the final answer as 1 10 (exact) or 0.0318 (3 s.f.). Note that the units in this case is metres per minute. 3 h

2026 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with comments _____________________________________________________________________________________ Page 4 of 26 OR  2 2 3 2 1 1Given 9 33 3 Differentiate w.r.t , d d d 3 2d d d dWhen 1, 0.5.d d d 0.5 3 2d d d 0.5 1 0.0318 (3 sf)d 5 10 V h h h h t V h hh ht t t Vh t h h t t h t                                          (b) [3] From the diagram, by Pythagoras Theorem,    22 2 2 2 2 2 3 3 6 Since 0, 6 d 1 1 3 6 2d 2 6 6 r h h h r r h h r h hh h h h h             2 d d d d d d 3 1 106 d 2 1 5When 1, 0.0285 (3 s.f)d 2510 5 5 5 r r h t h t h h h rh t               The radius of water surface is increasing at a rate of 0.0285 mper minute. Common mistake: Equating the volume of water to that of the volume of a hemisphere.  3 22 1 93 3V r h h    The shape of the lower part of the container is not a hemisphere. Many students tried to apply similar triangles (though incorrectly) to find the relation between r and h. See alternative for the correct version.

2026 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with comments _____________________________________________________________________________________ Page 5 of 26 OR From the diagram, by Pythagoras Theorem,     22 2 2 3 3 Differentiate w.r.t , d d2 2 3 ( 1)d d d d 3d d When 1, 5 5 (since 0). d 15 (2)d 10 d 1 5 0.0285 (3 s.f)d 255 5 r h t r hr ht t r hr ht t h r r r r t r t                    Alternative method to obtain relation between r and h (via similar triangles): ABC is similar to BDC . By similar triangles, 2 2 6 (6 ) 6 h r r h r h h h h       A B C D

2026 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with comments _____________________________________________________________________________________ Page 6 of 26 4 With reference to the origin O, three distinct points A, B and C are such that OA a, OB b and OC c. The mid-point of AC is D. (a) Express OD  in terms of a and c. [1] (b) Given that a c , 0b and BD is perpendicular to AC, show that OB is also perpendicular to AC by using a suitable scalar product. [3] (c) Given further that point D lies on line segment OB, 2a , 6b and 60AOC   , find the exact area of quadrilateral OABC. [2] Solution Comments (a) [1] 2OD a c  Note: 1 2OD AC    (b) [3] 2BD  a c b  Since BD is perpendicular to AC, 0BD AC       2 2 02 1 1 1 1 02 2 2 2 1 1 1 1 02 2 2 2 0 ( 0                                          a c b c a a c a a c c c a b c b a a c a c a c b c b a b c b a b a c) Since b 0 and a c , OB is perpendicular to AC. (shown) Alternative: OB AC OD BD AC OD AC BD AC                           As a c , AOC is isosceles and since the mid-point of AC is D, OD is perpendicular to AC. Thus 0OD AC     . Note: 2 2, cos AOC      a a = a c c = c a c = a c a c It is incorrect to assume OB k BD    (or O, B and D are collinear) even though OD and BD are both perpendicular to AC. This information is

2026 H2 Math Year 6 Preliminary Examination Paper 1: Solutions wit