2025 RI Math Prelim P1 solutions with comments

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 Page 1 of 32 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with comments 1 A curve has equation 2 1 cy ax b x    , where a, b and c are real constants. It is given that the curve crosses the x-axis at 2x . The normal to the curve at the point  0, 3 meets the x-axis at 7.5x . Find the values of a, b and c. [4] Solution Comments [4] 2 1 cy ax b x    Since  2, 0 and  0, 3 lies on the curve, 12 0 --- (1)3a b c   3 --- (2)b c    22 d 2 d 1 y cxax x   Gradient of normal at  0, 3 3 0 2 0 7.5 5   . Gradient of the tangent to the curve at (0, 3) is 5 2 and thus 5 --- (3)2a Using GC, 5, 3 and 62a b c   . There are quite many sign/conceptual errors in the attempt to find d d y x. Do note that the gradient of the normal to the curve at the point (0, 3) means that we need to substitute 0x in d d y x, and get 1 a as the gradient of the normal. The two points (0, 3) and (7.5, 0) gives the gradient of the normal as 2 5 .

2025 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with Comments _____________________________________________________________________________________ Page 2 of 32 2 The point P travels along the curve C with equation 1sin , 1 1y x x x    . Let the gradient of the curve C at the point P be m. If the x-coordinate of P is increasing at the rate of 9 units per second when 1 2x , find the exact value of the rate at which m is changing at this instant. [4] Solution Comments [4] 1sin , 1 1y x x x    Differentiate with respect to x: 1 2 d sind 1 y xm xx x     Differentiate with respect to x again:       2 2 22 2 2 2 2 2 2 32 d 11 21 2 1 11 11 1 1 1 2 1 d x x x x xx x x x x m x x x x                 When 1,2x 3 3 12d d 4 d d 1 d 9 1 1 4 4 d m t t m x x            Therefore, required rate is 14 3. Most students did well but many did not simplify the answer to the lowest terms. Answer mark was not awarded to non-exact answers and unsimplified answers. Some students substituted x=p, assuming the parameter x takes the value of p at point P, note that p will be taken as a constant in this case and we shall not differentiate with respect to p.

2025 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with Comments _____________________________________________________________________________________ Page 3 of 32 3 Relative to the origin O, the points A and C have position vectors a and 2b such that 5 2 4p p p  a i j k and 2 2  b i j k , where p is a positive constant. It is given that 2a b . (a) Find the exact value of p. [2] (b) Evaluate 2 2  a b a b . [1] (c) Evaluate a b and hence find the area of triangle OAC. [3] (d) Use a geometrical reason to explain why   1 2 24   a b a b a b . [2] Solution Comments (a) [2] Since 2a b , 2 2 2 2 2 2(5 ) ( 2 ) (4 ) 2 1 ( 2) 2 3 5 6 (given 0) 2 2 5 55 p p p p p p            Most students did well but many did not simplify the answer or made careless mistakes, which affected the rest of the parts, too. (b) [1]  2 2 sin 2 2 2 2 4 4 c 2e 0            a b a b a.a a.b b.a b.b a b a b Most students did well. (c) [3] 5 1 5 1 4 2 4 52 2 2 2 6 3 54 2 4 2 8 4 p p p p p                                                                     a b = Area of triangle OAC     2 22 1 22 4 5 4 1452 3 45 5           a b a b Some students did not evaluate a bto answer the first part of the question. Many students did not rationalise the denominator or simplify the answer.

2025 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with Comments _____________________________________________________________________________________ Page 4 of 32 (d) [2] Observe that 2OD a b  and 2CA a b   . From (b), OD is perpendicular to CA. In addition, CA and OD, which are the diagonals of the parallelogram OADC, bisect each other. From (c), area of triangle OAC  a b. But area of triangle OAC is also     1 1 2 2 1 2 24 CA OD       a b a b Hence   1 2 24   a b a b a b . Marks are awarded only if there are: 1. correct discussion on the perpendicular property of 2a b and 2a b . 2. correct discussion on the relationship between the areas, building upon Point 1. O A C D a 2b

2025 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with Comments _____________________________________________________________________________________ Page 5 of 32 4 (a) Find    2 9 d2 1 1 x xx x   . [4] (b) (i) Differentiate 2 1 1x  with respect to x. [1] (ii) Differentiate 2ln 1x  with respect to x. [1] (iii) Hence find   2 22 ln 1 d1 x x xx    . [4] Solution Comments (a) [4] Let    2 2 9 2 1 1 ( 1)2 1 1 x A B C x x xx x       . 29 ( 1) (2 1)( 1) (2 1)x A x B x x C x       Sub. 1 2x , 2A . Sub. 1x , 3C . Compare the coefficient of 2x , 2 0 1A B B    .    2 2 9 2 1 3d d 2 1 1 ( 1)2 1 1 3ln 2 1 ln 1 , 1 2 1 3ln 1 1 x x x x x xx x x x c c x x cx x                      Most students did well but many made careless mistakes, which affected the rest of the parts. A handful is still not familiar with splitting into appropriate partial fractions in order to integrate. (b) (i) [1]   22 2 d 1 2 d 1 1 x x x x       Most students did well but some made careless mistake.s (b) (ii) [1]  2 2 2 d d 1ln 1 ln 1d d 2 1 xx xx x x         Most students did well but again some made careless mistakes particularly for

2025 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with Comments _____________________________________________________________________________________ Page 6 of 32 those who did not simplify first using the property of logarithm. (b) (iii) [4]         2 22 2 2 2 2 2 2 2 2 22 2 2 2 2 2 2 ln 1 d1 1 2 ln 1 d2 ( 1) 1 1 1 ln 1 d2 1 1 1 ln 1 1 d22 1 1 ln 1 1 2 1 4 1 ln 1 1 4 1 x x xx x x xx xx xx x x x x xx x x cx x x cx                                       ---integrate by parts   2 2 ln 1 1ln 12 u x x     2 2 d 2 d ( 1) v x x x   2 d d 1 u x x x  2 1 1v x  Those students who could not do this part fail to observe the role of the previous two parts and the word “hence”. As a result, they could not select the correct terms for integration by parts. Still there are many who made careless mistakes in the integrations.

2025 H2 Math Year 6 Preliminary Examination Paper 1: Solutions with Comments _____________________________________________________________________________________ Page 7 of 32 5 (a) The diagram below shows the sketch of the graph of fy x . The curve passes through the points with coordinates  1, 0 and  3, 0, and has turning points at  1, 0 and  4, 4. The asymptotes are 0x , 2x and 2y . Sketch on separate diagrams, the graphs of (i) fy x , [3] (ii)   1 fy x , [3] showing clearly the main features of the graphs. (b) Describe a sequence of transformations which transforms the graph of 2 1xy x  to the gr