RI Yr 5 H2 Math TP 2025 (Soln w comment) - updated

RAFFLES INSTITUTION 2025 Year 5 H2 Mathematics Timed Practice Paper Questions and Solutions with comments ______________________ 2025 Yr 5 H2 Math Timed Practice Paper Solution with Comments General Comments  A small number of students wrote the wrong Exam Index Number on their scripts. For example, 260227 was mistakenly written as 262027, which belongs to another actual student. 跚跛跜距 Please double-check your Index Number before writing it down in future assessments.  A significant number of scripts had solutions written outside the designated boxes for each part of the question. 굓굔굕굖 Please note that for the actual A -level exams, your scripts will be digitally scanned, and any writing outside the boxes may not be captured. 跚跛跜距 Always keep your work within the provided boundaries.  A small number of students wrote their solutions for one question in the space of another question. 跚跛跜距 Do not do this— there may be different markers assigned to different questions , and your answer may be missed if it is in the wrong place.  A significant number of scripts that used Additional Page to write solutions did not indicate on the original allocated solution space to inform the marker to refer to the Additional Page. 跚跛跜距 If you use an Additional Page, always write “See Additional Page” clearly in the allocated space, or the marker may miss your continued work.  Most scripts were well presented, with clear and logical mathematical arguments.踀踁 However, in some cases: o the handwriting was difficult to read, or o the writing was too small, making it illegible, or o the ink was too light, making answers illegible  Reminder: Markers can only award marks for work they can read. Please ensure your handwriting is neat and legible, and that you write at a reasonable size. Solutions 1 Given that the curve y ax b x  passes through ( 1, 4)  and intersects another curve 3y cx  at 3x and 3x , find the values of a, b and c. [4] [4] Substituting ( 1, 4)  into y ax b x  , 4 4 (1)a b a b       At points of intersection, 3ax b x cx   When 3x , 3 3 3 3 1 (2) a b c a b c          When 3x , 3 3 3 3 c 1 (3) a b c a b         Using GC, 3a , 1b , 3c Most students were able to solve this question and arrive at the correct answer. However, a significant number chose to solve the simultaneous equations by hand, and some made careless mistakes in the process.

2 ______________________ 2025 Yr 5 H2 Math Timed Practice Paper Solution with Comments 2 Referred to an origin O, the position vectors of three points A, B and C are 3 2 5 i j k , 4 2 i j k and 2 + 4 i j k respectively. (a) Determine the value of  for which points O, A, B and C are coplanar. [2] (b) Given that OD OA OB        , find the value of  for which OD   is perpendicular to the y – axis. [2] (c) Find the exact area of triangle OAD. [2] (a) [2] Method 1 A vector normal to plane OABC 3 1 16 2 4 11 5 2 14 OA OB                                    So, 16 16 11 2 11 0 14 4 14 16 22 56 0 39 8 OC                                          Method 2 Given the points are coplanar, 3 1 2 2 4 4 5 2 s t                               , where s and t are constants. 3s t   --- (1) 2 2 4 s t  --- (2) 4 5 2s t  --- (3) Using GC to solve the above 3 eqns, we get 39 5 9, , 8 4 8s t    . Some students misunderstood the term “coplanar,” mistaking it for “collinear,” which led to incorrect interpretations in their answers. Note that when we say “four points are coplanar”, it means that there is a plane that the four points can lie on. (b) [2] 3 1 2 4 5 2 OD OA OB                               A direction vector of the y-axis is 0 1 0         . If OD   is perpendicular to the y – axis, This part is well done.

3 ______________________ 2025 Yr 5 H2 Math Timed Practice Paper Solution with Comments 0 1 0 0 3 0 2 4 1 0 5 2 0 2 4 0 1 2 OD                                        (c) [2] 3 1 3.5 1 1 2 4 02 2 5 2 4 OD OA OB                                       Area of OAD 1 2 3 3.5 8 1 1 2 0 5.52 2 5 4 7 1 64 30.25 492 1 5734 OA OD                                        The part is well done

4 ______________________ 2025 Yr 5 H2 Math Timed Practice Paper Solution with Comments 3 The function f is defined by   f : 3 1 , where , 1x x x x x     . (a) Find   1f x and state the domain of 1f . [4] (b) Sketch on the same diagram the graphs of  fy x and   1fy x  , giving the coordinates of any points where the curves cross the axes. [3] The function g is defined by g : e , where , 0xx x x     . (c) Show that the composite function fg exists and state the range of fg. [2] (a) [4]          2 2 2 3 1 , 1 2 3 1 4 4 1 y x x x x x x x                  2 4 1 1 4 y x x y       Since 1x , 1 4x y     1f 1 4x x    From the graph of f,  fR , 4  Domain of 1f  fR ,4   A few students did not recall that if 2 a b , then .a b So, they wrote 1 4x y   . Similar Qn  Tut 4 Sect A Qn 2(a), Sect B Qn 3(ii) (b) [3] Recall that  the graphs of f and 1f  are symmetrical about y = x.  the x- and y-scale should be the same  the question ask for the intercepts in coordinates-form  it is important to write down the coordinates of the endpoints on the graphs, so that one will know where does the graph starts from/ends at. Similar Qn  Tut 4 Sect B Qn 5(iii) O x O y y x (3,0)

5 ______________________ 2025 Yr 5 H2 Math Timed Practice Paper Solution with Comments (c) [2] For fg to exist, g fR D .    g fR 0,1 , D ,1   Since g fR D , fg exists. Using the graphs of f and g,       g f g g fgD 0, 0,1 3, 4 R R       Alternative Method to find fgR : Consider the graph of fg,     fg 3 e e 1 , 0x xx x       fgR 3, 4  Most students are able to recall the condition for composite function to exist. Similar Qn  Tut 4 Sect B Qn 2, Qn 6(ii) x y O 4 x y O

6 ______________________ 2025 Yr 5 H2 Math Timed Practice Paper Solution with Comments 4 It is given that 2 i is a root of the equation 4 3 2 2 15 0z az bz az     . (a) Find the values of the real numbers a and b , and without the use of a calculator, find the other roots of the equation. [6] (b) Deduce the roots of the equation 4 3 22 i i 15 0w a w bw a w     . [2] (a) [6] Method 1 Since all coefficients of the equation are real and 2 i is a root, then 2 i is also a root. The quadratic factor of the equation is             2 2 2 i 2 i 2 i 2 i 2 1 4 5 z z z z z z z                           Let 4 3 22 15z az bz az       2 2 4 5 2 3z z z cz           4 3 22 8 7 4 5 12 15z c z c z c z        Comparing coefficients of 3z and z , 8 5 12 4 20 5 c a c c c        Thus 5 8 13a   and  7 4 5 27b              4 3 2 2 2 2 13 27 13 15 0 4 5 2 5 3 0 2 i 2 i 2 1 3 0 z z z z z z z z z z z z                      Hence the other roots are 12 i, and 3.2 