2024 TJC JC2 Prelim Exam H2 Maths Paper 1 (Solutions)

2024 TJC Preliminary Exam H2 Mathematics Paper 1 (Suggested solutions) 1 A ball is rolling in a straight line such that its distance away from the starting point, s cm, can be modelled using the equation 4 bs at c t     , where t is the time taken in seconds, and a, b and c are real constants. The ball is at the starting point when 0t  , and moved 10 cm in the first 5 seconds. It moved another 9 cm in the next 16 seconds. Find the ball’s distance away from the starting point when 50.t  [4] [Solutions] Remarks 1When 0, 0, 0 (1) 2 1When 5, 10, 10 5 (2) 3 1When 21, 19, 19 21 (3) 5 t s b c t s a b c t s a b c                  Solving (1), (2) & (3) using GC, 10.08333 or 12 11557.5 or 2 11528.75 or 4 a b c     1 115 115 12 4 2 4 1 115 115When 50, (50) 25.09212 4 2 50 4 s t t t s            Thus the distance away from starting point is 25.1 cm (3 s.f.) - 3 unknowns need 3 equations to solve. To simplify equations to as shown. - To read key word such as “another”, “next” and “starting point”. - To watch presentation of labeling “when….” and to number all equations - Use GC to solve! - Always read back to see objective in this case find s when t=50.

2 On a single diagram, s ketch the graphs of 2 and y x p y qx   where the following conditions are satisfied, indicating the axial intercepts.  p and q are constants, p > 1 and q > 0, and  the graphs have only one point of intersection. [2] (a) State the least value of q. [1] (b) Solve the inequality 2 ,x p qx  leaving your answer in terms of p and q. [2] [Solutions] Remarks Useful Tips: When dealing with modulus curve, it is advisable to label the positive/negative equations on the diagram. Note that 2 , 22 (2 ), 2 px p x x p px p x         When sketching the graphs, ensure that  x and y intercepts are clearly indicated  the graph of 2y x p  is symmetrical about 2 px  the line y qx should be steeper than 2y x p  in order to have only one intersection point  the equation of each graph must be clearly labelled (a) For the graphs to have only one point of intersection, the line y qx has the same or a greater gradient than the line 2y x p  , i.e q ≥ 2. Least value of q = 2 (b) At the intersection point,  2 xx qp    2 2 q x p px q     For 2 ,x p qx  2 px q Note: When solving inequalities using graphical method, we should always attempt to first find intersection points (if any). From (a), intersection point occurs at 2 px . Hence when finding the intersection point, we should equate y = (2 )x p  with y = qx instead of 2y x p  with y = qx. y x O y = qx y = 2x  p y = (2x  p) p Use graph sketched earlier!

Alternative method for finding intersection point 2x p qx  2x p qx  or