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 2x p qx  2x qx p  2x qx p   2 q x p   2 q x p  2 px q  2 px q  (rejected 02 p q  since q ≥ 2) For students who used the alternative method, do pay attention to the correct reasons for rejecting the other answer.

3 Find (a)  2tan 1 dx x , [2] (b) 1sin 2 dx x  . [3] [Solutions] Remarks (a)       2 2 tan 1 d sec 1 1 d tan 1 x x x x x x c           - Students to note how to handle trigo with power 2 and linear angle! 1) 2 2sec , cosec : can integrate directly! 2) 2 2tan , cot : use trigo identity to convert to (1)! 3) 2 2sin ,cos : use double angle formula - Students must remember the trigo identity formula correctly! (b)       1 1 2 2 sin 2 d si 2 n 2 d 1 x x x x x x x             1 1 2 2 1 2 2 1 1 2 1 2 1sin 2 1 4 d 4 1 41sin 2 4 1sin 2 1 4 8 2 x x x x x x x c x x x c x                  - Students pls note that : “Inverse  Reciprocal” - Do not mix up 1d sind xx  with 1sin d x x  where we need to use by part for integration! - Must remember chain rule for differentiation and note the angle for trigo when applying the formula “   1d sind 2x x   2 2 1 2x ” - Students to remember the following formula 1for n   , identify carefully what is the f(x) that will affect the f ( )x and to divide by the new power: 1[f ( )][f ( )] 1f ( ) nn xx c nx     I C   I D           2 2 2 2 2 sin d cos d cosec d sec d cot d ax b x ax b x ax b x ax b x ax b x          

4 Do not use a calculator in answering this question. (a) It is given 3 iw  . (i) Find arg w. [1] (ii) Express 8iw in the form ier  where r > 0 and .     [3] (b) (i) It is given that   2 1 i 3 4ia    . Find the value of the real constant a. [2] (ii) Hence solve the equation     22 3 2i 1 i 0z z      . [3] [Solutions] Remarks (a)(i) 3 iw  (2nd quadrant) 1tan 3 6 y x       arg w 5 6 6     As calculators are not allowed, detailed workings should be given. Note that 1 1tan 3         gives .55, , ,....6 6 ..6    The principal argument is the angle such that .     (ii)   2 23 1 2w    88 8i i 1 2 256w w          8arg i arg i 8arg 582 6 43 7 6 ,6 6 w w                     Principal argument 5 6  Thus 8 5 i6i 256ew   Alternatively, from (i), 5 i62ew   8 8 8 5 40 2 i i i6 6 3 2e 2 e 256 ew           8 2 32 2 i3 7 i i6 5 i 6 i2i 256e 256e 256e e e 256 w                  i 1 arg i 2     Note that 43 5 6 6   r is the modulus and must be a positive real number. - What about 8iw ? Note that   i 2i = 1 e  - What is the geometrical relationship between i and iz? α 1

(b)(i)   2 2 1 i 3 4i 1 2 i 3 4i a a a        Comparing imaginary parts, 2 = 4a  a = 2 [Check: real parts = 2 21 1 2 3a    ] (ii)     22 3 2i 1 i 0z z      Using the quadratic formula,       2 3 2i 3 2i 4(2) 1 i 2(2) 3 2i 3 4i i 4 3 2i 4 11 i 2 1 2i using result n or i) ( z               Hence means to use the result in part (i)     2 1 2i 3 4i 3 4i 1 2i          to obtain the roots of the equation. Use earlier result in (i)!

5 The points A and B have position vectors a and b respectively. C is the point on line OB such that AC is perpendicular to OB. (a) By using a suitable scalar product, or otherwise, show that   2OC  a b b b    . [3] (b) Give a geometrical interpretation of a b b  . [1] (c) It is given that 3 3 1        a and 1 0 h        b . Given also that the length of the line segment AB is 5 units and angle AOB is an obtuse angle, find the exact value of h. [4] [Solutions] Remarks (a) Since C is a point on line OB, OC b     for some   AC is perpendicular to OB 0AC b       0b a b     0b b a b       2 b a b     2 a b b      Thus 2 a bOC b b        (shown) It is important to not just read the question but also to process the information provided. A good practice is to annotate on the question what each key piece of information translates to as you read the question e.g. … C is the point on line OB… (jot down OC b     ) … AC is perpendicular to OB… (jot down 0AC OB       ) (b) a b bOC b b               (1)a b a ba b bOC b b b b                 This is the length of projection of a onto b . OR this is the length OC. Need to be careful of the term used. Length of projectio