2024 Y6 Timed Prac Rev Paper 5 (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2023 Year 6   ________________________________________________ 2023 Y6 H2 Math Common Test: Solutions with comments Page 1 of 23     2023 Year 6 H2 Mathematics Common Test: Solutions with Comments 1 The points A, B and C have position vectors ,a b and c respectively. The points A and B are fixed while C varies. (a) Given that ,a b and c are non-zero vectors such that () ()ca b b ca , find the relationship between ()ca  and b . [2] (b) Hence describe geometrically the set of all possible positions of the point C. [2] Solutions Comments (a) [2] Since  () ()bc a c ab  , () () ca b b ca   () ()ca b ca b  2( )  ca b 0   ()ca b 0 Hence, () ca and b are parallel or ca . ()ca b 0 implies sin 0ca b  , where  is the angle between ()ca and b. Thus either 0ca or sin 0  (Note b0 ) (b) [2] From (a), since ca or () ca and b are parallel. () , , kk kk     ca b ca b   Point C lies on the line passing through the point A and the line is parallel to the vector b. Note that ca when 0k .     A 2024 Y6 H2 Math Timed Practice Revision Paper 5 Solution

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________   ________________________________________________ 2023 Y6 H2 Math Common Test: Solutions with comments Page 2 of 23       2 Complete the square for 229 5 .xx    [1] The function f is defined by 2f : 2 9 5 , for .xx x x   (a) Sketch the graph of f( ),yx stating the coordinates of any turning points and points of intersection with the axes. [2] (b) If the domain of f is further restricted to 1, xk state with a reason the largest value of k for which the function 1f exists. [2] In the rest of the question, the domain of f is ,1 2 .xx  (c) Find the solution set of 1f( ) f ( )xx  . [1] The function g is defined by 1g: x x , for x  , 05 x .  (d) State whether the composite function 1fg exists, justifying your answer. [2] (e) If the domain of g is further restricted to pxq and the composite function gg exists, find an equation involving p and q. [2] Solutions Comments [1] 2 2 94 129 5 2 48xx x     (a) [2] Take note of details when drawing a graph. 5.125 > 5, so the turning point should be slightly higher than the y- intercept (0,5) Graph should also be symmetrical about 2.25x (0.649, 0) (3.85, 0) or 94 1,48  

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________   ________________________________________________ 2023 Y6 H2 Math Common Test: Solutions with comments Page 3 of 23     (b) [2] Largest value of k is 9 4 . Since the line 41,2 8ym m  , cuts the graph of f( )yx at exactly one point, it is a 1-1 function and so f -1 exists. (c) [1] Note : since 12 x , f -1 exists. Method 1 The solution set of 1f( ) f ( )xx  is the same as that of f( )xx . 2(2 9 5)xx x   228 5 0xx     0.775 or 3.22 (3 s.f.)x   No solution for 12 x . The solution set is {} or  . Method 2 The solution set of 1f( ) f ( )xx  is the same as that of f( )xx . Since the graph of f do not intersect the line yx for 12 x , the solution set is {} or  . While we can solve the quadratic equation to get 0.775,3.22x , we need to check them against the domain. Note that set of no solution is either {} or  , known as the empty set. (d) [2] For 1fg to exist, we need 1gf fRD R  . Since g 1R, 5   and fR [2, 5] , 1gf fRD R  , and so 1fg does not exist. Take note that the domain of f is ,1 2 .xx  (e) [2] Let Dg = [p, q]. Then g 11R, qp  . For the composite function gg to exist, we need to have gg 11RD , , pqqp    . Therefore 1 pq  and 1 qp  1 and 1pq pq  so 1pq , where 05 pq . Note that g is a decreasing function    [ [ ] ]

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________   ________________________________________________ 2023 Y6 H2 Math Common Test: Solutions with comments Page 4 of 23     3 Cadence, the pedal speed in cycling, is measured in pedal stroke revolutions per minute (RPM). For example, a cadence of 60 RPM means that one pedal makes a complete revolution 60 times in one mi nute. Every complete revolution of the pedal stroke of a stationary bicycle is equivalent to cycling an approximate distance of 0.006 km. Lisa signed up for a beginner spinning class. The workout routine requires her to start cycling at 30 RPM consistently for the first minute and progressively increase the cadence by 2.5 RPM for every subsequent minute. (a) At which minute will Lisa cycle at 120 RPM? [2] (b) Find Lisa’s average speed of cycling, in km/h, for the first 15 minutes. [3] Celina signed up for an intermediate spinning class. The workout routine requires her to start cycling at 25 RPM consistently for the first minute and increase the cadence by 15% for every subsequent minute. (c) At which minute does Celina’s cadence first exceed 120 RPM? [2] (d) If Lisa and Celina start their workout routine at the same time, at which minute does Celina’s total distance cycled first exceed Lisa’s total distance cycled? [3] Solutions Comments (a) [2] AP: 30, 2.5ad 120 30 ( 1)(2.5) 37 n n   Lisa will cycle at 120 RPM at the 37th minute. (b) [3] 15 15 2(30) (14)(2.5)2 712.5 revolutions S     15Average speed 712.5 0.006 60 17.1 km/h   

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________   ________________________________________________ 2023 Y6 H2 Math Common Test: Solutions with comments Page 5 of 23     Alternative Method using mean PRM Mean PRM for first 15 minutes 1 2(30) (14)(2.5) 47.52    Average speed 47.5 0.006 60 17.1 km/h    (c) [2] GP: 25, 1.15ar   1 25 1.15 120 1 11.223 12.223 n n n         Celina’s cadence will first exceed 120 RPM at the 13th minute.  (d) [3] Let the time taken by Celina’s total distance to first exceed Lisa’s to be n. 25(1.15 1) 0.006 2(30) 2.5 1 0.0061.15 1 2 n n n      25 (1.15 1) (57.5 2.5 ) 00.15 2 n n n     Let 25 (1.15 1) (57.5 2.5 )0.15 2 n nYn   , by GC, 5.8627n Or use the table of values: n Y 5 6.44 0 6 1.34 > 0 Celina’s total distance cycled first exceed Lisa’s total distance cycled at the 6 th minute. Reminder: when using GC, you still need to provide working, such as the inequalities. Table of values can be used if the unknown to be solved is an integer and you need to show the change in values, in this case < 0 and > 0.

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________   ________________________________________________ 2023 Y6 H2 Math Common Test: Solutions with comments Page 6 of 23     4 A curve C has parametric equations  1 22 2x at , lny t , for 0 ta  and 1.a (a) C crosses the x-axis at the point P. Find the x-coordinate of P in terms of a. [1] (b) Show that d 0d y x  for 0 ta . [3] (c) Describe the behaviour of the tangent to C as 0t . [1] (d) Sketch the graph of C, stating the equations of any asymptotes and the coordinates of the points where the curve meets the axes. [2] (e) The tangent at P cuts the y-axis at the point Q and the normal at P cuts the y-axis