TMJC 2026 Prelim P2 Ans

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/02)/Math Dept Page 1 of 20 2026 H2 MATH (9758/02) JC 2 PRELIMINARY EXAMINATION – SOLUTIONS Qn Solution 1 Inequalities (a) (b) From the graph, to find intersection:     2 2 2 1 3 2 1 3 3 2 0 x x x x x x x            3 9 4 2 2 3 17 2 x       From graph, 3 17 3 17 or 2 2x x      y x O a −1 1 a Since there are 2 graphs, you need to label the graphs “Show clearly the axial intercepts” Make sure you label your intercepts and use a for y x a  Note a > 1 so (0, a) is above (0, 1)   , , x a x a y x a x a x a        Note: y x a  intersects   2 1y x  when x < a Make sure to answer the question   2 1 3x x   Note: the keyword is OR It is not comma or “and”

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/02)/Math Dept Page 2 of 20 Qn Solution 2 Vectors (a) 25 2 5OD OB OD   b  2 5AD OD OA    b a       Equation of line AD: OA AD r     2 , 5       r a b a  (shown) (b) OC kOA OC k   a  Equation of line BC: OB BC r       , k    r b a b  To find intersection between line AD and line BC,       2 5 21 1 5 k k                    a b a b a b a b b a Since a and b are non-zero, non-parallel vectors, 1 1 --- (1) 2 1 (2)5 k k               2 1 1 --- from (1)5 2 31 5 5 3 5 2 k k k               Substituting 35 2 k   into line BC,     3 5 2 3 3 15 2 5 2 1 3 2 25 2 OE k k k k k k kk                   b a b a b a b   Alternative 5 5 5 2 k k   Need to use a different parameter from (a) Equate lines AD and BC Compare coefficients of a and b Solve for ߤ in terms of k OK to stop here

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/02)/Math Dept Page 3 of 20 Substituting 5 55 2 kk   into line AD,   5 5 2 5 2 5 3 2 2 5 2 5 2 1 3 2 25 2 kOE k k k k k k kk                            a b a a b a b   (c) Area of triangle OEB 12 OE OB                1 1 3 2 22 5 2 1 3 2 22 5 2 3 ,2 5 2 k kk k kk k k                 a b b a b b b a b b b 0 Note that since 0 1, we have 5 2 0k k    (d) As         3 13 1 11, f2 5 2 2 2 2 5 2 1kk k k        a b a b Geometrical Interpretation of   1 1lim f 2k k      a b a b which refers to the area of triangle OAB. Alternative (by observation, without solving (c)) As 1,k Point E tends to Point A, hence area of triangle OEB tends to area of triangle OAB. Geometrical Interpretation of   1 lim f k k      a b is the area of triangle OAB. OK to stop here Use distributive property (rainbow) of cross product: Answer the question: Geometrical interpretation

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/02)/Math Dept Page 4 of 20 Qn Solution 3 Application of Differentiation (Maxima / Minima) (a) Base area (made up of 6 equilateral triangles with side x m)    21 3 36 sin 602 2x x x      Volume of pyramid 21 3 3 75 33 2 2 x h      2 2 75 75 --- (1) x h h x   By Pythagoras theorem,   2 2 2 2 2 3 04 xk x xk k        By Pythagoras theorem,   2 2 2 2 2 2 3 4 3 04 xl h xl h l           2 2 2 2 22 2 2 4 2 16 2 1 3 6 2 4 3 3 4 3 75 9 4 27 50625 (shown)4 S xl xx h xx h xx x x x                        x 2 x k h l 23 4 x This is 1 of the 6 equilateral triangles of hexagonal base.

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/02)/Math Dept Page 5 of 20 (b) 14 2 3 2 3 d 1 27 50625 101250 27d 2 4S x xx x x             For minimum S, dlet 0dSx  . 1 4 2 3 2 3 3 3 6 1 27 50625 101250 27 02 4 10125027 0 3750 3.9416 (5 s.f.) 3.94 m (3 s.f.) x xx x x x x x x                  When 3.9416x , 2 69.913 (5 s.f.) 69.9 m (3 s.f.) S   Apply chain rule to differentiate S. 1 4 2 2 27 50625 4 x x       is the denominator, hence we can cross multiply and the equation will simplify to 3 3 10125027 0x x  .

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/02)/Math Dept Page 6 of 20 Qn Solution 4 Graphing Techniques, Definite Integral (a) Solution: Thought process Step 1: Sketching the piecewise graph for 0 2 x  :   πsin for 0 1,f 2 2 for 1 2 x a xx x a x                When 0x ,  sin 0 a a  When 1x ,   πf 1 sin 1 2 a a       Note that 1x gives the maximum point of the sine curve. When 1x , 1 2 1 a a     (The graph is continuous at 1x ) When 2x ,  f 2 2 2 a a    O 1 2 3 4 5 6 a (0, a) a + 1 y x ݕ= f(ݔ) O 1 2 3 4 5 6 a (0, a) a + 1 (6, a) y x ݕ= f(ݔ) This is a piecewise function. Since a is an unknown positive real number, students can use 1a to visualise the shape of the graph on the GC, before computing the actual values. Most students showed understanding but were unable to sketch the graph accurately to get the full 3 marks. Students are also reminded to utilise the space efficiently and sketch a bigger graph. Before sketching the shape in, ensure that you have enough space for the entire graph. The spacing for one unit should be consistent throughout the x-axis.

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/02)/Math Dept Page 7 of 20 Step 2: Sketching the periodic part:    f f 2x x  for all real values of x means that the graph is periodic with period 2. Notice that the graph is continuous at 2x , since  f 2 a , which is the same y-value as the starting point of the sine curve. Repeat the basic shape shown on the GC 2 more times the graph of  fy x for 0 6.x  Useful tip: Student should use measure and use construction lines so that graph is sketched to scale. This ensures that the periodic graph looks repeated. For instance: - The three straight lines should be parallel to one another Note that a and a + 1 should be labelled on the should be labelled on the y-axis. The dotted lines are not asymptotes in this case. Step 3: Check ASMILE Note: Ensure open circle and closed circle are drawn clearly at the end point. (b) O 1 2 3 4 5 6 a (0, a) a + 1 (6, a) y x ݕ= f(ݔ) O 1 2 3 4 5 6 a (0, a) a + 1 (6, a) y x ݕ= f(ݔ)   3 0 f dx x is the area under the curve  fy x from 1x to 3x . As this is a periodic function, the two blue areas are the same.

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/02)/Math Dept Page 8 of 20                 3 0 1 2 3 0 1 2 1 2 0 0 1 0 1 0 f d f d f d f d 2 f d f d π 12 sin d 1 2 2 2 π 12 cos 2 1π 2 2 2 12 π 2 4 13 π 2 x x x x x x x x x x x x x a x a a x ax a a a a                                                    1 3 0 2 f d f dx x x x  from the graph. The equation πsin 2xy a       is for 0 1 x  . Do not use the equation πsin 2xy a       for 2 3 x  . From graph,   2 0 f dx x can be found using the formula for area of trapezium. Alternatively:   2 2 0 0 22 1 f d 2 d 22 1 2 x x x a x x x ax a               

TMJC/2026 JC2 Prel