TMJC 2026 Prelim P1 Ans

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/01)/Math Dept Page 1 of 15 2026 H2 MATH (9758/01) JC 2 PRELIMINARY EXAMINATION SOLUTIONS Qn Solution 1 Implicit Differentiation, Normal (a) 2 22 14x xy y   Differentiating w.r.t. x: d d4 2 0 d dy yx x y yx x       n d 2 4d d 4 d 2 show y x y x yx y x y x x y      (b) Since gradient of L is undefined at P d 0dyx  at P 4y x  Substituting 4y x into 2 22 14x xy y       222 4 4 14x x x x     2 1 1 or 1x x x    Since P is in the second quadrant, 1x and 4y Coordinates of P  1, 4 . (c)  Equation of L at P is 1x Substituting 1x into 2 22 14x xy y   , We get 2 12 0y y   Using GC, 4 or 3y y  (reject 4 y which represents pt P) Coordinates of Q  1, 3  . Comments: Generally well done Gradient of NORMAL is undefined  Gradient of TANGENT is zero Substitute condition into the equation of C. Q1 Q2 Q3 Q4 Student needs to be aware of which is the 2nd quadrant. Since gradient of normal is undefined, the normal is a vertical line. Visualisation of the question 1) Give a rough sketch (You need not know exactly how the graph looks like) 2) We can sketch the normal (a vertical line) and question states that it will cut the curve again.

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/01)/Math Dept Page 2 of 15 Qn Solution 2 Integration Techniques (a) 3 2 d1 3 duu x x x    When 2x , 9u When 0x , 1u   2 5 3 0 3 3 9 1 93 2 1 2 0 2 9 1 d 1 1 1 1 1 d3 1 2 23 3 1 4 40123 3 9 d 1 d3 x x x x x x x u u u u u u u u                            (b) 1 1 2 1 2 2 1 1 2 sin d 1sin d 1 sin d 1 1 1sin 12 2 s 1 2 2 in 1 x x x x x x x xx x x x xx x c x x x c                                    2 1d d 3x x u  1 2 sin 1 d 1 dd 1 u x v u v x xx x           2f 1 f ' 2 x x x x    1) Differentiate the substitution 2) Group all x to one side and all u to the other side to make it easier to do the substitution Substitution from x to u should be completed in 1 step:  Expression  dx  Upper and lower limits Recall: න ݑ௡ dݑ= ݑ௡ାଵ ݊+ 1 Keyword: algebraically  Should not use GC and show all working clearly  Final answer is exact Recall:       1 f f ' f d 1 n n xx x x n        Since it is indefinite integral, need the “+ c”

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/01)/Math Dept Page 3 of 15 Qn Solution 3 Differential Equations    2 2 2 2 2 d1 1 (1) d tan 1 sec (2) d d sec (3)d d yx y y x y y y x x               Sub (2) and (3) into (1):   2 2 2 2 2 1 d1 sec tan sec d d tan (shown)d 1 1cot d d 1 ln sin tan x x x x xx x c                    Since 0 2  ,   1ln sin tan x c   2 tan sin 1 yy y      1 2 ln tan 1 y x c y           1 2 1When 0, 1, ln 2 2 1ln tan ln 2. 21 x y c y x y               y 1 ඥ1 + ݕଶ sin ߠ= ݕ ඥ1 + ݕଶ Alternative (longer method) 2 2 tan Differentiate w.r.t , d secd Using chain rule, d d d d secd d d d y y y y x x x            Use “Shown” result to continue attempting the question. DO NOT SKIP when you cannot prove the result. Remember to put modulus for the result of cot d  . Remember to put “ + C ” Must change solution back to involving x and y Oppositetan Adjacent 1 yy 

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/01)/Math Dept Page 4 of 15 Qn Solution 4 Summation and Definite Integral (a)                     4 2 4 2 4 4 3 2 2 1 1 3 2 1 2 7 3 2 3 7 32 25 3 7 2 32 1 2 1 14 22 76 2 11 2 1 3 22 7 283 2 n r n r n n r r n r r r r r r r r nr r n n nn n n n n n n n                                                (b)(i) 1 1 11 ...2 3nS n     (b)(ii)   1 1 1 1 1 11 ...2 3 1lim d lim ln lim ln 1 n n n n n S xx x n                  As   1 1,ln 1 1 ... 2 3n n S        is divergent. Remember to include a bracket after the summation sign. Rewrite 2 4 n r r   as 4 1 2 2 1 1 n r r r r      to change the lower limit to 1r . You can then apply the given formula for 2 1 n r r   . Note:   4 3 7 n r r   is an AP sum where First term (when 4r ) = 3 7(4) 25  , Last term (when r n ) = 3 7 n , No of terms = upper limit – lower limit + 1 = 4 1 3n n    Therefore, using nS formula for AP    no of terms 3First term + Last term 25 3 72 2 n n     1 12 13 1n Note: The width of each of the n rectangles is 1 unit. To find nS , add together the area of the n rectangles. Note: Since the sum of rectangles is an overestimate of the area under the curve 1y x , 1 1 1 1 11 ... lim d2 3 nnS x x     

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/01)/Math Dept Page 5 of 15 Qn Solution 5 Graphing Techniques (a)    2 2 2 13 2 42 2 2 4 2 2 2 8 8 2 x kx Ay x x x x x A x x x A x               Comparing expressions, 8k  (shown) Alternative         2 2 2 4 2 2 13 2 4 4 13 4 8 8 5 x x x kx x x k x x k x              8 0 8 (shown) k k     (b) (c)   22 2 2 2 2 1 2 8 13 12x x xa b x         refers to the intersection between curve C and hyperbola with equation   2 2 2 2 2 1x ya b   . Finding oblique asymptotes of hyperbola, y x ݕ= 2ݔ − 4 ݔ= 2 ൬0, − 132 ൰ (0.419, −6.32) (3.58, 6.32) ݕ= 2ݔଶ − 8ݔ+ 13ݔ − 2 Note: This is a SHOW question, students need to show clearly that 22 13 2 42 2x kx Ay x x x      before comparing the expression. Note: The asymptotes intersect at (2,0). This relationship must be shown clearly on the graph. Intercepts must be given in coordinates form. Turning points should be in 3 s.f. (Use GC to find, you do not need to use differentiation)

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/01)/Math Dept Page 6 of 15 Qn Solution       2 2 2 2 22 2 2 2 0 2 2 2 x y a b xy b a y x b a by x a        Comparing asymptotes of hyperbola and oblique asymptote of curve C, to have 2 intersections (2 real roots) between the graphs, gradient of the asymptote with the positive gradient must be more than 2. 2 2 b a b a    (d) For 1a and 4b , equation of hyperbola   22 2 1 16yx   .     22 2 2 1 16 16 2 16 yx y x       Using GC, 0.107 or 3.89 (3 s.f.)x x  Alternative For 1a and 4b ,   22 2 1 2 8 132 1 16 2 x xx x          Using GC, 0.107 or 3.89 (3 s.f.)x x  Use the given values of a and b and GC to find the intersection between the hyperbola and C.

TMJC/2026 JC2 Preliminary Examination Solutions/H2 Math (9758/01)/Math Dept Page 7 of 15 Qn Solution 6 Graphing Techniques + Definite Integral (a) (b)                       0 0 22 2 1 1 1 0 02 2 2 2 1 1 1 0 0 2 3 2 2 1 1 1 2 3 2 2 2 Volume e 1 π 1 d π e 1 d e 1 π 1 d π e 2e 1 d 1 1e 1 π π e 2e3 2 1 1 1e 1 π 1 1 π e 2e 2 13 2 2 2 1 5e 2e 1 π e 2e π3 2 2 1= 6 x x x x x x x x x x x x x x                                                                        2 2 11e e π3 6      x y O Note that the curves intersect at (0, e – 1) and (–1,0). The asymptote y = – 1 should lie above the min point since – 1 > 1 – e (which is -1.72 to 3 s.f.) There is no need to expand (e − 1)ଶ because it’s a CONSTANT!