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RAFFLES INSTITUTION H2 Mathematics (9758) 2026 Year 5 _____________________________________________________ 2026 H2 Math Y5 Promotion Exam: Qns, Solns and Mark Scheme Page 1 of 35 2026 H2 Math Promotion Examination: Questions, Solutions and Comments 1 [It is given that a sphere of radius r units has surface area 24 r .] The diagram shows a sphere inscribed in a cylinder with base radius r units. The sphere remains inscribed within the cylinder as the volume of the cylinder increases at a rate of 2 units3 per minute. Find the exact rate of change of the surface area of the sphere when the volume of the cylinder is 16 units3. [4] Solution Comments Similar question for reference: C4B Additional Practice Q9b [4] Volume of the cylinder,  2 32 2V r r r   Differentiate wrt t, 2 2 2 2 d d 6d d d 1 d 2 1 d since 2d 6 d 6 3 d V r rt t r V V t r t r r t           Surface area of sphere, 24S r  Differentiate wrt t, 2 d d 8 88d d 3 3S r r rt t r r     When 16V  , 3 32 16 8 2r r r        d 8 4 d 3 2 3 S t   Exact rate of change of surface area of sphere is 43 units2 per minute. A lot of students misread the qn and took 16 units3 to represent volume of the sphere rather than that of the cylinder. This led to an incorrect expression for ddrt and an incorrect value of r. A lot of students also attempted to find d d S V by first finding S in terms of V , i.e. 2 34π 2πVS       . This approach was unnecessarily tedious and often led to careless algebraic errors. Another common error was differentiating V w.r.t r without first expressing V in terms of r only, i.e. 2 2 dπ 2π 4πdVV r h rh r r    In this case, h is treated as a constant, which is incorrect, since h varies with r. Students should first use the relationship 2h r to write V in terms of r only before differentiating wrt r. r

Raffles Institution H2 Mathematics 2026 Year 5 ________________________________________________________________________________________ _____________________________________________________ 2026 H2 Math Y5 Promotion Exam: Qns, Solns and Mark Scheme Page 2 of 35 2 (a) Solve the inequality 22 5 7 2 7x x x    using an algebraic method. [4] (b) Hence, solve the inequality 22 5 7 2 7x x x    . [2] Solution Comments Similar question for reference: (a) C3 Tutorial Section B Q4 (b) C3 Tutorial Section B Q2 (a) [4]            2 2 22 2 2 2 2 2 2 5 7 2 7 2 5 7 2 7 2 7 ( 1) 2 7 2 7 (( 1) 1) 0 2 7 ( 2) 0 3.5 or 0 2 x x x x x x x x x x x x x x x x                       Alternative:          2 2 2 2 5 7 2 7 (*) 2 5 7 2 7 and 2 5 7 2 7 2 7 0 and 2 7 2 0 7 70 or and 22 2 7 or 0 22 x x x x x x x x x x x x x x x x x x                            Some students who attempted this method only squared the RHS to give   222 5 7 2 7x x x    . Others who squared both sides correctly later divided the inequality by   2 2 7x without considering the case 7 2x  . Most students attempted the alternative method. Many students used ‘or’ instead of ‘and’ but still went on to find the intersecting region for the final solution. Instead of x a a x a     , some had x a x a x     . When finding the intersection of the inequalities, many of the students missed out 7 2x  . (b) [2] Replace x with x , and note that the inequality sign is now a strict inequality, there is no equality case at 3.5x   . 0 2 2 0.x x      The most common mistake was not observing the change in the inequality sign from  to  . –3.5 0 2

Raffles Institution H2 Mathematics 2026 Year 5 ________________________________________________________________________________________ _____________________________________________________ 2026 H2 Math Y5 Promotion Exam: Qns, Solns and Mark Scheme Page 3 of 35 3 (a) It is given that e e 4 2x y y   and that 0y when 0x . (i) Show that   2 2 2 d de e 4 e 0d dx y y y yx x       . [2] (ii) Find in exact form, the first two non-zero terms of the Maclaurin expansion of y . [2] (b) Given that  is sufficiently small, show that   2 sin 2 14 1 tan 2 a b c           , where , and a b c are real constants to be determined. [4] Solution Comments Similar question for reference: (a) C7C Tutorial Section B Q6 (b) C7C Tutorial Section B Q7b (a)(i) [2] e e 4 2x y y   Differentiate wrt x:   d de e 4 0 (*)d d de e 4 0 d x y x y y y x x y x        Differentiate wrt x:     22 2 d de e 4 e 0 d dx y y y yx x         (shown) Generally well done. Common mistakes include:  d d4 4d d yy yx x   d de ed d y y yyx x   d 2 2dx  (a)(ii) [2] When 0x , 0 (given)y , d 1d 5yx  2 2 d 26 or 0.208d 125yx   , First two non-zero exact terms of the Maclaurin expansion of y are 21 135 125x x  . Generally well done. Mistakes include 0e 0 , 2y  (even though it was already given that 0y  in the question), d 1d 5yx  , 2 2 d 26 125d y x  .

Raffles Institution H2 Mathematics 2026 Year 5 ________________________________________________________________________________________ _____________________________________________________ 2026 H2 Math Y5 Promotion Exam: Qns, Solns and Mark Scheme Page 4 of 35 (b) [4]          2 1 2 2 2 sin 24 1 tan sin cos 2 cos sin 24 4 1 tan 1 11 2 2 1 since is small22 1= 1 2 2 1 2 1 1 3 , where 1 and 3 2 a c b                                               Some students applied small angle approximation for cos and sin instead. Some students did not know that tan  for small . They rewrote sintan as cos  but were unable to simplify further. Many students did not continue with binomial expansion of   11   , or applied binomial expansion wrongly to get either 1 or 1  .

Raffles Institution H2 Mathematics 2026 Year 5 ________________________________________________________________________________________ _____________________________________________________ 2026 H2 Math Y5 Promotion Exam: Qns, Solns and Mark Scheme Page 5 of 35 4 A curve C has equation 3 3 3x y xy  . (a) Use calculus to find the exact equations of tangents to C that are parallel to the x-axis. [4] (b) Show that C is symmetrical about the line y x . Hence write down the equations of tangents to C that are parallel to the y-axis. [2] (c) Describe geometrically the region enclosed by the tangents found in parts (a) and (b). [1] Solution Comments Similar question for reference: C4B Additional Practice Q2 (a) [4] 3 3 3 (*)x y xy   Differentiate wrt x: 2 2 2 2 d d3 3 3 d d d d y yx y x y x x y x y x x y          2 2dWhen 0, dy x y y xx     Substitute into (*): 3 6 3 3x x x    3 3 2 0x x   30 or 2x x  Since 2y x , 30 or 4 respectivelyy y  Equations of tangents that are parallel to x-axis are 30 and 4y y  . Most students can differentiate reasonably well to get an expression for d d y x . Those who did not get the correct expression made algebraic errors. A number of students stopped at 2y x , thinking that that is the final answer. However, a horizontal tangent needs to be of the form y k , where k is a constant. (b) [2] 3 3: 3 (*)C x y xy   Replacing y by x and x by y: equation of C becomes 3 3 3y x yx  which remains unchanged. C is symmetrical about the line y x . Hence, equation