2021 H2 Math RI Prelim Paper 2 Solns

2021 Y6 H2 Math Preliminary Paper 2: Solutions with comments Page 1 of 28 2021 Year 6 H2 Math Preliminary Paper 2: Solutions with Comments Section A: Pure Mathematics [40 marks] 1 Functions f and g are defined by 2( 1)f: e , , 1g : , , 1 2.2 xxx x xx x − ∈ ∈ ≤<−   (i) Sketch the graph of f( )yx= . [1] (ii) If the domain of f is restricted to xk≥ , state with a reason the least value of k for which the function 1f− exists. [2] In the rest of the question, the domain of f is xk≥ , using the value of k found in part (ii). (iii) Find 1g () x− and show that the composite function 11gf−− exists. [4] (iv) Find the range of 11gf−− . [1] Solutions Comments (i) [1] The graph and its properties can be easily obtained from the GC. However, a good number of students did not label either or both the y- intercept and minimum point. Students should also note that the graph is symmetrical about the line 1x= . (ii) [2] For 1f − to exist, f must be a one-one function. Least value of 1k = . Many students elaborated the horizontal line test instead of stating that the condition is for f to be a 1-1 function. In this case, students have to take note of the precise phrasing “every horizontal line [ ), 1,y kk= ∈∞ , cuts the graph of f at one and only one point.” 2( 1)e xy −=

Raffles Institution H2 Mathematics 2021 Year 6 __________________________________________________________________________________________ 2021 Y6 H2 Math Preliminary Paper 2: Solutions with comments Page 2 of 28 (iii) [4] 1Let 2 12 12 y x x y x y = − −= = − 1 1g () 2x x − = − For 11gf−− to exist, 11fgRD−− ⊆ . [ )1 ffR D 1,− = = ∞ [ )1 ggD R 1,− = = ∞ Since 11fgR = D−− , ∴ 11gf−− exists. Most students managed to find 1g() x− correctly, with a very small number leaving their answers as 1 1g () 2x y − = − , which is incorrect. Most students could remember the condition to check for 11gf−− to exist. Often mistakes were made in either finding 1gD − or keeping fD =  (unrestricted). (iv) [1] 11 gf[1, ) [1, ) [1, 2) −− ∞ → ∞ → [ )11gfR 1, 2−− = Note: [ )1 ffD R 1,− = = ∞ , [ )1 ffR D 1,− = = ∞ OR Since 11fgRD−− = , 11 1 ggf gR R D [1, 2)−− − = = = Most students who used the arrow diagram successfully found 11gfR −− . Students have to note the order of the functions involved and the horizontal asymptote in the graph of 1g− . A number of students also managed to recognize the relationship in alternative method, thus were able to state the answer correctly. 0 0

Raffles Institution H2 Mathematics 2021 Year 6 __________________________________________________________________________________________ 2021 Y6 H2 Math Preliminary Paper 2: Solutions with comments Page 3 of 28 2 (a) Three consecutive terms of a decreasing geometric progression has a product of 5832. If the first number is reduced by 24, these 3 numbers in the same order will form an arithmetic progression. Find the three terms of the geometric progression. [5] (b) The fractal called Sierpiński Triangle is depicted below. Fig. 1 shows an equilateral triangle of side 1. In stage 1, the triangle in Fig. 1 is divided into four smaller identical equilateral triangles and the middle triangle is removed to give the triangle shown in Fig. 2. In stage 2, the remaining three equilateral triangles in Fig. 2 are each divided into four smaller identical equilateral triangles and the middle triangles are removed to give the triangle shown in Fig. 3 and the process continues. Let nT be the total area of triangles removed after n stages of the process. (i) Show that 1 3 16T = . [1] (ii) Find 10T . [3] (iii) State the exact value of lim nn T →∞ . [1] Solution Comments (a) [5] Let the 3 numbers be x r , x and xr , where x is the middle term and r is the common ratio. ( )( ) 3 5832 5832 18 x x xrr x x = = = If the first number is reduced by 24, it is now 24x r − . Since the 3 numbers now form an AP, 24x x x xrr  − −=− Substitute 18x= into the above equation, Most students were able to solve this part successfully, but the efficiency depends on what they wrote as the first 3 terms. Quite many let the 3 numbers be nxr , 1nxr + , 2nxr + , and could not solve due to the extra variable n. They were able to get 1 18nxr + = but Fig. 1 Fig. 2 Fig. 3 … Stage 1 Stage 2

Raffles Institution H2 Mathematics 2021 Year 6 __________________________________________________________________________________________ 2021 Y6 H2 Math Preliminary Paper 2: Solutions with comments Page 4 of 28 ( )( ) 2 18 24 18 18 18 3 10 3 0 31 3 0 rr rr rr  − −=− − += − −= 1 3r∴= or 3r = (rejected  it is a decreasing GP, i.e. 01 r< < ) Thus, the original 3 numbers are 54, 18, 6. could not solve it due to failure to recognise this as the second term. (b) (i) [1] Let A be the original area of the triangle in Fig. 1. ( )( )1 1 1 sin 602 3 4 A= ° = 1 13 4 16TA= = (shown) Students who could not solve this part generally were not able to sieve out the information that there are 4 smaller identical triangles in Figure 2. Many students resorted to finding the height of the triangle in Figure 1 to find the area, when a simple application of the formula suffice. (ii) [3] 1 1 2 1 Area of triangle removed in stage 1 3Area of triangles removed in stage 2 4 3Area of triangles removed in stage 3 4 Area of tria T T T = = =  1 1 3ngles removed in stage 4 n nT − = Students who were able to identify a GP generally managed to get the answer. Many students sort of recognise that there is some addition of the area of triangles, and tried to use the sum of a GP

Raffles Institution H2 Mathematics 2021 Year 6 __________________________________________________________________________________________ 2021 Y6 H2 Math Preliminary Paper 2: Solutions with comments Page 5 of 28 10 29 11 1 1 10 1 Total area of triangles removed after 10 stages, 33 3 44 4 31 4 31 4 0.409 (3 s.f.) T TT T T T  =+ + ++     −  = − =  formula. Not many left the part blank. (iii) [1] 1 3lim 3 41 4 nn TT →∞ = = − OR 3lim 4 nn TA →∞ = = It is good to see that many students, while not able to solve the previous part, make an attempt at part (iii). 3 It is given that ln 1 8e . xy= + (i) Show that ( ) dln 4 e .d xyyy x = [1] (ii) Show that the value of 2 2 d d y x when 0x= is 368e27 . [4] (iii) Hence find the Maclaurin series for 1 8ee x+ up to and including the term in 2x . [2] (iv) Denoting the answer found in part (iii) as ( ) ,g x find the set of values of x for which ( )g x is within 0.5± of the value of 1 8e .e x+ [3] Solution Comments (i) [1] ln 1 8e xy= + Differentiate w.r.t x, ( ) ( ) 1d 1 8ed 2 1 8e 1 d 4e d ln dln 4 e (shown)d x x x x y yx y yx y yyy x = + = = As this is a show question, clear working is vital. Students who did this question in another way should consider these 2 solutions provided as they are the most efficien