SOTA 2022 Year 6 MAA HL Prelim Paper 2 Solutions

Yr6/5/MATAA/HP2/Aug2022 18 pages © School of the Arts, Singapore Year 6 Mathematics: analysis and approaches Higher level Paper 2 Preliminary Examinations Tuesday 30 August 2022 2 hours Instructions to candidates • Do not open this examination paper until instructed to do so. • A graphic display calculator is required for this paper. • Section A: answer all questions. Answers must be written within the answer boxes provided. • Section B: answer all questions on the answer sheets provided. Write your name and class on each answer sheet, and attach them to the examination paper. • Unless otherwise stated in the question, all numerical answers should be given exactly or correct to three significant figures. • A clean copy of the mathematics: analysis and approaches formula booklet is required for this paper. • The maximum mark for this examination paper is [110 marks]. Section A Section B Question Marks Question Marks 1 10 2 11 3 12 4 5 6 7 8 9 Name: 110 Class: Index: Solution

- 2 - Yr6/5/MATAA/HP2/Aug2022 Full marks are not necessarily awarded for a correct answer with no working. Answers must be supported by working and/or explanations. Solutions found from a graphic display calculator should be supported by suitable working. For example, if graphs are used to find a solution, you should sketch these as part of your answer. Where an answer is incorrect, some marks may be given for a correct method, provided this is shown by written working. You are therefore advised to show all working. Section A Answer all questions. Answers must be written within the answer boxes provided. Working may be continued below the lines, if necessary. (a) Or (b) Mtd 1 Yes. From (a), when , Mtd 2 Let Then and . Thus is a solution to (a). x=3−λ,y=10−2λ,z=λλ∈!λ=−1(x,y,z)=(4,12,−1)x=3−λ=4⇒λ=−1y=10−2(−1)=12z=λ=−1(4,12,−1) 1. [Maximum mark: 5] (a) Solve the system of linear equations: [3] ()(b) (b) Determine whether is a solution to (a). [2] 4x−3y−2z=−18y+2z=10x+z=3(4,12,−1)

- 3 - Yr6/5/MATAA/HP2/Aug2022 Turn over (a) (shown) (b) = 2.206356… f(w)dw05∫=1kw2+25dw05∫=1k[15tan−1(w5)]05=115k[(π4)−0]=1k=20π 2. [Maximum mark: 5] The probability density function of the continuous random variable is defined by (a) Show that . [3] (b) Find , the expected value of . [2] Wf(w)=kw2+250≤w≤5,0otherwise.⎧⎨⎪⎩⎪k=20πE(W)W

- 4 - Yr6/5/MATAA/HP2/Aug2022 3. [Maximum mark: 7] A company manufactures washing detergent. They gave a number of identical coloured shirts to some customers and asked them to keep a record of how often the shirts were washed. After four months, the company went back to the customers and note down how many times the shirts had been washed, and the colour intensity of the shirt. Number of washes, N 5 7 9 4 13 8 7 Colour Intensity, I 17 14.2 9 15.4 8.3 9.4 9 The relationship between N and I can be modelled by the regression line with equation Find the value of a and of b. (ii) Write down the value of Pearson’s product-moment correlation coefficient, r, for these data. [4] Interpret, in context, the value of a found in part (a)(i). [1] A customer washed a shirt ten times. (c) Use your regression line to estimate the colour intensity of the shirt. [2] Q3(a)(i) (ii) –0.781 b In every additional one wash, the colour intensity decreases/weaken by 0.967. c 9.41 (b) (a) (i)

- 5 - Yr6/5/MATAA/HP2/Aug2022 Turn over (a) No. of arrangements ways (b) 6 consonants D, N, C, F, L, M 3 vowels A, E, I Case Selections A) 3C 1V B) 4C 0V Total no. of selections = 60 + 15 = 75. "EA"1way!"#XXXXXXX7!ways!"##$##=1×7!=504063⎛⎝⎜⎞⎠⎟×31⎛⎝⎜⎞⎠⎟=20×3=6064⎛⎝⎜⎞⎠⎟×30⎛⎝⎜⎞⎠⎟=15×1=15 4. [Maximum mark: 5] Consider the nine letters from the two words “DANCE” and “FILM”. (a) Find the number of 9-letter arrangements which begins with “EA” in that order. [2] (b) Four letters are selected at random. Find the number of selections which contain more consonants than vowels. [3]

- 6 - Yr6/5/MATAA/HP2/Aug2022 5. [Maximum mark: 6] The probability of a bolt being faulty is 0.2. A random sample of 25 bolts is tested. (a) Find the probability that there are more than 3 faulty bolts in the sample. [3] These bolts are sold in bags of 25. Glenn buys 15 bags. Find the probability that exactly 8 of these bags contain more than 3 faulty bolts. [3] Q5a Let X be the number of faulty bolts. b (b)

- 7 - Yr6/5/MATAA/HP2/Aug2022 Turn over 6. [Maximum mark: 8] The diagram above shows an isosceles triangle ABC where AB = BC = 4 cm and AC = 6 cm. The arc XC is part of a circle with centre A and radius 6 cm. The arc YC is also part of another circle with centre B and radius 4 cm. Point A, B, X and Y are on a straight line. (a) Show that angle CBY is 1.45 radians. [3] Find the perimeter of the shaded region. [5] Q4a b (b) A B X Y 4 cm 6 cm C 4 cm diagram not to scale

- 8 - Yr6/5/MATAA/HP2/Aug2022 Required area x=3+2y−y2=0⇒y=−1or3=xdy03∫+(−x)dy34∫=(3+2y−y2)dy03∫+(−3−2y+y2)dy34∫=[3y+y2−13y3]03+[−3y−y2+13y3]34=9+73=343units2 7. [Maximum mark: 5] Find the exact area of the region enclosed by the curve with equation and the y-axis for . x=3+2y−y20≤y≤4

- 9 - Yr6/5/MATAA/HP2/Aug2022 Turn over (a) Let Mtd 1 By cover-up rule Mtd 2 Comparison method (b) 5−3𝑥!3−5𝑥!−2𝑥" =5−3𝑥!(3+𝑥!)(1−2𝑥!) =23+𝑥!+11−2𝑥! =23(1+𝑥!3)#$+(1−2𝑥!)#$ =23[1−𝑥!3+(𝑥!3)!−+𝑥!3,"+⋯]+[1+2𝑥!+(2𝑥!)!+⋯] =!"+#$%𝑥&+##'&(𝑥)+⋯(up to x4) (=1.667 + 1.778 x + 4.074 x4 +…) f(x)=5−3x3−5x−2x2=5−3x(3+x)(1−2x)5−3x(3+x)(1−2x)=A3+x+B1−2xA=5−3(−3)1−2(−3)=2B=5−3(12)3+(12)=15−3x(3+x)(1−2x)=A3+x+B1−2x=(A+3B)+(B−2A)x(3+x)(1−2x)A+3B=5B−2A=−3⎧⎨⎪⎩⎪ ⇒A=2 and B=1f(x)=23+x+11−2x 8. [Maximum mark: 7] Let 𝑓(𝑥)= !"#$#"!$"%$!. (a) Express into partial fractions. [3] (b) Hence, or otherwise, expand !"#$!#"!$!"%$" in ascending powers of x up to and including the term in 𝑥). [4] f(x)

- 10 - Yr6/5/MATAA/HP2/Aug2022 9. [Maximum mark: 8] A particle moves along the axis with velocity at time t seconds given by for . (a) Find the acceleration of when t = 4. [2] (b) The particle is instantaneously at rest when t = T1 and t = T2 where . Find and . [2] (c) Show that passes the starting point twice and find the distance travelled by at these instants. [4] (a) Or Acceleration = (b) T1 = 0.136 s, T2 = 18.1 s (c) Displacement at time P passes starting point displacement = 0. Mtd 1 (Graphical) Solving gives T = 0.28492 s or T = 24.344 s Mtd 2 (nsolve) nsolve() gives T = 0.28492 s nsolve() gives T = 24.344 s Distance travelled for Distance travelled for Px−vms−1v(t)=2−cot2t+13⎛⎝⎜⎞⎠⎟0≤t≤25PPT1<T2T1T2PP 0.0803ms−2(correct to 3s.f.) =s(T)=v(t)dt0T∫⇒ s(T)=0v(t)dt0T∫=0,T,0.2v(t)dt0T∫=0,T,250≤T≤0.28492=|v(t)|dt00.28492∫=0.04659m0≤T≤24.344=|v(t)|dt024.344∫=54.728m