SJI 2022 Year 6 MAAHL Prelim Paper 3 Solutions

2022 HLMAA P3 ms 1 1a Using scalar product, 1bi Negative 1bii 1biii From (ii), m=2.27, 25.7. But when m is positive, is positive, implying that p is acute. Hence, there will be no m such that p is obtuse. 1c Smallest angle = 17.2o m = 5.57 1di 1−23⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟·3−2m⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟=1−23⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟3−2m⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟cosp1(3)+(−2)(−2)+3(m)=12+−2()2+3232+−2()2+m2cosp7+3m=1413+m2cospcosp=7+3m1413+m2p=arccos7+3m1413+m2⎛⎝⎜⎞⎠⎟cos150°=7+3m1413+m2−32=7+3m1413+m234=7+3m()21413+m2()!3m2−84m+175=07+3m1413+m2 limm→∞7+3m1413+m2=limm→∞7m+31413m2+1=0+3140+1=314

2022 HLMAA P3 ms 2 1dii 1diii or 1e 1f limm→∞p=limm→∞arccosgm()()⎡⎣⎤⎦=arccoslimm→∞gm()⎡⎣⎤⎦=arccos314⎡⎣⎢⎤⎦⎥=36.7°limm→−∞p=180°−36.7°=143.3°arccos−314⎛⎝⎜⎞⎠⎟=143.3° 17.2≤p<143.31−23⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟×3−2m⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟=1−23⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟3−2m⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟sinp−2m+6−m−9()4⎛⎝⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟=1413+m2sinp−2m+6()2+9−m()2+42=1413+m2sinp!sinp=5m2−42m+1331413+m2()p=arcsin5m2−42m+1331413+m2()⎛⎝⎜⎜⎞⎠⎟⎟

2022 HLMAA P3 ms 3 1g 1h Part (e) = Part (g) = Functions reason Range of arccos(x) is while range of arcsin(x) is . Hence, the upper bound for part (e) is different from part (g). OR Vectors reason Defn of angle between 2 vectors allows for angles to lie in quadrants 1,2. Hence, scalar product finds the true angle between 2 vectors. While cross product defn is only for acute angle between 2 vectors. 2a 2b LHS = Fo = 3 RHS = F1 – 2 = 5 – 2 = 3 So, LHS = RHS 2c n = 2 means . So, LHS = 3(5) = 15 = 17 – 2 = RHS n = 3 means So, LHS = 3(5)(17) = 255 = 257 – 2 = RHS 17.2≤p≤9017.2°≤p<143.3°17.2°≤p≤90°0°,180°⎡⎣⎤⎦−90°,90°⎡⎣⎤⎦F0=3,F1=5,F2=17,F3=257F0F1=F2−2F0F1F2=F3−2

2022 HLMAA P3 ms 4 2d Let P(n) be the statement where When n = 1, LHS = Fo = 3, RHS = F1 – 2 = 5 – 2 = 3 So, LHS = RHS, and P(1) is true. Assume P(k) is true for some . Ie Consider n = k+1, LHS = RHS So, P(k+1) is true. Since P(1) is ture, and P(k) true imples P(k+1) true, by MI, P(n) is true for all positive integers n. 2ei Since , then Hence, p is a factor of . 2eii Since , p is also a factor of Fm – 2. 2f p is a factor of both Fm and Fm – 2. 2gi Part (f) is impossible as Fm and Fm – 2 are consecutive odd numbers. So no numbers other than 1 can be a common factor for both Fm and Fm – 2. 2gii By proof by contradiction, p cannot be a common factor for two Fermat numbers. OR no two Fermat numbers have a common factor greater than 1. 2h Factors for F1 = 5 are 1, 5 Factors for F2 = 17 are 1, 17. 2i 1 2j Each distinct prime factor can only appear in one and only one Fermat number. 2k Since there are infinitely many Fermat numbers, there must be infinitely many prime numbers. F0F1F2...Fn−1=Fn−2Fn=22n()+1,n∈!+k≥1,k∈ZF0F1F2...Fk−1=Fk−2=F0F1F2...Fk−1Fk=Fk−2()Fk=22k()+1−2⎡⎣⎢⎤⎦⎥22k()+1⎡⎣⎢⎤⎦⎥=22k()⎡⎣⎢⎤⎦⎥22k()⎡⎣⎢⎤⎦⎥−1=22k()+2k()−1=22k+1()−1=Fk+1−2Fl=apF0F1...Fl...Fm−1=F0F1...ap()...Fm−1F0F1...Fl...Fm−1F0F1...Fl...Fm−1=Fm−2