A Level H3 Mathematics Solutions (2017-2023 and specimen)

A-Level H3 Mathematics Solutions (2017-2026) Thang Pang Ern October 7, 2024 Contents 1 2017 Specimen Paper Solutions 2 2 2017 Paper Solutions 10 3 2018 Paper Solutions 16 4 2019 Paper Solutions 23 5 2020 Paper Solutions 30 6 2021 Paper Solutions 39 7 2022 Paper Solutions 47 8 2023 Paper Solutions 55 9 2024 Paper Solutions 61 10 2025 Specimen Paper Solutions 62 11 2025 Paper Solutions 69 12 2026 Paper Solutions 69 1

1 2017 Specimen Paper Solutions Question 1 (a) Using the Cauchy-Schwarz inequality, " x y 2 + y z 2 + z x 2 # " y z 2 + z x 2 + x y 2# ≥ x y y z

+ y z z x

+ z x x y 2 " x y 2 + y z 2 + z x 2 #2 ≥ x z + y x + z y 2 x y 2 + y z 2 + z x 2 ≥ x z + y x + z y so we have proven the upper bound for x z + y x + z y. Next, using the AM-GM inequality, we have x z + y x + z y ≥ 3 3 s x z y x z y

= 3 so we have proven the lower bound for x z + y x + z y. (b) (i) By definition of the scalar product, for any two vectors a =   a1 a2 a3   and b =   b1 b2 b3   , we have a •b = |a||b|cos θ, where θ is the angle between the two vectors. Since|cos θ | ≤1, then a •b ≤ |a||b|, which implies that   a1 a2 a3   •   b1 b2 b3   ≤

  a1 a2 a3  

  b1 b2 b3  

and the result follows. For equality to hold, we must have ai = kbi for all 1 ≤ i ≤ 3 and some k ∈ R\ {0}. (ii) Using the Cauchy-Schwarz inequality, " x√y + z 2 + y√z + x 2 + z√x + y 2# h√y + z 2 + √ z + x 2 + √x + y 2i ≥ (x + y + z)2 x2 y + z + y2 z + x + z2 x + y

(y + z + z + x + x + y) ≥ (x + y + z)2 2 x2 y + z + y2 z + x + z2 x + y

≥ x + y + z and equality holds if and only if x = y = z. Question 2 (i) Using the substitution u = x2, the integral becomes Z 3 2 x2 x − 1 dx = Z 3 2 x + 1 + 1 x − 1 dx = ln2 + 7 2 . The substitution is motivated by the presence of the square root in the denominator and the fact that 4 and 9 are square numbers. 2

(ii) Using the substitution y = xu, we have dy dx = x du dx + u 1 x dy dx = du dx + u x so the differential equation becomes du dx = f (u) . (iii) Dividing both sides by x, we have 1 x dy dx = r x y − x y + y x2 so f y x

= r x y − x y . As such, f (u) = 1√u − 1 u = √u − 1 u . The differential equation becomes du dx = √u − 1 u . From (i), we have 2u√u + 3u + 6√u + 6ln |√u − 1| 3 = x + c, where c is a constant. As the solution curve passes through 1 3 , 4 3

, then u = 4. Substituting these into the above equation yields c = 13. Hence, 2 y x r y x + 3 y x

+ 6 r y x + 6ln

r y x − 1 = 3x + 39. When y = 9x, we have 60 + 6ln2 = 3x, so x = 20 + 2ln2, which is the required x-coordinate. Question 3 (i) (a) Let S = {a,2a, . . .(p − 1) a}. For all 1 ≤ i ≤ p −1, none of the ia ∈ S is divisible by p because a is not divisible by p. Suppose ai ≡ a j (mod p). Then, there exists λ ∈ Z such that ai = λ p + a j, so a (i − j) =λ p. However, p does not divide a so p must divide i − j. That is, i ≡ j (mod p). As 1 ≤ i, j ≤ p − 1, then i = j so all the elements in S are distinct. In mod p, the elements in S are a permutation of T , where T = {1,2, . . . ,p − 1}. (b) In mod p, the product of the elements in S is congruent to the product of the elements in T . That is, a · 2a · 3a · (p − 1) a ≡ 1 · 2 · 3 · (p − 1) (mod p). So, ap−1 ≡ 1 (mod p). (ii) By the binomial theorem, (x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5 = x5 + y5 + 5k where k ∈ Z. So, x5 + y5 = (x + y)5 − 5k. As x5 + y5 ≡ 0 (mod5 ), then (x + y)5 ≡ 0 (mod5 ). We use the method of contraposition to prove that x + y ≡ 0 (mod5 ). That is to say, given that x + y is not a multiple of 5, we wish to prove that (x + y)5 is also not a multiple of 5. We write x + y = 5k + r, where 1 ≤ r ≤ 4. So, (x + y)5 = (5k + r)5 = 3125k5 + 3125k4r + 1250k3r2 + 250k2r3 + 25kr4 + r5 so (x + y)5 ≡ r5 (mod5 ) but because 1 ≤ r ≤ 4, then r5 is not a multiple of 5 and so we have proven that x + y ≡ 0 (mod5 ). As such, there exists α ∈ Z such that x + y = 5α. Then, x = 5α − y. Hence, x5 + y5 = (5α − y)5 + y5 = 3125α5 − 3125α4y + 1250α3y2 − 250α2y3 + 25αy4. It follows that x5 + y5 is divisible by 25. 3

Remark for Question 3: This deals with a well-known result in number theory called Fermat’s little theorem. It states that if gcd (a, p) =1 (i.e. a is not divisible by p), then ap−1 ≡ 1 (mod p). An alternative representation says that for any integer a, ap ≡ a (mod p). Our method of proving Fermat’s little theorem was using modulo inverse. Question 4 (i) 5n (ii) (a) B1 = 5 and B2 = 24; B2 can be calculated easily by considering the complement of the event ‘never chooses Scrambled eggs on consecutive days’ so B2 = 52 − 1. (b) We consider two cases. • Case 1 (Scrambled eggs on the 1st day): On the 2nd day, she has 4 choices remaining. There would be no restrictions on what she has on the remaining n − 2 days. This contributes to 4Bn−2. • Case 2 (no Scrambled eggs on the 1st day): On the 1st day, she has 4 choices. Thereafter, she has no restrictions on what she has on the remaining days. This contributes to 4Bn−1. Since the 2 cases are mutually exclusive, the result follows. (c) Let Pk be the proposition that B3k+1 ≡ 0 (mod5 ) for all k ∈ Z≥0. When k = 0, we have B1 = 5, which is divisible by 5. So, P0 is true. Assume Pr is true for some r ∈ Z≥0. Then, B3r+1 ≡ 0 (mod5 ). We are required to show B3r+4 ≡ 0 (mod5 ). Using the relation in (iib), as Bk = 4Bk−1 + 4Bk−2, then B3r+4 = 4B3r+3 + 4B3r+2 = 4 (4B3r+2 + 4B3r+1) +4B3r+2 = 20B3r+2 + 16B3r+1 ≡ 16B3r+1 (mod5 ) ≡ 0 (mod5 ) by induction hypothesis Since P0 is true and Pr is true implies Pr+1 is true, by mathematical induction, Pk is true for all k ∈ Z≥0. 4

Question 5 (i) (a) Consider the following graph of y = xp for p < 0 and x > 0 (we set i = 2 here but actually, i is arbitrary): 1 1.5 2 2.5 3 0.5 1 1.5 y = xp x = i x = i + 1 x y Z i+1 i xp dx denotes the area bounded by the curve, the x-axis and the ordinates x = i and x = i + 1. We construct the rectangle above which has a base of 1 unit and a height of (i + 1)p. Its area is (i + 1)p units2, which is less than the given integral. (b) It suffices to prove that Z i+1 i xp dx < ip + (i + 1)p 2 . Naturally, we would think of the right side of the inequality as the area of another figure other than a rectangle. Consider the following graph: 1 1.5 2 2.5 3 0.5 1 1.5 y = xp x = i x = i + 1 x y We construct a trapezium bounded by thex-axis and the ordinates x = i and x = i +1. Its area is ip + (i + 1)p 2 . The integral is less than the area of the trapezium and the result follows. (ii) Using (ia), (i + 1)p < Z i+1 i xp dx 2p + 3p + . . .+ np < Z 2 1 xp dx + Z 3 2 xp dx + . . .+ Z n n−1 xp dx n ∑ k=1 kp < 1 + Z n 1 xp dx The required sum is ∞ ∑ k=1 kp so as n → ∞, we have ∞ ∑ k=1 kp < 1 + xp+1 p + 1 ∞ 0 = lim n→∞

1 + np+1 − 1 p + 1

= 1 − 1 p + 1 = p 1 + p . 5

(iii) In (ii), we used (ia) to show that 2p + 3p + . . .+ np < Z n 1 xp dx. Considering the integral on the right side of the equation, we have Z n 1 xp dx = np+1 − 1 p + 1 . Adding 1p = 1 to both sides, we establish an upper bound for 1p + 2p + 3p + . . .+ np. Using (ib), we have Z i+1 i xp dx < ip + (i + 1)p 2 Z 2 1 xp dx + Z 3 2 xp dx + . . .+ Z n n−1 xp dx < 1p + 2p 2 + 2p + 3p 2 + . . .+ (n − 1)p + np 2 Z n 1 xp dx < 1p 2 + np 2 + n−1 ∑ k=2 kp np+1 − 1 p + 1 < 1p 2 + np 2 + n−1 ∑ k=2 kp 1p + np 2 + np+1 − 1 p + 1 < n ∑ k=1 kp so we have established a lower bound for 1p + 2p + 3p + . . .+ np. Therefore, 1 + np 2np+1 + np+1 − 1 np+1 (p + 1) < 1p + 2p + 3p + . . .+ np np+1 < 1 np+1 + np+1 − 1 np+1 (p + 1) 1 2np+1 + 1 n + 1 p + 1 − 1 np+1 (p + 1) < 1p + 2p + 3p + . . .+ np np+1 < 1 np+1 + 1 p + 1 − 1 np+1 (p + 1) As p > −1, then p + 1 > 0. As n → ∞ on both sides, by the squeeze theorem, the upper and lower bounds will tend to 1 p + 1. Therefore, lim n→∞ 1p + 2p + 3p + . . .+ np np+1

= 1 p + 1 . Remark for Question 5: There is a formula for the sum 1 p + 2p + . . .+ np which is known as Faulhaber’s formula. It states that n ∑ k=1 kp = 1 p + 1 p ∑ r=0 p + 1 r

Brnp−r+1, where Br denotes the sequence of Bernoulli numbers. Question