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 pe