AQA A Level Mathematics Paper 1 (Pure)
Practice Questions
Question 1: Trigonometric Equations & Proof
Solve the equation:
sin(2𝜃) = 𝑘cos𝜃 for 0 ≤ 𝜃 ≤ 2𝜋
where 𝑘 is a constant.
(a) Find all solutions in terms of 𝑘.
(b) Determine the number of distinct solutions when 𝑘 = 3.
𝜋
(c) If 𝜃 = 6 is a solution, find the value of 𝑘.
Mark Scheme:
(a)
o Use identity: sin(2𝜃) = 2sin𝜃cos𝜃.
o Rewrite: 2sin𝜃cos𝜃 − 𝑘cos𝜃 = 0 ⇒ cos𝜃(2sin𝜃 − 𝑘) = 0.
o Solutions:
𝜋 3𝜋
cos𝜃 = 0 ⇒ 𝜃 = 2 , .✓
2
𝑘 𝑘 𝑘
sin𝜃 = 2 ⇒ 𝜃 = arcsin (2) , 𝜋 − arcsin (2). ✓✓
𝜋 3𝜋 𝑘 𝑘
o Final answer: 𝜃 = 2 , , arcsin (2) , 𝜋 − arcsin (2).
2
(b)
3
o For 𝑘 = 3: sin𝜃 = 2, which has no real solutions. ✓
𝜋 3𝜋
o Only solutions: 𝜃 = 2 , . Total = 2 solutions. ✓
2
(c)
𝜋
o Substitute 𝜃 = 6 :
𝜋 𝜋 √3 √3
sin ( 3 ) = 𝑘cos ( 6 ) ⇒ =𝑘⋅ ⇒ 𝑘 = 1. ✓✓
2 2
Question 2: Calculus – Differential Equations
A curve satisfies the differential equation:
𝑑𝑦 𝑦 2 + 2𝑥𝑦
= (𝑥 > 0)
𝑑𝑥 𝑥2
(a) Use the substitution 𝑦 = 𝑣𝑥 to show that:
𝑑𝑣 𝑣 2 + 𝑣
=
𝑑𝑥 𝑥
(b) Solve the differential equation to find 𝑦 in terms of 𝑥.
, (c) Given that 𝑦 = 1 when 𝑥 = 1, determine the particular solution.
Mark Scheme:
(a)
𝑑𝑦 𝑑𝑣
o Substitute 𝑦 = 𝑣𝑥 ⇒ 𝑑𝑥 = 𝑣 + 𝑥 𝑑𝑥. ✓
o Original equation:
𝑑𝑣 (𝑣𝑥)2 +2𝑥(𝑣𝑥)
𝑣 + 𝑥 𝑑𝑥 = = 𝑣 2 + 2𝑣. ✓
𝑥2
𝑑𝑣
o Rearranged: 𝑥 𝑑𝑥 = 𝑣 2 + 𝑣. ✓
(b)
o Separate variables:
1 1
∫ 𝑑𝑣 = ∫ 𝑑𝑥
𝑣(𝑣 + 1) 𝑥
1 1
Partial fractions: 𝑣 − 𝑣+1. ✓
o Integrate:
ln ∣ 𝑣 ∣ −ln ∣ 𝑣 + 1 ∣= ln ∣ 𝑥 ∣ +𝐶. ✓
𝑣
Simplify: 𝑣+1 = 𝐴𝑥. ✓
𝑦
o Substitute 𝑣 = 𝑥 :
𝑦 𝐴𝑥 2
= 𝐴𝑥 ⇒ 𝑦 = 1−𝐴𝑥. ✓
𝑦+𝑥
(c)
o Substitute 𝑥 = 1, 𝑦 = 1:
𝐴 1
1 = 1−𝐴 ⇒ 𝐴 = 2. ✓
1 2
𝑥 𝑥2
o Final solution: 𝑦 = 2
1 = 2−𝑥. ✓
1− 𝑥
2
Question 3: Vectors & 3D Geometry
The line 𝐿 has equation:
1 2
𝐫 = (−2) + 𝜆 ( 1 )
3 −1
The plane 𝛱 has equation 2𝑥 − 𝑦 + 3𝑧 = 7.
(a) Find the coordinates of the point where 𝐿 intersects 𝛱.
(b) Calculate the acute angle between 𝐿 and 𝛱.
(c) Find the shortest distance from the origin to the line 𝐿.
Mark Scheme:
(a)
o Parametrize 𝐿: 𝑥 = 1 + 2𝜆, 𝑦 = −2 + 𝜆, 𝑧 = 3 − 𝜆. ✓
Practice Questions
Question 1: Trigonometric Equations & Proof
Solve the equation:
sin(2𝜃) = 𝑘cos𝜃 for 0 ≤ 𝜃 ≤ 2𝜋
where 𝑘 is a constant.
(a) Find all solutions in terms of 𝑘.
(b) Determine the number of distinct solutions when 𝑘 = 3.
𝜋
(c) If 𝜃 = 6 is a solution, find the value of 𝑘.
Mark Scheme:
(a)
o Use identity: sin(2𝜃) = 2sin𝜃cos𝜃.
o Rewrite: 2sin𝜃cos𝜃 − 𝑘cos𝜃 = 0 ⇒ cos𝜃(2sin𝜃 − 𝑘) = 0.
o Solutions:
𝜋 3𝜋
cos𝜃 = 0 ⇒ 𝜃 = 2 , .✓
2
𝑘 𝑘 𝑘
sin𝜃 = 2 ⇒ 𝜃 = arcsin (2) , 𝜋 − arcsin (2). ✓✓
𝜋 3𝜋 𝑘 𝑘
o Final answer: 𝜃 = 2 , , arcsin (2) , 𝜋 − arcsin (2).
2
(b)
3
o For 𝑘 = 3: sin𝜃 = 2, which has no real solutions. ✓
𝜋 3𝜋
o Only solutions: 𝜃 = 2 , . Total = 2 solutions. ✓
2
(c)
𝜋
o Substitute 𝜃 = 6 :
𝜋 𝜋 √3 √3
sin ( 3 ) = 𝑘cos ( 6 ) ⇒ =𝑘⋅ ⇒ 𝑘 = 1. ✓✓
2 2
Question 2: Calculus – Differential Equations
A curve satisfies the differential equation:
𝑑𝑦 𝑦 2 + 2𝑥𝑦
= (𝑥 > 0)
𝑑𝑥 𝑥2
(a) Use the substitution 𝑦 = 𝑣𝑥 to show that:
𝑑𝑣 𝑣 2 + 𝑣
=
𝑑𝑥 𝑥
(b) Solve the differential equation to find 𝑦 in terms of 𝑥.
, (c) Given that 𝑦 = 1 when 𝑥 = 1, determine the particular solution.
Mark Scheme:
(a)
𝑑𝑦 𝑑𝑣
o Substitute 𝑦 = 𝑣𝑥 ⇒ 𝑑𝑥 = 𝑣 + 𝑥 𝑑𝑥. ✓
o Original equation:
𝑑𝑣 (𝑣𝑥)2 +2𝑥(𝑣𝑥)
𝑣 + 𝑥 𝑑𝑥 = = 𝑣 2 + 2𝑣. ✓
𝑥2
𝑑𝑣
o Rearranged: 𝑥 𝑑𝑥 = 𝑣 2 + 𝑣. ✓
(b)
o Separate variables:
1 1
∫ 𝑑𝑣 = ∫ 𝑑𝑥
𝑣(𝑣 + 1) 𝑥
1 1
Partial fractions: 𝑣 − 𝑣+1. ✓
o Integrate:
ln ∣ 𝑣 ∣ −ln ∣ 𝑣 + 1 ∣= ln ∣ 𝑥 ∣ +𝐶. ✓
𝑣
Simplify: 𝑣+1 = 𝐴𝑥. ✓
𝑦
o Substitute 𝑣 = 𝑥 :
𝑦 𝐴𝑥 2
= 𝐴𝑥 ⇒ 𝑦 = 1−𝐴𝑥. ✓
𝑦+𝑥
(c)
o Substitute 𝑥 = 1, 𝑦 = 1:
𝐴 1
1 = 1−𝐴 ⇒ 𝐴 = 2. ✓
1 2
𝑥 𝑥2
o Final solution: 𝑦 = 2
1 = 2−𝑥. ✓
1− 𝑥
2
Question 3: Vectors & 3D Geometry
The line 𝐿 has equation:
1 2
𝐫 = (−2) + 𝜆 ( 1 )
3 −1
The plane 𝛱 has equation 2𝑥 − 𝑦 + 3𝑧 = 7.
(a) Find the coordinates of the point where 𝐿 intersects 𝛱.
(b) Calculate the acute angle between 𝐿 and 𝛱.
(c) Find the shortest distance from the origin to the line 𝐿.
Mark Scheme:
(a)
o Parametrize 𝐿: 𝑥 = 1 + 2𝜆, 𝑦 = −2 + 𝜆, 𝑧 = 3 − 𝜆. ✓
