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Title: Edexcel AS Level Further Maths Core Pure Revision Tests

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Title: Edexcel AS Level Further Maths Core Pure Revision Tests Description: Enhance your exam preparation with this comprehensive set of revision tests designed for Edexcel AS Level Further Maths Paper 1 Core Pure. These practice questions come complete with detailed solution outlines and clear rationale, covering key topics such as complex numbers, sequences, algebra, induction, differentiation, and binomial expansion. Use these tests to reinforce your understanding and boost your confidence for your upcoming exams! Hashtags: #FurtherMaths #Edexcel #ASLevel #Revision #ExamPrep #CorePure #MathsPractice

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Subido en
18 de marzo de 2025
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Escrito en
2024/2025
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Below is a set of sample revision test questions modeled in the style of an
Edexcel AS Level Further Maths Paper 1 Core Pure exam. Each question is
followed by its solution outline and a brief rationale explaining the purpose of
the question and the techniques it reinforces.


Revision Test: Edexcel AS Level Further Maths – Paper 1 Core Pure



Question 1: Complex Numbers

Problem:
Find all complex solutions of the equation

z4=16.z^4 = 16.z4=16.

Solution Outline:

1. Express 16 in polar form: 16=16(cos⁡0+isin⁡0).16 = 16\big(\cos 0 + i\sin
0\big).16=16(cos0+isin0).

2. In polar form, the general solution for zzz is given by: z=164(cos⁡2πk4+isin⁡2πk4),k=0,1,2,3.z =
\sqrt[4]{16}\left(\cos\frac{2\pi k}{4} + i\sin\frac{2\pi k}{4}\right), \quad k = 0, 1, 2, 3.z=416
(cos42πk+isin42πk),k=0,1,2,3.

3. Since 164=2\sqrt[4]{16} = 2416=2, the solutions are:

o For k=0k=0k=0: z=2(cos⁡0+isin⁡0)=2z = 2\big(\cos 0 + i\sin 0\big) = 2z=2(cos0+isin0)=2.

o For k=1k=1k=1: z=2(cos⁡π2+isin⁡π2)=2iz = 2\big(\cos\frac{\pi}{2} +
i\sin\frac{\pi}{2}\big) = 2iz=2(cos2π+isin2π)=2i.

o For k=2k=2k=2: z=2(cos⁡π+isin⁡π)=−2z = 2\big(\cos\pi + i\sin\pi\big) = -
2z=2(cosπ+isinπ)=−2.

o For k=3k=3k=3: z=2(cos⁡3π2+isin⁡3π2)=−2iz = 2\big(\cos\frac{3\pi}{2} +
i\sin\frac{3\pi}{2}\big) = -2iz=2(cos23π+isin23π)=−2i.

Rationale:
This question tests your ability to convert a number to polar form and apply De Moivre’s theorem to
extract roots. Mastery of these techniques is essential for handling complex number problems in the
exam.



Question 2: Sequences and Series

Problem:
Consider the arithmetic sequence defined by

, an=3n+2.a_n = 3n + 2.an=3n+2.

Find the sum of the first 20 terms.

Solution Outline:

1. Identify the first term: a1=3(1)+2=5.a_1 = 3(1) + 2 = 5.a1=3(1)+2=5.

2. Find the 20th term: a20=3(20)+2=60+2=62.a_{20} = 3(20) + 2 = 60 + 2 = 62.a20
=3(20)+2=60+2=62.

3. Use the arithmetic series sum formula: S20=202(a1+a20)=10(5+62)=10×67=670.S_{20} =
\frac{20}{2}(a_1 + a_{20}) = 10(5 + 62) = 10 \times 67 = 670.S20=220(a1+a20
)=10(5+62)=10×67=670.

Rationale:
This question reinforces the use of arithmetic sequence formulas. It tests both your ability to extract
sequence terms and apply the formula for the sum of an arithmetic series—a key skill for the paper.



Question 3: Algebra and Factorisation

Problem:
Factorize completely:

2x3−8x2+8x.2x^3 - 8x^2 + 8x.2x3−8x2+8x.

Solution Outline:

1. Factor out the common factor 2x2x2x: 2x3−8x2+8x=2x(x2−4x+4).2x^3 - 8x^2 + 8x = 2x(x^2 - 4x +
4).2x3−8x2+8x=2x(x2−4x+4).

2. Recognize that x2−4x+4x^2 - 4x + 4x2−4x+4 is a perfect square: x2−4x+4=(x−2)2.x^2 - 4x + 4 = (x
- 2)^2.x2−4x+4=(x−2)2.

3. Thus, the complete factorisation is: 2x(x−2)2.2x(x - 2)^2.2x(x−2)2.

Rationale:
This problem is designed to test your factorisation techniques including factoring out common factors
and recognizing perfect square trinomials—skills that are often examined in core pure papers.



Question 4: Proof by Induction

Problem:
Prove by mathematical induction that for all natural numbers n≥1n \ge 1n≥1,

1+3+5+⋯+(2n−1)=n2.1 + 3 + 5 + \dots + (2n - 1) = n^2.1+3+5+⋯+(2n−1)=n2.

Solution Outline:

1. Base Case (n=1n = 1n=1): 1=12(true).1 = 1^2 \quad \text{(true)}.1=12(true).
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