Cambridge AS Level Mathematics 9709

By Azhar ul Haque Sario

This is a reference book titled "Cambridge AS Level Mathematics 9709" written by author Azhar ul Haque Sario. It is a comprehensive guide that aids students in their revision for the Cambridge AS Level Mathematics 9709 exam. This book is intended to supplement the official textbooks and study materials and provides a concise and accessible summary of key concepts in a well-organized format.

The book covers the complete syllabus of AS Level Mathematics 9709 and is organized into three sections, Pure Mathematics 1, Pure Mathematics 2, and Pure Mathematics 3, each of which corresponds to a different paper in the exam. This book is intended for the preparation of Paper 1, 2, and 3.The headings and subheadings used in this book have been taken from the official Cambridge AS Level Mathematics 9709 syllabus as a guide for the content and organization of the book.

The author developed this reference book based on their interpretation of the syllabus and past exam questions and did not copy any content or material from other sources. The content within this book is entirely the author's work and includes headings, material, and computations covering all the topics in AS Level Mathematics 9709.

It is important to note that this is a reference book and not intended to replace official textbooks or study materials. The author wrote this book for their own personal reference and revision, and it is a supplement to the official materials provided by Cambridge. Therefore, it is designed to help students in their exam preparation journey and not a guaranteed source of accuracy or completeness of information.

Hence, "Cambridge AS Level Mathematics 9709" is a comprehensive reference book that completes the syllabus of AS Level Mathematics 9709. The author wrote it to supplement official study materials and organized the content based on the official Cambridge AS Level Mathematics 9709 syllabus. Therefore, this book is a reliable reference material for the students appearing in the exam in 2023, 2024, and 2025.

• Language: English
• Publisher: Azhar ul Haque Sario
• Release date: Nov 5, 2023
• ISBN: 9798223891390

Azhar ul Haque Sario

Hello, my name is Azhar ul Haque Sario, and I am excited to introduce myself to you. I have a strong educational background, having studied O and A levels before pursuing an MBA. I am also a certified project manager and hold Google certifications in digital marketing and e-commerce. Aside from my professional experience, I am also passionate about investing. As an investor, I have developed a keen eye for spotting profitable opportunities and have a track record of making sound investment decisions. I believe that investing is an essential component of building long-term wealth and financial security, and I am committed to helping others achieve their investment goals as well. In my free time, I love sharing my insights and knowledge with others. You can find me posting daily articles on my LinkedIn profile, where I share tips and advice on everything from investing to marketing and beyond. I am always looking for ways to learn, grow, and make a positive impact, and I look forward to connecting with you soon.

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Cambridge AS Level Mathematics 9709

For Exam in 2023, 2024, and 2025

Azhar ul Haque Sario

Azhar ul Haque Sario

Copyright © 2023 Azhar ul Haque Sario

Cambridge AS Level Mathematics 9709

© 2023 Azhar ul Haque Sario

All rights reserved. This book or any portion thereof may not be reproduced or used in any manner whatsoever without the express written permission of the publisher except for the use of brief quotations in a book review.

This book is a reference book developed solely for the purpose of aiding students' revision for the Cambridge AS Level Mathematics 9709

exam. The headings and subheadings used as an idea in this book are taken from the official Cambridge AS Level Mathematics 9709

syllabus as a guide and organization for the content.

The author of this book is not affiliated with or endorsed by the Cambridge, which does not endorse or sponsor this product. The content within this book is based on the author's understanding and interpretation of the syllabus and past exam questions.

The author did not copy any content or material from other sources and wrote their own material to supplement the official textbooks and study materials.

This reference book is not intended to replace the official textbooks and study materials but rather to supplement and provide a summary of key concepts in a concise and accessible format.

The author cannot guarantee the accuracy or completeness of the information contained in this book and shall not be held responsible for any errors or omissions.

Please note that this is reference book.

Disclaimer: This book, titled Cambridge AS Level Mathematics 9709, is a reference book developed solely for the purpose of aiding students in their revision for the Cambridge AS Level Mathematics 9709 exam. It is designed to provide a concise and accessible summary of key concepts, based on the author's understanding and interpretation of the syllabus and past exam questions. The headings and subheadings in the book are taken from the official Cambridge AS Level Mathematics 9709

as a guide and organization for the content, but no content or material has been cited from Cambridge official work. Therefore, it is a study guide and reference of the

Cambridge AS Level Mathematics 9709

subject which includes heading, material, and computations, that are entirely the author's work. The author wrote this book for their own personal reference and revision, and it is not intended to replace official textbooks or study materials. Additionally, it should be noted that this is reference book for Cambridge AS Level Mathematics 9709.

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Contents

Title Page

Copyright

Dedication

Section I -  Pure Mathematics 1

(for Paper 1)

Quadratics

Step 1: Solve the linear equation for one variable.

Step 3: Solve the quadratic equation.

Step 4: Find the value of x.

Functions

Coordinate Geometry

Circular Measure

Trigonometry

Series

Differentiation

Integration

Section II - Pure Mathematics 2

(for Paper 2)

Algebra

Logarithmic and Exponential Functions

Step 1: Shift the term not containing x to the other side of the inequality:

Step 2: Take the logarithm of both sides:

Step 3: Simplify the equation using logarithmic rules:

Step 4: Simplify and solve for x:

Trigonometry

Differentiation

Integration

Numerical Solution of Equations

Section III - Pure Mathematics 3

(for Paper 3)

Algebra

Logarithmic and Exponential Functions

Trigonometry

Differentiation

Integration

Numerical Solution of Equations

Vectors

Differential equations

Complex Numbers

About The Author

Section I -  Pure Mathematics 1

(for Paper 1)

Quadratics

Completing the square is a mathematical technique used to convert a quadratic polynomial, represented by the equation ax2 + bx + c, into its equivalent squared form. This technique is useful when we want to derive important information about the graph of the quadratic function, such as the position of the vertex or to sketch the graph.

To perform the process of completing the square, we first take the coefficient of the x term, which is represented by b, and divide it by 2. We then square the result of this division and add it to both sides of the equation.

For example, let's complete the square for the quadratic equation y = 2x2 + 8x + 3.

First, we divide the coefficient of the x term, which is 8, by 2 to get 4. We then square 4 to get 16.

Next, we add 16 to both sides of the equation:

y + 16 = 2(x2 + 4x + 8)

Now, we can write the squared form of the expression inside the parentheses by taking half of the coefficient of x, which is 4, and squaring it to get 16:

y + 16 = 2(x + 2)2 + 7

This is the completed square form of the original quadratic equation. By examining this form, we can see that the vertex of the graph of this function is located at the point (-2,7).

To graph the quadratic function, we can use the completed square form, which tells us that the vertex is located at (-2, 7) and that the shape of the graph is a parabola that opens upwards.

Sample Question:

Let's say you have a quadratic equation y = x2 + 6x + 11. Use the process of completing the square to find its vertex and then sketch the graph.

Solution:

First, we divide the coefficient of the x term, which is 6, by 2 to get 3. We then square 3 to get 9 and add it to both sides of the equation:

y + 9 = (x+3)2 + 2

Now we have the completed square form of the quadratic equation. By comparing this form to y = a(x-h)2 + k, we can identify the coordinates of the vertex, which is located at the point (-3,2).

To sketch the graph, we can use the completed square form and plot the vertex at (-3, 2). We can then use the quadratic formula to find the x-intercepts and plot those points on the graph as well. From there, we can draw the parabola to create a visual representation of the function.

Interpretation:

Completing the square is a powerful tool in mathematics that allows us to transform a quadratic equation into its equivalent squared form. This technique is very useful when we want to derive important information about the graph of a quadratic function, such as the position of the vertex or to sketch the graph. By using Completing the square technique, we can easily find important information about the quadratic equation such as Vertex and axis of symmetry.

here are 10 sample problems based on the information provided about Completing the Square:

1. Complete the square for the quadratic equation y = x^2 + 4x - 1

Solution:

First, we divide the coefficient of the x term, which is 4, by 2 to get 2. We then square 2 to get 4 and add it to both sides of the equation:

y + 4 = (x+2)^2 - 5

Now we have the completed square form of the quadratic equation. By comparing this form to y = a(x-h)^2 + k, we can identify the coordinates of the vertex, which is located at the point (-2,-5).

2. Use the process of completing the square to find the vertex of the quadratic equation y = 3x^2 - 12x - 5

Solution:

First, we divide the coefficient of the x term, which is -12, by 2 to get -6. We then square -6 to get 36 and add it to both sides of the equation:

y + 36 = 3(x-2)^2 - 11

Now we have the completed square form of the quadratic equation. By comparing this form to y = a(x-h)^2 + k, we can identify the coordinates of the vertex, which is located at the point (2,-11/3).

3. Find the vertex and sketch the graph of the quadratic equation y = -2x^2 + 12x - 19

Solution:

First, we divide the coefficient of the x term, which is 12, by 2 to get 6. We then square 6 to get 36 and add it to both sides of the equation:

y + 36 = -2(x-3)^2 -1

Now we have the completed square form of the quadratic equation. By comparing this form to y = a(x-h)^2 + k, we can identify the coordinates of the vertex, which is located at the point (3, -1).

To sketch the graph, we can use the completed square form and plot the vertex at (3, -1). We can then use the quadratic formula to find the x-intercepts and plot those points on the graph as well. From there, we can draw the parabola to create a visual representation of the function.

4. Find the vertex and axis of symmetry for the quadratic equation y = 2x^2 - 16x + 7

Solution:

First, we divide the coefficient of the x term, which is -16, by 2 to get -8. We then square -8 to get 64 and add it to both sides of the equation:

y + 64 = 2(x-4)^2 - 1

Now we have the completed square form of the quadratic equation. By comparing this form to y = a(x-h)^2 + k, we can identify the coordinates of the vertex, which is located at the point (4, -1/2).

The axis of symmetry is the vertical line that passes through the vertex, which is x = 4.

5. Find the vertex and sketch the graph of the quadratic equation y = 5x^2 - 30x + 11

Solution:

First, we divide the coefficient of the x term, which is -30, by 2 to get -15. We then square -15 to get 225 and add it to both sides of the equation:

y + 225 = 5(x-3)^2 - 4

Now we have the completed square form of the quadratic equation. By comparing this form to y = a(x-h)^2 + k, we can identify the coordinates of the vertex, which is located at the point (3,-4/5).

To sketch the graph, we can use the completed square form and plot the vertex at (3,-4/5). We can then use the quadratic formula to find the x-intercepts and plot those points on the graph as well. From there, we can draw the parabola to create a visual representation of the function.

6. Find the vertex and axis of symmetry for the quadratic equation y = x^2 - 8x + 15

Solution:

First, we divide the coefficient of the x term, which is -8, by 2 to get -4. We then square -4 to get 16 and add it to both sides of the equation:

y + 16 = (x-4)^2 - 1

Now we have the completed square form of the quadratic equation. By comparing this form to y = a(x-h)^2 + k, we can identify the coordinates of the vertex, which is located at the point (4,-1).

The axis of symmetry is the vertical line that passes through the vertex, which is x = 4.

7. Find the vertex and sketch the graph of the quadratic equation y = 2x^2 - 12x - 9

Solution:

First, we divide the coefficient of the x term, which is -12, by 2 to get -6. We then square -6 to get 36 and add it to both sides of the equation:

y + 36 = 2(x-3)^2 - 3

Now we have the completed square form of the quadratic equation. By comparing this form to y = a(x-h)^2 + k, we can identify the coordinates of the vertex, which is located at the point (3,-3).

To sketch the graph, we can use the completed square form and plot the vertex at (3,-3). We can then use the quadratic formula to find the x-intercepts and plot those points on the graph as well. From there, we can draw the parabola to create a visual representation of the function.

8. Find the vertex and axis of symmetry for the quadratic equation y = -4x^2 + 20x - 15

Solution:

First, we divide the coefficient of the x term, which is 20, by 2 to get 10. We then square 10 to get 100 and add it to both sides of the equation:

y + 100 = -4(x-5)^2 + 5

Now we have the completed square form of the quadratic equation. By comparing this form to y = a(x-h)^2 + k, we can identify the coordinates of the vertex, which is located at the point (5,5).

The axis of symmetry is the vertical line that passes through the vertex, which is x = 5.

9. Find the vertex and sketch the graph of the quadratic equation y = -2x^2 + 16x + 5

Solution:

First, we divide the coefficient of the x term, which is 16, by 2 to get 8. We then square 8 to get 64 and add it to both sides of the equation:

y + 64 = -2(x-4)^2 + 73

Now we have the completed square form of the quadratic equation. By comparing this form to y = a(x-h)^2 + k, we can identify the coordinates of the vertex, which is located at the point (4,73).

To sketch the graph, we can use the completed square form and plot the vertex at (4,73). We can then use the quadratic formula to find the x-intercepts and plot those points on the graph as well. From there, we can draw the parabola to create a visual representation of the function.

10. Find the vertex and axis of symmetry for the quadratic equation y = 3x^2 + 24x - 2

Solution:

First, we divide the coefficient of the x term, which is 24, by 2 to get 12. We then square 12 to get 144 and add it to both sides of the equation:

y + 144 = 3(x+4)^2 + 10

Now we have the completed square form of the quadratic equation. By comparing this form to y = a(x-h)^2 + k, we can identify the coordinates of the vertex, which is located at the point (-4,10/3).

The axis of symmetry is the vertical line that passes through the vertex, which is x = -4.

To find the discriminant of a quadratic polynomial ax2 + bx + c, we use the formula:

b2 - 4ac

where a, b, and c are the coefficients of the quadratic polynomial.

If the discriminant is positive, then the quadratic polynomial has two distinct real roots.

If the discriminant equals zero, then the quadratic polynomial will have one real root that is repeated.

If the discriminant is negative, then the quadratic polynomial has two complex roots.

For example, let's say we have the quadratic polynomial 2x2 + 5x + 3.

Using the formula, we can find the discriminant as:

b2 - 4ac

= (5)2 - 4(2)(3)

= 25 - 24

= 1

As the discriminant is positive, the quadratic polynomial will have two separate and real roots.

If the discriminant were 0, for example in the quadratic polynomial x2 + 4x + 4, then we would have a repeated root because the discriminant is 0, which indicates that the quadratic polynomial has one repeated root.

Here are 10 mathematical problems related to the discriminant of a quadratic polynomial, along with their solutions and explanations:

Problem 1:

Find the discriminant of the polynomial 3x^2 + 4x - 2 and determine the nature of its roots.

Solution:

The coefficients of the polynomial are a = 3, b = 4, and c = -2. So, applying the formula,

discriminant = b^2 - 4ac

= 4^2 - 4(3)(-2)

= 16 + 24

= 40

Since the discriminant is positive, the polynomial has two distinct real roots.

Problem 2:

Determine the values of k for which the equation (k+1)x^2 - 4x + 3k = 0 has only one solution.

Solution:

For the equation to have only one solution, the discriminant must be zero. So, applying the formula,

discriminant = b^2 - 4ac

= (-4)^2 - 4(k+1)(3k)

= 16 - 12k - 12k - 12

= 4 - 24k

For the discriminant to be zero, we have 4 - 24k = 0

k = 1/6

So, the equation has only one solution when k = 1/6.

Problem 3:

Find the roots of the equation x^2 + 5x + 6 = 0.

Solution:

The coefficients of the polynomial are a = 1, b = 5, and c = 6. So, applying the formula,

discriminant = b^2 - 4ac

= 5^2 - 4(1)(6)

= 1

The fact that the discriminant is positive indicates that the polynomial possesses two genuine real roots.

Using the formula for quadratic equation, we get the roots:

x = (-b ± sqrt(discriminant)) / 2a

= (-5 ± sqrt(1)) / 2(1)

= -3, -2

So, the roots of the equation are -3 and -2.

Problem 4:

Find the values of k for which the equation x^2 + kx + 4 = 0 has no real roots.

Solution:

If the discriminant is negative, it shows that the equation will have no real roots.

So, applying the formula,

discriminant = b^2 - 4ac

= k^2 - 4(1)(4)

= k^2 - 16

For the discriminant to be negative, we have k^2 - 16 < 0

k^2 < 16

|k| < 4

So, the equation has no real roots for values of k such that |k| < 4.

Problem 5:

Determine the values of x for which the expression x^2 - 6x + 8 is positive.

Solution:

We can factorize the expression as (x - 4)(x - 2). So, the expression is positive when both factors are positive or both factors are negative.

When x > 4, both factors are positive, so the expression is positive.

When x < 2, both factors are negative, so the expression is positive.

So, the expression is positive for x < 2 or x > 4.

Problem 6:

Find the values of k for which the equation x^2 + 2kx + k + 1 = 0 has real roots that are equal.

Solution:

For the equation to have real roots that are equal, the discriminant must be zero. So, applying the formula,

discriminant = b^2 - 4ac

= (2k)^2 - 4(1)(k+1)

= 4k^2 - 4k - 4

= 4(k^2 - k - 1)

For the discriminant to be zero, we have k^2 - k - 1 = 0. Using the formula for roots of a quadratic equation, we get:

k = (1 ± sqrt(5)) / 2

So, the equation has real roots that are equal when k = (1 ± sqrt(5)) / 2.

Problem 7:

Determine the values of p for which the polynomial x^2 + px + 4 has a repeated root.

Solution:

For the polynomial to have a repeated root, the discriminant must be zero. So, applying the formula,

discriminant = b^2 - 4ac

= p^2 - 4(1)(4)

= p^2 - 16

For the discriminant to be zero, we have p^2 - 16 = 0. This gives us p = ±4.

So, the polynomial has a repeated root when p = ±4.

Problem 8:

Determine the values of k for which the equation 3x^2 - 2kx + 1 = 0 has no real roots.

Solution:

For the equation to have no real roots, the discriminant must be negative. So, applying the formula,

discriminant = b^2 - 4ac

= (-2k)^2 - 4(3)(1)

= 4k^2 - 12

For the discriminant to be negative, we have 4k^2 - 12 < 0. This gives us k^2 < 3. So,

-k sqrt(3).

So, the equation has no real roots when k lies in the interval (-sqrt(3), sqrt(3)).

Problem 9:

Find a quadratic polynomial whose roots are 2 and -3.

Solution:

We know that if alpha and beta are the roots of a quadratic polynomial, then the polynomial is given by:

(x - alpha)(x - beta)

Using this formula, we get:

(x - 2)(x + 3)

= x^2 + x - 6

So, the quadratic polynomial whose roots are 2 and -3 is x^2 + x - 6.

Problem 10:

Determine the range of k for which the equation 2x^2 + kx + 1 = 0 has real roots.

Solution:

The equation will possess real roots if the value of the discriminant is either zero or positive. So, applying the formula,

discriminant = b^2 - 4ac

= k^2 - 8

For the discriminant to be non-negative, we have k^2 - 8 >= 0. This gives us k >= 2sqrt(2) or k <= -2sqrt(2).

So, the equation has real roots when k lies in the interval (-∞, -2sqrt(2)] U [2sqrt(2), ∞).

Solving Quadratic Equations and Inequalities:

A quadratic equation is an equation of the form ax^2 + bx + c = 0, where a, b, and c are constants, and x is the variable we are solving for. A quadratic inequality is an inequality of the form ax^2 + bx + c > 0 or ax^2 + bx + c < 0, where a, b, c, and x are the same as in the quadratic equation.

There are three main methods for solving quadratic equations: factorising, completing the square, and using the quadratic formula.

Factorising:

In order to factorise a quadratic equation, we need to find two numbers that multiply to give us the constant term c, and add to give us the coefficient of the x term b. Once we have these two numbers, we can express the quadratic equation in the form (x + p)(x + q) = 0, where p and q are the two numbers we found. We can then solve for x by setting each factor equal to zero. If the quadratic equation cannot be factorised easily, we can still use other methods such as completing the square or the quadratic formula.

Completing the Square:

To complete the square, we add and subtract (b/2a)^2 to the quadratic equation, where a, b, and c are as in the original equation. We then express the quadratic equation in the form a(x + (b/2a))^2 + c - (b^2/4a) = 0. We can then solve for x by taking the square root of both sides, and isolating x.

Using the Quadratic Formula:

The quadratic formula is given by x = (-b ± sqrt(b^2 - 4ac)) / 2a. We can use the quadratic formula directly to find the roots of the quadratic equation. We simply substitute the values of a, b, and c into the formula, and simplify. Once we have the roots, we can solve for x by setting each root equal to x.

To solve a quadratic inequality, we need to find the values of x that satisfy the inequality. We can use the same methods as for solving quadratic equations, but we need to be careful about the inequality sign. For example, if we are solving the inequality ax^2 + bx + c > 0, we need to find the values of x for which the expression on the left is greater than zero. We can find these values by using similar methods as for solving quadratic equations, but we need to pay attention to the signs of the factors.

Sample Question:

Solve the quadratic equation 2x^2 + x - 3 = 0.

Solution:

First, we can try to factorise the quadratic equation. We need to find two numbers that multiply to give -6 and add up to give 1. The numbers are 3 and -2. So, we can express the quadratic equation as (2x + 3)(x - 1) = 0. Therefore, either 2x + 3 = 0 or x - 1 = 0. Solving for x, we