15 STEP Exam Questions And Answers

The '15 STEP Exam Questions and Answers' provide a solid foundation for mastering various mathematical fields, including algebra, geometry, calculus, and discrete mathematics.

Each section thoroughly covers essential concepts and problem-solving techniques, such as quadratic equations, triangle properties, and function differentiation. This comprehensive guide is designed to help students prepare for challenging exams and improve their overall mathematical skills.

Whether you're a student looking to strengthen your foundational knowledge or a professional aiming to refresh your skills, these insights will be incredibly valuable.

Each topic is interconnected, allowing you to develop a cohesive understanding of advanced mathematics.

Algebra and Functions

Let's solve the quadratic equation \(3x^2 – 12x + 9 = 0\) by using the quadratic formula. This formula, given by \(x = rac{-b \pm \sqrt{b^2 – 4ac}}{2a}\), is a reliable method for finding the roots of quadratic equations.

First, we need to identify the coefficients: \(a = 3\), \(b = -12\), and \(c = 9\). Substituting these values into the quadratic formula allows us to calculate the possible values of \(x\).

Understanding how to solve quadratic equations is crucial because they frequently appear in various fields, including physics, engineering, and economics. Grasping this concept can help you tackle more complex mathematical problems in the future.

Solve the quadratic equation \(3x^2 – 12x + 9 = 0\) using the quadratic formula

To solve the quadratic equation \(3x^2 – 12x + 9 = 0\) using the quadratic formula, we start by identifying the coefficients: \(a = 3\), \(b = -12\), and \(c = 9\).

The quadratic formula is \(x = rac{-b \pm \sqrt{b^2 – 4ac}}{2a}\). Substituting these values in, we get:

\[x = rac{12 \pm \sqrt{144 – 108}}{6}\]

Simplifying further:

\[x = rac{12 \pm \sqrt{36}}{6}\]

This results in:

\[x = rac{12 \pm 6}{6}\]

So, the solutions are:

\[x = rac{18}{6} = 3\]

and

\[x = rac{6}{6} = 1\]

Thus, the solutions to the quadratic equation are \(x = 3\) and \(x = 1\).

This method is important because it provides a reliable way to find the roots of any quadratic equation, a fundamental skill in mathematics.

Geometry and Trigonometry

In the study of Geometry and Trigonometry, a crucial skill is calculating the area of a triangle using Heron's formula. For example, to find the area of a triangle with sides measuring 8, 15, and 17 units, we need to compute the semi-perimeter first and then apply Heron's formula to determine the precise area.

This method is vital because it allows us to calculate the area without needing to know the height of the triangle, which is often not readily available. Using Heron's formula simplifies the process and provides an accurate result based solely on the lengths of the sides.

To begin, the semi-perimeter (s) is calculated by adding all the sides of the triangle and dividing by two:

\[ s = rac{(8 + 15 + 17)}{2} = 20 \]

Next, we apply Heron's formula:

\[ ext{Area} = \sqrt{s(s – a)(s – b)(s – c)} \]

Plugging in the values:

\[ ext{Area} = \sqrt{20(20 – 8)(20 – 15)(20 – 17)} \]

\[ ext{Area} = \sqrt{20 imes 12 imes 5 imes 3} \]

\[ ext{Area} = \sqrt{3600} \]

\[ ext{Area} = 60 ext{ square units} \]

Thus, the area of the triangle is 60 square units. This approach is not only straightforward but also efficient, making it a valuable tool in solving geometric problems.

Calculate the area of a triangle with sides of length 8, 15, and 17 using Heron's formula

Using Heron's formula is a reliable way to find the area of a triangle with side lengths of 8, 15, and 17.

First, we need to determine the semi-perimeter (s) of the triangle. We do this by adding the side lengths and dividing the sum by two:

\[ s = rac{8 + 15 + 17}{2} = 20 \.

Next, we apply Heron's formula:

\[ ext{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]

Here, \( a = 8 \), \( b = 15 \), and \( c = 17 \). Substituting these values into the formula, we get:

\[ ext{Area} = \sqrt{20(20-8)(20-15)(20-17)} = \sqrt{20 \cdot 12 \cdot 5 \cdot 3} \]

Calculating this, we find:

\[ ext{Area} = \sqrt{3600} = 60 \]

Therefore, the area of the triangle is 60 square units.

This method is efficient and helps in understanding the geometric properties of triangles.

Statistics and Probability

Understanding probability is essential for grasping how likely different outcomes are in random events.

Let's take an example where a bag contains 5 red, 4 blue, and 3 green balls.

If we want to determine the probability that two randomly picked balls are both green, we need to calculate the combined probability of these specific events.

This exercise is fundamental in mastering the concepts of statistics and probability.

A bag contains 5 red, 4 blue, and 3 green balls. If two balls are picked at random, what is the probability that both are green?

To determine the probability that both randomly picked balls are green, we need to use combinatorics and probability principles. The bag contains a total of 12 balls: 5 red, 4 blue, and 3 green.

Firstly, the total number of ways to choose 2 balls out of 12 is calculated using the combination formula:

\[ inom{12}{2} = 66. \]

Next, we find the number of ways to pick 2 green balls from the 3 available:

\[ inom{3}{2} = 3. \]

Thus, the probability that both picked balls are green is the ratio of these combinations:

\[ rac{3}{66} = rac{1}{22}. \]

This straightforward approach ensures an accurate understanding of combinatorial probability, which is crucial for mastering the topic and applying it effectively.

Calculus

In this section on Calculus, we will concentrate on differentiating polynomial functions.

Specifically, we will work through the differentiation of the function \(f(x) = x^3 – 4x^2 + 5x – 2\).

This skill is essential for understanding how rates of change and slopes behave, which is a cornerstone of calculus.

Mastering differentiation of polynomials will equip you to handle more complex problems as you progress in your studies.

Differentiate the function \(f(x) = x^3 – 4x^2 + 5x – 2\)

To differentiate the function \(f(x) = x^3 – 4x^2 + 5x – 2\), we use the power rule on each term. In calculus, the power rule states that for any term \(ax^n\), the derivative is \(a \cdot n \cdot x^{n-1}\).

Starting with \(x^3\), the derivative is \(3x^2\). For \(-4x^2\), it becomes \(-8x\). The derivative of \(5x\) is \(5\). Constant terms like \(-2\) have a derivative of zero.

Putting all this together, the derivative of the function, \(f'(x)\), is \(3x^2 – 8x + 5\).

This step-by-step approach ensures accuracy and clarity, which is essential for anyone studying calculus.

Discrete Mathematics

In the field of Discrete Mathematics, understanding planar graphs is essential.

For example, let's look at a simple graph \( G \) with 8 vertices and 15 edges. To find out how many regions would be formed if \( G \) is planar, we can use Euler's formula: \( V – E + F = 2 \). Here, \( V \) stands for vertices, \( E \) for edges, and \( F \) for faces or regions.

Euler's formula is particularly important because it provides a straightforward way to analyse planar graphs, which are graphs that can be drawn on a plane without any edges crossing. This concept is not only fundamental to graph theory but also has practical applications in areas such as computer network design, geography, and even biology.

By applying Euler's formula to our specific example, with \( V = 8 \) and \( E = 15 \), we can rearrange it to solve for \( F \):

\[ F = 2 – V + E \]

\[ F = 2 – 8 + 15 \]

\[ F = 9 \]

Therefore, if the graph \( G \) is planar, it would create 9 regions. Understanding these relationships helps us make sense of complex networks and their properties, which is a key element in many scientific and practical problems.

A simple graph G has 8 vertices and 15 edges. Find the number of regions formed if G is planar

To determine the number of regions formed if a simple graph \( G \) with 8 vertices and 15 edges is planar, we can use Euler's formula for planar graphs.

Euler's formula states that for a planar graph, \( V – E + R = 2 \), where \( V \) stands for vertices, \( E \) for edges, and \( R \) for regions.

Given that \( V = 8 \) and \( E = 15 \), we substitute these values into the formula:

\[ 8 – 15 + R = 2 \]

Simplifying this equation, we get:

\[ -7 + R = 2 \]

Thus:

\[ R = 9 \]

So, if graph \( G \) is planar, it will form 9 regions.

This method ensures precise calculation and helps in understanding the structural properties of planar graphs.

STEP Exploration

When exploring the sum of the first *n* terms of an arithmetic sequence with an initial term of 5 and a common difference of 3, you need to use the formula for the sum of an arithmetic series.

This formula, \( S_n = rac{n}{2} (2a + (n-1)d) \), helps you calculate the sum by substituting the given values.

Understanding and applying this formula is essential for accurately solving related problems in the STEP exam.

Investigate the sum of the first n terms of an arithmetic sequence where the first term is 5 and the common difference is 3

To explore the sum of the first \(n\) terms of an arithmetic sequence with a first term of 5 and a common difference of 3, we use the formula for the sum of an arithmetic sequence.

The sum \(S_n\) of the first \(n\) terms is given by:

\[ S_n = rac{n}{2} imes (2a + (n – 1)d) \]

Here, \(a\) is the first term, and \(d\) is the common difference.

Substituting \(a = 5\) and \(d = 3\), we get:

\[ S_n = rac{n}{2} imes (2 imes 5 + (n – 1) imes 3) = rac{n}{2} imes (10 + 3n – 3) = rac{n}{2} imes (3n + 7) \]

This formula allows us to quickly calculate the sum of the first \(n\) terms in the sequence.

Understanding this is important because it simplifies the process of finding the sum, making it easier to tackle related mathematical problems.

Exam Specifics: No Calculator

When solving the equation \(2^x = 32\) without a calculator, it is crucial to understand the properties of exponents.

Recognising that \(32\) can be written as \(2^5\) reveals that \(x\) must be \(5\).

This method shows how fundamental mathematical principles can simplify seemingly difficult problems, which is very useful during the STEP exam.

Solve for \(x\) in the equation \(2^x = 32\) without using a calculator

To find the value of \(x\) in the equation \(2^x = 32\) without using a calculator, you need to recognise that 32 is a power of 2.

Here's a step-by-step explanation:

• \(2^1 = 2\)
• \(2^2 = 4\)
• \(2^3 = 8\)
• \(2^4 = 16\)

So, \(32\) can be written as \(2^5\).

Therefore, we equate the exponents in \(2^x = 2^5\), which means \(x = 5\).

Thus, \(x\) equals 5.

Higher Level Topics: Complex Numbers

In this section, we'll look at how to transform complex numbers from their standard form to polar form, using the example \(7 – 24i\). This process involves determining the modulus and the argument of the complex number, which are key components for expressing it in polar coordinates. Understanding this conversion is vital for tackling more advanced problems in complex number theory effectively.

First, let's calculate the modulus, which is the distance from the origin to the point representing the complex number on the complex plane. For \(7 – 24i\), the modulus is:

\[

\sqrt{7^2 + (-24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25

\]

Next, we'll find the argument, which is the angle the complex number makes with the positive real axis. The argument \( heta \) can be found using the arctangent function:

\[

heta = an^{-1}\left(rac{-24}{7}\right)

\]

By calculating this, we can find the angle in radians or degrees, depending on the context.

Once we have the modulus \( r = 25 \) and the argument \( heta \), we can write the complex number in polar form as:

\[

25 ext{cis}(heta)

\]

Here, \(ext{cis}(heta)\) is shorthand for \(\cos(heta) + i\sin(heta)\).

Being able to switch between standard and polar form allows for easier multiplication, division, and exponentiation of complex numbers, making it an essential skill in advanced mathematics.

Express the complex number \(7 – 24i\) in polar form

To express the complex number \(7 – 24i\) in polar form, start by determining its magnitude and argument.

The magnitude \(r\) can be found using the formula \(r = \sqrt{a^2 + b^2}\), where \(a = 7\) and \(b = -24\). This gives us \(r = \sqrt{7^2 + (-24)^2} = 25\).

Next, we calculate the argument, denoted as \(heta\). The argument is given by \(heta = an^{-1}\left(rac{b}{a}\right)\). Substituting the values, we get \(heta = an^{-1}\left(rac{-24}{7}\right) \approx -1.29\) radians.

Therefore, the polar form of the complex number \(7 – 24i\) is \(25 ext{cis}(-1.29)\).

• Ensure the magnitude calculation is accurate.
• Double-check the angle using a calculator.
• Practise converting various complex numbers to polar form.

This approach ensures accurate and efficient solutions, especially useful for exams like the STEP.

Functions

Understanding the concept of function inverses is crucial in advanced mathematics.

To find the inverse of the function \(f(x) = 2x – 5\), we need to solve for \(x\) in terms of \(y\) and then express \(y\) as a function of \(x\).

This process is essential because it allows us to reverse the effects of a function, making it easier to work with in various mathematical scenarios.

Find the inverse of the function \(f(x) = 2x – 5\)

Finding the inverse of the function \(f(x) = 2x – 5\) is a key skill in mathematics, especially useful for STEP exam preparation. Here's how to do it:

1. Start by replacing \(f(x)\) with \(y\): \(y = 2x – 5\).
2. Rearrange the equation to solve for \(x\): \(x = rac{y + 5}{2}\).
3. Finally, swap \(y\) with \(x\): \(f^{-1}(x) = rac{x + 5}{2}\).

This process helps you understand how to reverse operations and solve for the original variable, a fundamental concept in algebra.

Sets, Relations, and Groups

When studying sets, relations, and groups, a fundamental concept is proving set identities.

A key example is showing that for any sets \(A\) and \(B\), the expression \(A \cup (B \cap A')\) simplifies to \(A \cup B\).

This exercise illustrates the use of set operations and helps deepen your understanding of how sets interact.

Prove that if A and B are sets, then \(A \cup (B \cap A') = A \cup B\)

To prove that \(A \cup (B \cap A') = A \cup B\), we need to demonstrate that each side of the equation is a subset of the other. Understanding these relationships between sets is fundamental in mathematics. Here's a clear, step-by-step proof:

1. First, consider any element \(x \in A \cup (B \cap A')\). This means \(x\) is either in \(A\) or in \(B \cap A'\).

If \(x\) is in \(A\), then it is obviously in \(A \cup B\.

If \(x\) is in \(B \cap A'\), it is in \(B\) but not in \(A\), which also means \(x\) is in \(A \cup B\.

2. Conversely, take any element \(x \in A \cup B\). This means \(x\) is either in \(A\) or in \(B\).

If \(x\) is in \(A\), then it is in \(A \cup (B \cap A')\.

If \(x\) is in \(B\) but not in \(A\), then \(x\) is in \(B \cap A'\) and therefore in \(A \cup (B \cap A')\.

Thus, we've shown that every element of \(A \cup (B \cap A')\) is in \(A \cup B\) and vice versa.

Therefore, \(A \cup (B \cap A') = A \cup B\).

Sequences and Series

In this section, we will explore how to calculate the sum of the first 20 terms of a geometric sequence, starting with the terms 2, 6, and 18. Knowing the formula for the sum of a geometric series is crucial for solving these types of problems efficiently. We'll break down each step to ensure clarity and accuracy.

Understanding geometric sequences is important as they frequently appear in various mathematical contexts, such as finance, physics, and computer science. By mastering the sum formula, you can tackle a wide range of problems with confidence.

Let's begin by identifying the key components of our sequence. The first term (a) is 2, and the common ratio (r) can be found by dividing the second term by the first term, which gives us 6/2 = 3.

The formula for the sum of the first n terms, S_n, of a geometric sequence is:

\[S_n = a rac{(1-r^n)}{1-r}\]

For our sequence, a = 2, r = 3, and n = 20. Substituting these values into the formula gives us:

\[S_{20} = 2 rac{(1-3^{20})}{1-3}\]

This simplifies to:

\[S_{20} = 2 rac{(1-3^{20})}{-2}\]

\[S_{20} = 1-3^{20}\]

Calculating \(3^{20}\) and simplifying will give us the final sum. This method ensures we accurately find the sum of the first 20 terms.

Find the sum of the first 20 terms of the geometric sequence 2, 6, 18#COMMA

To find the sum of the first 20 terms of the geometric sequence 2, 6, 18, we need to use the formula for the sum of a geometric series.

The formula is \( S_n = a rac{r^n – 1}{r – 1} \), where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.

In this sequence:

• The first term \(a = 2\)
• The common ratio \(r = 3\)

Using the formula, we get:

1. Identify the first term (\(a\)).
2. Determine the common ratio (\(r\)).
3. Apply the formula \( S_n = a rac{r^n – 1}{r – 1} \).
4. Perform the calculations step-by-step.

This approach helps us efficiently determine the sum of the sequence.

Vectors

In this section, we will delve into the calculation of the dot product of two given vectors.

Specifically, we will use the vectors \(\vec{a} = 3\vec{i} – 2\vec{j} + \vec{k}\) and \(\vec{b} = -\vec{i} + 4\vec{j} – 5\vec{k}\) to determine \(\vec{a} \cdot \vec{b}\).

This exercise will help you understand vector operations better and see their practical applications in mathematical contexts.

Given vectors \(\vec{a} = 3\vec{i} – 2\vec{j} + \vec{k}\) and \(\vec{b} = -\vec{i} + 4\vec{j} – 5\vec{k}\), calculate \(\vec{a} \cdot \vec{b}\)

To find the dot product \(\vec{a} \cdot \vec{b}\), you need to multiply the corresponding components of each vector and then sum the results. Given \(\vec{a} = 3\vec{i} – 2\vec{j} + \vec{k}\) and \(\vec{b} = -\vec{i} + 4\vec{j} – 5\vec{k}\), the steps are as follows:

\[

\vec{a} \cdot \vec{b} = (3 imes -1) + (-2 imes 4) + (1 imes -5) = -3 – 8 – 5 = -16

\]

Understanding how to calculate the dot product is crucial, especially for exams like the STEP. Here's why:

• It helps you grasp vector components better.
• It improves your proficiency with dot product calculations.
• Ensuring accuracy in these calculations is vital for exam success.

Keep practising these operations to build confidence and accuracy. Your commitment to mastering such mathematical concepts will greatly benefit your academic and professional journey.

Integration

Integration is a key concept in calculus that involves determining the area under curves.

For example, let's evaluate the integral \(\int (3x^2 – 2x + 1) dx\), which deals with a polynomial function.

This exercise will strengthen your understanding of the power rule for integration, an essential technique for solving these types of problems.

Evaluate the integral \(\int (3x^2 – 2x + 1) dx\)

To evaluate the integral \(\int (3x^2 – 2x + 1) \, dx\), we apply basic integration rules to each term of the polynomial individually. This process is a key concept in mathematics, particularly relevant for university-level exams.

By solving this integral, students can improve their problem-solving skills, which are essential for academic success.

Here's how to integrate each term:

1. Integrate \(3x^2\): \(\int 3x^2 \, dx = x^3\)
2. Integrate \(-2x\): \(\int -2x \, dx = -x^2\)
3. Integrate \(1\): \(\int 1 \, dx = x\)

Combining these results, the integral is \(x^3 – x^2 + x + C\), where \(C\) is the constant of integration.

This step-by-step approach ensures that the solution is clear and precise, aiding students in their learning process.

Probability Distributions

Probability distributions are essential for understanding random variables in statistics.

For example, consider the expected value of a discrete random variable \(X\) with the probability mass function \(P(X=x) = rac{1}{2^x}\) for \(x = 1, 2, 3, \ldots\).

Calculating this expected value helps us understand the central tendency of \(X\), giving us insight into its probabilistic behaviour.

What is the expected value of a discrete random variable X with the probability mass function \(P(X=x) = rac{1}{2^x}\), for \(x = 1, 2, 3, \ldots\)?

To find the expected value of the discrete random variable \(X\) with the given probability mass function \(P(X=x) = rac{1}{2^x}\) for \(x = 1, 2, 3, \ldots\), we need to calculate the sum of the series \(E(X) = \sum_{x=1}^{\infty} x \cdot rac{1}{2^x}\). This series converges to a finite value by utilising the properties of infinite geometric series and differentiation techniques often used in mathematics.

Here's a step-by-step breakdown:

1. Understand the Probability Mass Function (PMF): The PMF \(P(X=x) = rac{1}{2^x}\) tells us the probability of each value \(x\). For \(x = 1, 2, 3, \ldots\), these probabilities decrease exponentially.
2. Calculate the Expected Value: The expected value \(E(X)\) is found by summing the series \( \sum_{x=1}^{\infty} x \cdot rac{1}{2^x} \). This involves a sum of terms where each term is the product of \(x\) and its corresponding probability.
3. Utilise Infinite Geometric Series: Recognising that this sum can be handled with techniques involving geometric series is crucial. By differentiating the sum of a geometric series, we can find the exact value.

Understanding these steps is essential, not just for solving this particular problem, but also for excelling in exams like STEP, which test your ability to apply these mathematical concepts.

If you're preparing for the STEP exam or helping others understand this topic, mastering these techniques will be invaluable.

Matrices

Understanding how to calculate a matrix determinant is essential for solving various linear algebra problems.

To illustrate, let us find the determinant of the matrix \(egin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}\).

This basic concept helps in determining properties like invertibility, which is crucial in many advanced mathematical applications.

Calculate the determinant of the matrix \(egin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}\)

Calculating the determinant of the given 2×2 matrix \(egin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}\) is a key skill in linear algebra, especially for those preparing for the STEP exam.

The determinant of a 2×2 matrix \(egin{pmatrix} a & b \ c & d \end{pmatrix}\) can be found using the formula \(ad – bc\).

Applying this to our matrix:

\[

ext{Det} \left( egin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix} \right) = (1 \cdot 4) – (2 \cdot 3) = 4 – 6 = -2

\]

Understanding how to calculate determinants is important for several reasons:

• Solving Linear Systems: Determinants help determine if a system of equations has a unique solution.
• Higher-Level Mathematics: They are foundational for topics like eigenvalues and eigenvectors.
• STEP Exam Preparation: Determinants frequently appear in exam questions.

Differential Equations

In this section, we will solve the differential equation \(rac{dy}{dx} = 3y\) with the initial condition \(y(0) = 2\).

Using standard methods for solving first-order linear differential equations, we will find the explicit form of \(y(x)\).

This process will demonstrate essential techniques that are useful for tackling similar problems in the STEP exam.

Solve the differential equation \(rac{dy}{dx} = 3y\) with the initial condition \(y(0) = 2\)

To solve the differential equation \(rac{dy}{dx} = 3y\) with the initial condition \(y(0) = 2\), we need to separate the variables and integrate both sides. This type of problem is common in advanced mathematics exams and helps to test your understanding of key methods.

First, we rearrange the equation to \(rac{1}{y} rac{dy}{dx} = 3\). By integrating both sides, we get \(\ln|y| = 3x + C\). To remove the logarithm, we exponentiate both sides, giving us \(y = e^{3x + C}\). This can be rewritten as \(y = Ce^{3x}\).

Next, we apply the given initial condition \(y(0) = 2\). Substituting \(x = 0\) and \(y = 2\) into the equation, we find that \(C = 2\). Therefore, the final solution to the differential equation is \(y = 2e^{3x}\).

• Initial Condition: \(y(0) = 2\)
• Separate Variables: \(rac{1}{y} rac{dy}{dx} = 3\)
• Integrate: \(\ln|y| = 3x + C\)

This process shows how to solve a first-order differential equation using separation of variables, a fundamental technique in calculus.

Probability

Understanding probability is essential for excelling in the STEP exam, especially when addressing scenarios involving coin flips.

To determine the probability of getting at least one head in four flips of a fair coin, a useful approach is to first calculate the probability of the opposite event—getting no heads at all.

By subtracting this probability from one, we can find the desired probability.

This method not only simplifies the problem but also strengthens your problem-solving skills in probability questions.

What is the probability of obtaining at least one head in four flips of a fair coin?

To determine the probability of obtaining at least one head in four flips of a fair coin, it is useful to first calculate the probability of the complementary event—getting no heads at all.

Since each flip of a fair coin has a probability of 0.5 for landing on heads, the probability of landing on tails in one flip is also 0.5. Therefore, the probability of getting tails in all four flips is \(0.5^4 = 0.0625\).

Consequently, the probability of obtaining at least one head is \(1 – 0.0625 = 0.9375\). This approach highlights the importance of understanding complementary events in probability.

• Calculating complementary events
• Understanding fair coin flips
• Applying probability rules

Geometry

In this section, we will explore a fundamental property of triangles: the relationship between the sides and their opposite angles.

Specifically, we aim to prove that in any triangle, the angle opposite the longest side is the largest angle.

This principle is crucial for understanding the intrinsic properties of triangles and their applications in various geometric problems.

Prove that in any triangle, the angle opposite the longest side is the largest angle

In any triangle, the angle opposite the longest side is the largest angle. This can be demonstrated using the properties of triangle inequalities and the Law of Sines.

First, according to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side. This implies that the longest side of the triangle is opposite the largest angle.

Second, the Law of Sines states that the ratio of the length of a side to the sine of its opposite angle is constant for all three sides of a triangle. Therefore, the longest side will correspond to the largest angle because a larger side length results in a larger sine value, and thus a larger angle.

To summarise:

• The triangle inequality theorem ensures the longest side is opposite the largest angle.

Understanding these principles helps in visualising and solving problems related to triangle properties.

Optimization

Optimisation problems often involve finding the maximum or minimum values of a function within specific constraints. For example, consider the task of determining the dimensions of a rectangle with the largest possible area that can be inscribed in a circle with a radius of 5. This requires applying principles of calculus and geometry. Such problems not only test your understanding of these concepts but also enhance your problem-solving skills, which are crucial for the STEP exam.

Understanding optimisation is important because it helps you solve real-world problems efficiently. By mastering these techniques, you can improve your performance on exams and in practical applications.

Find the dimensions of the rectangle with the maximum area that can be inscribed in a circle of radius 5

To determine the dimensions of the rectangle with the maximum area that can be inscribed in a circle of radius 5, we need to apply some geometric principles and a bit of calculus. Here's how we can approach it:

• Visualise the Problem: Imagine a rectangle perfectly inscribed within a circle. The diagonal of this rectangle will be equal to the circle's diameter, which is 10 units (since the radius is 5).
• Use the Pythagorean Theorem: The relationship between the sides of the rectangle and its diagonal (the diameter of the circle) can be expressed using the Pythagorean theorem.
• Formulate the Area Equation: Express the area of the rectangle in terms of one variable by substituting into the Pythagorean theorem.

Here's a more detailed breakdown:

1. Setting Up the Problem:
• Let the lengths of the rectangle be \(x\) and \(y\).
• The diagonal \(d\) of the rectangle is 10 units (diameter of the circle).
• According to the Pythagorean theorem: \(x^2 + y^2 = 10^2\).
2. Express the Area:
• The area \(A\) of the rectangle is \(A = x \cdot y\).
3. Substitute:
• From the Pythagorean relationship: \(y = \sqrt{100 – x^2}\).
• Therefore, the area can be written as \(A = x \cdot \sqrt{100 – x^2}\).
4. Differentiate to Optimise:
• To find the maximum area, we need to differentiate \(A\) with respect to \(x\), and then set the derivative to zero to find the critical points.
• Solving for \(x\) and \(y\) will give us the dimensions of the rectangle that maximise the area.

Through this method, we find that the rectangle with the maximum area inscribed in a circle is actually a square. Each side of this square has a length of \(\sqrt{50}\) (approximately 7.07 units), as the square's diagonal (which is the diameter of the circle) is 10 units. This square configuration ensures the largest possible area within the given constraints.

Statistical Analysis

In statistical analysis, knowing how to calculate the mean, median, and mode of a set of numbers is fundamental.

These measures of central tendency offer valuable insights into the distribution of the data, which is essential for interpreting results accurately.

Mastering these concepts is crucial for anyone preparing for the STEP exam, as they form the basis for understanding more complex statistical methods.

Given a set of numbers {3, 7, 8, 5, 12, 14, 21, 13, 18}, calculate the mean, median, and mode

Calculating the mean, median, and mode of the set {3, 7, 8, 5, 12, 14, 21, 13, 18} involves basic statistical techniques essential for understanding data distributions. Here's how you can do it:

• Mean: To find the mean, add all the numbers together and divide by the total count, which is 9.
• Median: To determine the median, first order the numbers and then select the middle value.

Detailed Steps

1. Mean:
• Add all the numbers: 3 + 7 + 8 + 5 + 12 + 14 + 21 + 13 + 18 = 101
• Divide by the number of values: 101 / 9 = 11.22
• So, the mean is 11.22
2. Median:
• Order the numbers: 3, 5, 7, 8, 12, 13, 14, 18, 21
• The middle value of this ordered set is 12 (since it's the fifth number in a set of nine)
• Therefore, the median is 12
3. Mode:
• Look for the number that appears most frequently. In this set, all numbers appear only once.
• Hence, there is no mode for this data set.

Understanding these calculations helps in interpreting and presenting data accurately, which is crucial for making informed decisions.