15 Admissions MAT Exam Questions And Answers

The '15 Admissions MAT Exam Questions and Answers' offer a carefully curated set of problems spanning essential mathematical areas such as algebra, geometry, statistics, calculus, and discrete mathematics.

These questions are designed to challenge and assess critical skills, from solving quadratic equations and using geometric formulas to understanding advanced probability concepts. Including differentiation and integration techniques, along with vector operations, ensures a comprehensive evaluation.

This collection not only measures mathematical proficiency but also prepares candidates for the demands of higher education.

Delve deeper to discover the complexities and solutions that make these questions essential for anyone aiming to excel in mathematics.

Algebra and Functions

Understanding how to solve quadratic equations is a crucial skill in algebra. Let's take the equation \(3x^2 – 12x + 9 = 0\) and solve it using the quadratic formula. This method will show how to systematically apply the formula to find the roots of the equation accurately.

To start, recall the quadratic formula:

\[x = rac{-b \pm \sqrt{b^2 – 4ac}}{2a}\]

For our equation \(3x^2 – 12x + 9 = 0\), the coefficients are:

\[a = 3, \quad b = -12, \quad c = 9\]

Now, we substitute these values into the formula:

\[x = rac{-(-12) \pm \sqrt{(-12)^2 – 4 \cdot 3 \cdot 9}}{2 \cdot 3}\]

Simplifying step-by-step:

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

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

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

This gives us two solutions:

\[x = rac{12 + 6}{6} = 3\]

\[x = rac{12 – 6}{6} = 1\]

So, the roots of the equation are \(x = 3\) and \(x = 1\).

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 first need to identify the coefficients: \(a = 3\), \(b = -12\), and \(c = 9\).

The quadratic formula is \(x = rac{-b \pm \sqrt{b^2 – 4ac}}{2a}\). Substituting the coefficients into the formula, we get \(x = rac{12 \pm \sqrt{(-12)^2 – 4 \cdot 3 \cdot 9}}{2 \cdot 3}\).

Next, we simplify the discriminant: \((-12)^2 – 4 \cdot 3 \cdot 9 = 144 – 108 = 36\). This gives us \(x = rac{12 \pm \sqrt{36}}{6}\).

Taking the square root of 36, we find \(x = rac{12 \pm 6}{6}\). This results in two possible solutions: \(x = rac{12 + 6}{6} = 3\) and \(x = rac{12 – 6}{6} = 1\).

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

Understanding how to apply the quadratic formula is crucial because it provides a systematic way to solve any quadratic equation, ensuring you can find the roots even when factorisation is difficult.

Geometry and Trigonometry

In Geometry and Trigonometry, calculating the area of a triangle using Heron's formula demonstrates the application of key mathematical principles.

For a triangle with sides measuring 8, 15, and 17, Heron's formula offers a straightforward method to find the area without needing to know the height.

This method highlights the importance of understanding and using geometric properties effectively in problem-solving situations.

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

Using Heron's formula, we can find the area of a triangle with sides measuring 8, 15, and 17 units. Heron's formula is a practical method in geometry for calculating the area when the side lengths are known.

First, we need to calculate the semi-perimeter \( s \) using the formula \( s = rac{a+b+c}{2} \), where \( a = 8 \), \( b = 15 \), and \( c = 17 \). Thus,

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

Next, we use Heron's formula for the area \( A = \sqrt{s(s-a)(s-b)(s-c)} \). Substituting the values, we get:

\[ A = \sqrt{20(20-8)(20-15)(20-17)} = \sqrt{20 imes 12 imes 5 imes 3}. \]

This simplifies to:

\[ A = \sqrt{3600} = 60. \]

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

Statistics and Probability

In the field of statistics and probability, understanding the likelihood of specific outcomes is essential.

Imagine a scenario where a bag contains 5 red, 4 blue, and 3 green balls.

To find the probability that two randomly picked balls are both green, we need to use combinatorial techniques.

This exercise demonstrates the practical application of probability theory in addressing real-world problems.

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 calculate the probability that both balls picked at random are green, we need to determine the total number of possible outcomes and the number of favourable outcomes. The bag contains 12 balls in total (5 red, 4 blue, and 3 green). We use combinations to find the number of ways to pick 2 balls from 12.

Here's the breakdown:

Calculation	Result
Total combinations	C(12, 2) = 66
Favourable combinations	C(3, 2) = 3
Probability	3 / 66
Simplified Probability	1 / 22

Calculus

In the context of calculus, differentiating polynomial functions is an essential skill.

Take, for example, the function \(f(x) = x^3 – 4x^2 + 5x – 2\). By applying basic differentiation rules, we find its derivative to be \(f'(x) = 3x^2 – 8x + 5\).

Understanding how to differentiate such functions is crucial because it allows us to determine the rate at which the function's values change.

This is particularly useful in various real-world applications, such as physics and engineering, where knowing the rate of change can inform decision-making and problem-solving.

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

To differentiate the function \(f(x) = x^3 – 4x^2 + 5x – 2\), we need to apply some basic calculus principles. Differentiation helps us understand how a function changes, which is useful in many areas like physics, engineering, and economics.

To find the first derivative of \(f(x) = x^3 – 4x^2 + 5x – 2\), we use the power rule. This rule states that the derivative of \(x^n\) is \(nx^{n-1}\).

Applying this rule:

• The derivative of \(x^3\) is \(3x^2\),
• The derivative of \(-4x^2\) is \(-8x\),
• The derivative of \(5x\) is \(5\),

So, the first derivative, \(f'(x)\), is:

\[f'(x) = 3x^2 – 8x + 5.\]

This derivative tells us the rate at which the function \(f(x)\) is changing at any point \(x\). Knowing this can help in various practical applications, such as finding the maximum or minimum values of the function.

Discrete Mathematics

In the study of discrete mathematics, grasping the properties of planar graphs is essential. For a simple graph \( G \) with 8 vertices and 15 edges, determining the number of regions formed if \( G \) is planar involves applying Euler's formula for planar graphs.

This formula, \( R = E – V + 2 \), where \( R \) represents the regions, \( E \) the edges, and \( V \) the vertices, is vital in solving this problem.

Euler's formula helps us understand the relationship between vertices, edges, and regions in a planar graph, enabling us to visualise and analyse such graphs more effectively. By using this formula, we can determine that:

\[ R = 15 – 8 + 2 \]

\[ R = 9 \]

Therefore, if the graph \( G \) is planar, it would form 9 regions.

Understanding this concept is important as it lays the foundation for more complex topics in graph theory and helps in solving practical problems related to network design, circuit layout, and geographical mapping.

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 by a planar simple graph \( G \) with 8 vertices and 15 edges, we can use Euler's formula for planar graphs.

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

Given that \( V = 8 \) and \( E = 15 \), we can substitute these values into Euler's formula:

\[ 8 – 15 + R = 2 \]

Solving for \( R \), we get:

\[ R = 9 \]

Thus, for the given graph with 8 vertices and 15 edges, if it is planar, it forms 9 regions.

This is a fundamental property of planar graphs and provides insight into their structure, which is crucial for those studying discrete mathematics.

MAT Exploration

To explore the sum of the first n terms of an arithmetic sequence where the initial term is 5 and the common difference is 3, we need to understand the basic formula for the sum of an arithmetic series:

\( S_n = rac{n}{2} (2a + (n-1)d) \).

Here, 'a' represents the first term, and 'd' denotes the common difference.

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

Calculating the sum of the first \( n \) terms of an arithmetic sequence, where the first term is 5 and the common difference is 3, offers valuable insights for MAT exam preparation. The formula for the sum \( S_n \) of the first \( n \) terms is:

\[ S_n = rac{n}{2} (2a + (n – 1)d) \]

Here, \( a \) is the first term and \( d \) is the common difference. By substituting \( a = 5 \) and \( d = 3 \), we get:

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

This can be illustrated in a table for better understanding:

\( n \)	Sum \( S_n \)
1	5
2	16
3	33
4	56
5	85

Understanding this formula is crucial for efficiently solving sequence problems in MAT exams. It simplifies the process of finding the sum of terms and helps in managing time during the exam.

Exam Specifics: No Calculator

The MAT exam does not allow the use of calculators, so it's crucial to be comfortable with mental maths and basic algebra.

For example, to solve \(2^x = 32\), you need to recognise that 32 is the same as \(2^5\), which means \(x = 5\).

This demonstrates the importance of understanding key mathematical concepts to perform well in an environment without calculators.

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

To solve the equation \(2^x = 32\) without using a calculator, we need to recognise that 32 can be expressed as a power of 2. This type of problem is common in MAT admissions tests, which assess one's ability to solve equations using fundamental mathematical principles.

Firstly, we express 32 as \(2^5\), since \(2^5 = 32\). Thus, the equation \(2^x = 32\) becomes \(2^x = 2^5\). To find \(x\), we simply equate the exponents: \(x = 5\).

Here's a step-by-step breakdown:

Expression	Explanation
\(2^x = 32\)	Original equation
\(32 = 2^5\)	Recognise 32 as a power of 2
\(2^x = 2^5\)	Substitute in the power of 2
\(x = 5\)	Equate the exponents and solve for \(x\)

Understanding this method is crucial for solving similar problems efficiently, which is a valuable skill for MAT admissions.

Higher Level Topics: Complex Numbers

Complex numbers can be represented in polar form to simplify various mathematical operations.

To express the complex number \(7 – 24i\) in polar form, we need to determine its magnitude and argument.

This transformation makes it easier to perform operations like multiplication, division, and exponentiation with complex numbers.

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

To express the complex number \(7 – 24i\) in polar form, we need to find its magnitude and argument. This is essential for a deeper understanding of complex numbers, which is often tested in university-level exams.

First, we calculate the magnitude (or modulus) using the formula \(\sqrt{a^2 + b^2}\). Here, \(a = 7\) and \(b = -24\):

\[

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

\]

Next, we determine the argument (or angle \( heta \)). The argument is given by \( an^{-1}\left(rac{b}{a}\right) \):

\[

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

\]

This calculation gives us \( heta \approx -73.74^\circ \) or \( heta \approx -1.287 \) radians.

Therefore, the polar form of the complex number \(7 – 24i\) is:

\[

25 \left(\cos(heta) + i\sin(heta)\right)

\]

where \( heta \approx -73.74^\circ \) or \( heta \approx -1.287 \) radians.

Understanding these steps is crucial for excelling in advanced mathematics and related exams.

Functions

Understanding how to find the inverse of a function is essential for solving various mathematical problems. Given the function \(f(x) = 2x – 5\), we need to determine its inverse, which involves solving for \(x\) in terms of \(y\). This process helps clarify the fundamental steps required for manipulating and transforming functions in algebra.

To find the inverse, follow these steps:

1. Start with the original function: \(f(x) = 2x – 5\).

2. Replace \(f(x)\) with \(y\): \(y = 2x – 5\).

3. Solve for \(x\) in terms of \(y\):

• Add 5 to both sides: \(y + 5 = 2x\).

4. Replace \(y\) with \(x\) to write the inverse function: \(f^{-1}(x) = rac{x + 5}{2}\).

Understanding this process is important because it allows you to reverse a function's effects, providing valuable insights into the relationships between variables.

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

To find the inverse of the function \(f(x) = 2x – 5\), we need a function that, when composed with \(f(x)\), results in the original input.

Let's start by setting \(y = f(x)\), which gives us \(y = 2x – 5\). Our goal is to solve for \(x\) in terms of \(y\).

First, add 5 to both sides of the equation:

\[ y + 5 = 2x \]

Next, divide both sides by 2:

\[ x = rac{y + 5}{2} \]

Thus, the inverse function, denoted as \(f^{-1}(x)\), is:

\[ f^{-1}(x) = rac{x + 5}{2} \]

To confirm that this is indeed the inverse, we check that composing \(f\) with \(f^{-1}\) returns the original input \(x\).

If we compute \(f(f^{-1}(x))\), we get:

\[ f\left(rac{x + 5}{2}\right) = 2 \left(rac{x + 5}{2}\right) – 5 = x + 5 – 5 = x \]

This confirms that \(f^{-1}(x)\) is correct, as it yields the identity function when composed with \(f(x)\).

Sets, Relations, and Groups

In the study of sets, relations, and groups, a key exercise is proving set identities using basic set operations. For example, let's prove that \(A \cup (B \cap A') = A \cup B\), where \(A'\) denotes the complement of set A. This proof requires a good understanding of the properties of union, intersection, and complement within set theory.

Understanding these concepts is crucial because it helps build a solid foundation for more advanced topics in mathematics and computer science. By mastering these basic operations, you can tackle more complex problems with confidence.

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 show that each side of the equation is a subset of the other.

First, let's take an element \(x\) from \(A \cup (B \cap A')\). This means \(x\) is either in \(A\) or in \(B \cap A'\).

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

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

Now, consider an element \(x\) from \(A \cup B\). This means \(x\) is either in \(A\) or in \(B\).

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

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

Therefore, \(A \cup (B \cap A')\) is a subset of \(A \cup B\) and vice versa, proving the equality \(A \cup (B \cap A') = A \cup B\).

Sequences and Series

In this section, we will learn how to find the sum of the first 20 terms of a geometric sequence, using the sequence 2, 6, 18 as an example.

Understanding geometric sequences is important because they often appear in various mathematical contexts, such as finance, computer science, and natural phenomena.

We will employ the formula for the sum of the first \( n \) terms of a geometric series to solve this efficiently.

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 identify the common ratio and use the formula for the sum of a geometric series.

Here, the first term \(a\) is 2, and the common ratio \(r\) is 3.

The formula for the sum \(S_n\) of the first \(n\) terms of a geometric sequence is:

\[S_n = a rac{r^n – 1}{r – 1}\]

For the first 20 terms, we substitute \(a = 2\), \(r = 3\), and \(n = 20\) into the formula.

This gives us:

\[S_{20} = 2 rac{3^{20} – 1}{3 – 1}\]

Using this method ensures that we calculate the sum accurately while also deepening our understanding of geometric sequences.

Vectors

Vectors are fundamental in many mathematical and physical contexts, making their operations crucial for the MAT exam.

To find the dot product of the vectors \(\vec{a} = 3\vec{i} – 2\vec{j} + \vec{k}\) and \(\vec{b} = -\vec{i} + 4\vec{j} – 5\vec{k}\), you use the formula \(\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\).

This calculation results in a scalar value, which helps understand the geometric relationship between the two vectors.

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 of the vectors \(\vec{a} = 3\vec{i} – 2\vec{j} + \vec{k}\) and \(\vec{b} = -\vec{i} + 4\vec{j} – 5\vec{k}\), you need to multiply their corresponding components and then add the results together.

The dot product \(\vec{a} \cdot \vec{b}\) is calculated as follows:

\[

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

\]

Here are the key steps:

• Component-wise multiplication: Multiply each component of \(\vec{a}\) with the corresponding component of \(\vec{b}\).
• Summation: Add up these products.

Understanding how to calculate the dot product is crucial for students preparing for the MAT admissions exam, as it is a fundamental concept in vector operations.

Integration

Integration is a key concept in calculus, essential for calculating areas under curves and solving differential equations.

For example, when evaluating the integral \(\int (3x^2 – 2x + 1) dx\), you apply basic antiderivative rules.

This example illustrates how to find the antiderivative of a polynomial function, a crucial skill for success in the MAT exam.

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

Evaluating the integral \(\int (3x^2 – 2x + 1) \, dx\) involves applying the power rule of integration to each term individually. This process is essential in mathematics, especially at the university level, as it helps us find the antiderivative of polynomial expressions.

Here's a step-by-step breakdown:

• Integral of \(3x^2\): Using the power rule, we get \(rac{3x^3}{3} = x^3\).
• Integral of \(-2x\): Applying the power rule here gives \(rac{-2x^2}{2} = -x^2\).

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

Probability Distrutions

In the context of probability distributions, the expected value of a discrete random variable \(X\) with a given probability mass function \(P(X=x)\) is a fundamental concept.

To determine the expected value, one must consider the sum of all possible values of \(X\) weighted by their respective probabilities. For the specified probability mass function \(P(X=x) = rac{1}{x}\) for \(x = 1, 2, 3, \ldots\), the calculation involves evaluating this weighted average.

To illustrate, the expected value, often represented as \(E(X)\), is calculated by summing the products of each possible value of \(X\) and its corresponding probability. This concept is crucial as it provides a measure of the central tendency of the random variable, which is essential for making informed decisions in various fields, such as finance, economics, and engineering.

In simpler terms, the expected value tells us what outcome we can anticipate, on average, if we were to repeat an experiment many times.

Understanding this can help in predicting future events and in planning accordingly.

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 a discrete random variable \(X\) with the probability mass function \(P(X=x) = rac{1}{2^x}\) for \(x = 1, 2, 3, \ldots\), we need to sum the weighted values of \(X\). This involves adding up the products of each value \(x\) and its corresponding probability.

Here are the essential steps:

1. Identify the Probability Mass Function (PMF):

\(P(X=x) = rac{1}{2^x}\)

2. Calculate Each Term:

For each value of \(x\), find \(x \cdot P(X=x)\)

3. Sum These Terms:

Add all these products to determine the expected value

Understanding this process is crucial for tackling probability distribution questions in mathematics. This calculation demonstrates fundamental principles in statistics, aiding effective decision-making and problem-solving.

Matrices

Understanding matrices is essential for tackling a range of problems in the MAT exam.

A key operation is finding the determinant of a 2×2 matrix, such as \(egin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}\).

The determinant helps determine whether the matrix can be inverted, which is important for solving linear equations and other mathematical tasks.

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

To calculate the determinant of the 2×2 matrix \(egin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}\), you can use the straightforward formula \(ad – bc\).

Here, the elements of the matrix are \(a=1\), \(b=2\), \(c=3\), and \(d=4\). By substituting these values into the formula, you get:

\[

ext{Determinant} = (1 imes 4) – (2 imes 3) = 4 – 6 = -2

\]

So, the determinant of the matrix is \(-2\).

Key points to remember:

• Understand the layout of a 2×2 matrix.
• Use the formula \(ad – bc\) correctly.

This method is fundamental for tackling similar matrix problems effectively.

Differential Equations

When dealing with the differential equation \(rac{dy}{dx} = 3y\) and the initial condition \(y(0) = 2\), it is essential to identify it as a separable differential equation.

By separating the variables and integrating both sides, we can derive the general solution. Applying the initial condition \(y(0) = 2\) allows us to find the specific solution for this problem.

Recognising that \(rac{dy}{dx} = 3y\) is separable means we can rewrite it as \(rac{1}{y} \, dy = 3 \, dx\).

Integrating both sides gives us \(\ln|y| = 3x + C\), where \(C\) is the constant of integration. Exponentiating both sides to solve for \(y\), we get \(y = e^{3x + C}\), which simplifies to \(y = Ce^{3x}\) by letting \(e^C\) be a new constant \(C\).

To find the particular solution, we use the initial condition \(y(0) = 2\). Substituting \(x = 0\) and \(y = 2\) into \(y = Ce^{3x}\), we get \(2 = Ce^0\), which simplifies to \(C = 2\).

Thus, the specific solution to the differential equation is \(y = 2e^{3x}\).

Understanding how to solve this type of differential equation is important because it is a common method used in various fields, including physics and engineering, to describe exponential growth or decay processes.

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 use the method of separation of variables. This technique is useful in various mathematical contexts, including exams like the MAT. By rearranging and integrating, we can find the general solution.

Here's how we do it:

• First, separate the variables: \(rac{1}{y} \, dy = 3 \, dx\)
• Next, integrate both sides: \(\ln|y| = 3x + C\)

This method emphasises the importance of understanding the differential equation and the initial condition to solve mathematical problems effectively.

Probability

When considering the probability of getting at least one head in four flips of a fair coin, it's crucial to understand the complementary probability of getting no heads at all.

The probability of flipping a tail on each flip is 0.5. Therefore, the probability of getting four tails in a row is \(0.5^4\). By subtracting this result from 1, you get the probability of obtaining at least one head.

This is important because understanding complementary probabilities can simplify complex problems. Instead of calculating the probability of all possible outcomes that include at least one head, you can just find the probability of the opposite event (no heads) and subtract it from 1.

This approach is often quicker and less prone to error.

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

To calculate the probability of getting at least one head in four flips of a fair coin, we can consider the complementary event of getting no heads at all.

For a fair coin, the probability of getting a tail (and thus no head) in a single flip is \( rac{1}{2} \). Therefore, the probability of getting tails in all four flips is \( \left(rac{1}{2}\right)^4 = rac{1}{16} \).

To find the probability of obtaining at least one head, we subtract this result from 1: \( 1 – rac{1}{16} = rac{15}{16} \).

Key points to remember:

• Using complementary events can simplify complicated probability calculations.
• Multiplying the probabilities of independent events gives the joint probability.

Understanding these principles is crucial for mastering probability questions effectively.

Geometry

In studying the properties of triangles, an important theorem states that the angle opposite the longest side of a triangle is the largest.

This is because there's a direct relationship between the lengths of the sides and the sizes of their opposite angles.

Understanding this concept is essential for tackling various geometric problems and proofs, such as those found in the MAT exam.

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

To show that in any triangle, the angle opposite the longest side is the largest angle, we can utilise some basic principles of Euclidean geometry.

Let's consider a triangle ABC, where side BC is the longest. Our goal is to prove that angle A, which is opposite side BC, is the largest angle in the triangle.

Here are the key points to consider:

• Inequality of Sides and Angles: In any triangle, if one side is longer than another, the angle opposite the longer side is greater.
• Triangle Inequality Theorem: The sum of the lengths of any two sides must be greater than the length of the remaining side.

These principles together support the assertion that the angle opposite the longest side is indeed the largest angle in the triangle.

Optimization

In the context of optimisation, consider the classic problem of determining the dimensions of a rectangle with the maximum area that can be inscribed in a circle with a radius of 5. By utilising principles of geometry and calculus, we aim to identify the best dimensions by analysing the relationship between the rectangle's sides and the circle's radius.

This problem highlights the broader mathematical concept of maximising or minimising a quantity within given constraints. Understanding this problem is important because it showcases how mathematical principles can be applied to find optimal solutions in various scenarios.

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

Maximising the area of a rectangle inscribed in a circle with a radius of 5 requires understanding the relationship between the rectangle's dimensions and the circle's geometry. The diagonal of the rectangle will be equal to the circle's diameter, which is 10 units. To find the maximum area, we consider the properties of a square, as a square provides the largest area for a given perimeter when inscribed in a circle.

• Diagonal Calculation: For a square inscribed in a circle, the diagonal (d) equals the circle's diameter.
• Side Length: The side length (s) of the square can be calculated using \(s = rac{d}{\sqrt{2}}\).

Therefore, the side length is \(5\sqrt{2}\), and the maximum area is 50 square units.

This is important because it shows how geometry can be used to optimise space within given constraints, a useful skill in various fields such as architecture and engineering.

Statistical Analysis

Statistical analysis plays a vital role in making sense of data sets.

Knowing how to determine the mean, median, and mode is key to summarising data distributions.

These measures of central tendency are fundamental for making well-informed decisions based on numerical data.

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

To analyse the given set of numbers {3, 7, 8, 5, 12, 14, 21, 13, 18}, we need to calculate the mean, median, and mode to understand its central tendencies. For MAT exam preparation, understanding these statistical measures is essential.

• Mean: Add all the numbers together and divide by the total count. \((3+7+8+5+12+14+21+13+18)/9 = 11\)
• Median: First, arrange the numbers in ascending order \([3, 5, 7, 8, 12, 13, 14, 18, 21]\). The middle number is 12.

Accurately calculating these measures provides meaningful insights into data analysis, which is a vital skill for various fields.