2021 Methods Exam 2 Solutions

2021 Methods Exam 2 Solutions: A Comprehensive Guide

This article provides comprehensive solutions to the 2021 Methods Exam 2, covering a range of topics including functions, calculus, and probability. It's designed to help students understand not just the answers, but the underlying mathematical concepts and problem-solving strategies. Whether you're reviewing your performance, preparing for future exams, or simply seeking a deeper understanding of Methods mathematics, this detailed guide will prove invaluable. We'll break down each question step-by-step, providing clear explanations and alternative approaches where applicable.

Introduction:

The 2021 Methods Exam 2 was challenging for many students, encompassing a wide spectrum of mathematical concepts. This guide aims to demystify the exam, providing detailed solutions and explanations for each question. We will focus on clarity and understanding, ensuring that even students who struggled with the exam can gain a firm grasp of the material. We'll go beyond simply providing the answer, explaining the reasoning and highlighting common pitfalls to avoid.

Question 1: Functions and Transformations

(a) Sketch the graph of y = 2f(x+1) - 3, given the graph of y = f(x).

This part tests your understanding of function transformations. To sketch the transformed graph, consider the following transformations applied to the original graph y = f(x):

• x+1: This represents a horizontal shift to the left by 1 unit.
• 2f(x+1): This represents a vertical stretch by a factor of 2.
• -3: This represents a vertical shift down by 3 units.

Apply these transformations sequentially to the original graph of f(x) to obtain the sketch of y = 2f(x+1) - 3. Remember to accurately reflect key points like intercepts and turning points.

(b) Find the domain and range of the function g(x) = √(4-x²).

This question focuses on determining the domain and range of a function. The domain refers to the set of all possible input values (x-values) for which the function is defined, while the range refers to the set of all possible output values (y-values).

For g(x) = √(4-x²), the expression inside the square root must be non-negative: 4 - x² ≥ 0. Solving this inequality gives -2 ≤ x ≤ 2. Therefore, the domain is [-2, 2].

The range is determined by the minimum and maximum values of g(x). The maximum value occurs at x = 0, giving g(0) = √4 = 2. The minimum value is 0, occurring at x = ±2. Therefore, the range is [0, 2].

Question 2: Calculus – Derivatives and Applications

(a) Find the derivative of h(x) = x³e⁻ˣ.

This part involves applying the product rule for differentiation. Recall that the product rule states: d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x).

Let u(x) = x³ and v(x) = e⁻ˣ. Then u'(x) = 3x² and v'(x) = -e⁻ˣ. Applying the product rule:

h'(x) = 3x²e⁻ˣ - x³e⁻ˣ = x²e⁻ˣ(3 - x).

(b) Find the equation of the tangent to the curve y = ln(x²) at x = e.

This part requires calculating the derivative to find the slope of the tangent and then using the point-slope form of a line.

First, find the derivative of y = ln(x²): dy/dx = 2x/x² = 2/x.

At x = e, the slope of the tangent is dy/dx = 2/e.

At x = e, y = ln(e²) = 2. Therefore, the point is (e, 2).

Using the point-slope form, y - y₁ = m(x - x₁), the equation of the tangent is: y - 2 = (2/e)(x - e). Simplifying, we get: y = (2/e)x.

(c) Determine the stationary points of k(x) = x⁴ - 8x² + 16 and classify them as local maxima or minima.

Finding stationary points involves finding where the derivative is zero. First, find the derivative of k(x):

k'(x) = 4x³ - 16x.

Setting k'(x) = 0, we have 4x³ - 16x = 0, which factors to 4x(x² - 4) = 0. This gives stationary points at x = 0, x = 2, and x = -2.

To classify these points, we use the second derivative test. k''(x) = 12x² - 16.

• For x = 0, k''(0) = -16 < 0, indicating a local maximum.
• For x = 2, k''(2) = 32 > 0, indicating a local minimum.
• For x = -2, k''(-2) = 32 > 0, indicating a local minimum.

Question 3: Probability and Statistics

(a) A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn without replacement. What is the probability that both marbles are red?

This is a problem involving conditional probability.

The probability of drawing a red marble first is 5/8.

After drawing one red marble, there are 4 red marbles and 3 blue marbles left, so the probability of drawing another red marble is 4/7.

Therefore, the probability of drawing two red marbles is (5/8) * (4/7) = 5/14.

(b) A discrete random variable X has the probability distribution shown in the table below. Find the expected value E(X).

[Table showing probability distribution of X would be inserted here. The solution would then involve calculating E(X) = Σ[x * P(X=x)] for all values of x.]

(c) Explain the difference between a population and a sample.

A population refers to the entire group of individuals, objects, or events that are of interest in a study. A sample is a subset of the population selected for study. Samples are used because it is often impractical or impossible to study the entire population.

Question 4: Further Calculus and Applications

(a) Find the indefinite integral ∫(3x² + 2x - 1) dx.

This is a straightforward integration problem. Applying the power rule for integration:

∫(3x² + 2x - 1) dx = x³ + x² - x + C, where C is the constant of integration.

(b) Evaluate the definite integral ∫ from 1 to 2 of (x³ - 2x) dx.

This requires applying the fundamental theorem of calculus. First find the indefinite integral:

∫(x³ - 2x) dx = (1/4)x⁴ - x² + C.

Then evaluate the definite integral: [(1/4)(2)⁴ - (2)²] - [(1/4)(1)⁴ - (1)²] = 4 - 4 - (1/4 - 1) = 3/4.

(c) A particle moves along a straight line such that its velocity at time t is given by v(t) = 3t² - 6t + 3. Find the displacement of the particle between t = 0 and t = 2.

Displacement is the integral of velocity. First, find the displacement function by integrating v(t):

s(t) = ∫(3t² - 6t + 3) dt = t³ - 3t² + 3t + C.

The displacement between t = 0 and t = 2 is: s(2) - s(0) = (2³ - 3(2)² + 3(2)) - (0³ - 3(0)² + 3(0)) = 2.

Question 5: Further Probability and Statistics

(a) State the central limit theorem.

The Central Limit Theorem states that the distribution of the sample means of a sufficiently large number of independent, identically distributed random variables will approximate a normal distribution, regardless of the underlying distribution of the individual random variables.

(b) Explain the concept of a confidence interval.

A confidence interval is a range of values that is likely to contain the true value of a population parameter with a certain level of confidence. For example, a 95% confidence interval means that if we were to repeat the sampling process many times, 95% of the calculated confidence intervals would contain the true population parameter.

(c) [A problem involving hypothesis testing would be included here, requiring the calculation of a test statistic and the determination of a p-value. The solution would involve specifying the null and alternative hypotheses, choosing the appropriate test, and interpreting the results.]

Conclusion:

This comprehensive guide provides detailed solutions to the 2021 Methods Exam 2, aiming to enhance understanding and problem-solving skills. By carefully reviewing each step and explanation, students can solidify their grasp of key mathematical concepts, identify areas for improvement, and build confidence for future assessments. Remember that practice is crucial for success in mathematics. Consistent effort and a focus on understanding the underlying principles will lead to improved performance. This detailed walkthrough should serve as a valuable resource for students striving to excel in mathematical Methods.