IB Math AA HL Paper 2 Question Bank

The IB Math AA HL Paper 2 Question Bank is a great way to prepare for your upcoming IB Math AA HL test. This question bank provides you with a wide range of questions that cover all the topics in the curriculum, so you can be sure that you are thoroughly prepared for the exam. The question bank is also a great way to familiarize yourself with the format and structure of the IB Math AA HL exam, so you can feel confident and comfortable when taking the test.

Section A

1.) The weight X of a particular animal is normally distributed with μ= 200kg and σ= 15kg. An animal of this population is overweight if it has a weight greater than 230 kg

a) Find the probability than an animal is overweight.
P (X>230) = 0.02275

b) We select 2 animals from this population. Find the probabilities that

i) both animals are overweight
(0.02275)2 = 0.000518

ii) only one animal is overweight.
0.02275)* (1-0.02275)* 2 = 0.0445

c) We select 7 animals of this population. Find the probability that exactly two of them are overweight.
Y follows B(n,p) with n = 7, p =0.02275
P(Y=2) = 0.00969

2.) Let f(x) = x3 – 3x2 + 2. Find the point of inflection of the graph of y = f(x).

To find the point of inflection, we need to find the critical points of the function and then determine the concavity. The critical points are found by setting the first derivative of the function to 0 and solving for x.
f'(x) = 3x2 – 6x

Setting f'(x) = 0 and solving for x, we find two critical points:
x = 0 and x = 2

To determine the concavity, we need to find the second derivative of the function and evaluate it at the critical points.
f”(x) = 6x – 6
f”(0) = -6 and f”(2) = 6

At x = 0, f”(x) is negative, so the graph is concave down. At x = 2, f”(x) is positive, so the graph is concave up. Since the concavity changes at x = 2, this is the point of inflection. Therefore, the point of inflection of the graph of y = f(x) is (2, f(2)) = (2, -2).

3.) A farmer has 600 meters of fencing and wants to enclose a rectangular area for his sheep. He wants the enclosed area to be as large as possible and has a stream running along one side of the rectangle that will serve as the fourth side.

Formulate the problem as a calculus problem and find the maximum area.

We can formulate this problem as an optimization problem using calculus. Let x be the length of the rectangle and y be the width. We know that the total fencing available is 600 meters and that the rectangle has two sides of length x and two sides of length y.
Therefore, the expression for the total fencing used is 2x + 2y = 600.

The area of the rectangle is given by A = xy. We want to maximize this area subject to the constraint
2x + 2y = 600.

We can start by isolating y in the constraint equation: y = -x + 300

Now we can substitute this expression for y into the area expression and get A = x(-x + 300) = -x2 + 300x
We can find the critical points by taking the derivative of the area expression with respect to x and setting it equal to 0:
A'(x) = -2x + 300 = 0
x = 150

Now we need to check if this critical point is a maximum or minimum. We can do this by taking the second derivative of the area expression and evaluating it at the critical point:
A”(x) = -2

Since the second derivative is negative, the critical point x = 150 is a maximum. We can substitute this value of x back into the constraint equation to find the corresponding value of y:
y = -x + 300 = -150 + 300 = 150

Therefore, the maximum area is 150*150 = 22500 square meters, and the dimensions of the rectangle are 150m by 150m.

4.) Consider the following data:

x1234
y2378

a) Find the mean and the variance for the values of x.
Mean= 2.5, Variance= 1.118032 = 1.25

b)Find the mean and the variance for the values of y.
Mean= 5, Variance= 2.5459512 = 6.5

c)Find the correlation coefficient r.
0.965

d)Describe the relation between x and y.
Strong positive

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5.) Let f’(x) = 2e-x + 10sin5x +1.

Find f(x), given that the curve of this function passes through the point A(0,5).
f(x) = -2e-x – 2cos5x + x + c
f(0) = 5
c= 9
f(x) = -2e-x – 2cos5x + x + 9

6.) A car rental company charges a flat rate of $20 per day plus $0.25 per mile driven. A customer rented a car for x days and drove it a total distance of y miles.

Formulate the cost function C(x,y) of the rental and use it to find the total cost of renting the car for 5 days and driving it 250 miles.

The cost function C(x,y) of the rental can be formulated using the information given in the problem. We know that the flat rate for renting the car is $20 per day and the cost for the distance driven is $0.25 per mile. Therefore, we can write the cost function as
C(x,y) = 20x + 0.25y

To find the total cost of renting the car for 5 days and driving it 250 miles, we substitute x = 5 and y = 250 into the cost function and evaluate:
C(5,250) = 20(5) + 0.25(250) = 100 + 62.5 = $162.50

Therefore, the total cost of renting the car for 5 days and driving it 250 miles is $162.50.

Section B

7.) A company has developed a new product and wants to determine the price that will maximize its profit. The company’s cost, C, in dollars, to produce x units of the product is modelled by the function C(x) = 2000 + 2x2 + 3x. The revenue, R, in dollars, from selling x units of the product is modelled by the function R(x) = px, where p is the unit price in dollars.

a) Write the profit function P(x,p).
The profit function, P(x,p), is the difference between the revenue, R(x), and the cost, C(x).
Therefore, P(x,p) = R(x) – C(x) = px – (2000 + 2x2 + 3x)

b) Determine the number of units, x, that must be sold to break even.
To determine the number of units that must be sold to break even, we need to find the number of units x that make P(x,p) = 0. So, we set P(x,p) = 0 and solve for x:
px – (2000 + 2x2 + 3x) = 0
px = 2000 + 2x2 + 3x
x = (p – 3)/(2p)

c) Determine the unit price, p, that will maximize the profit.
To determine the unit price that will maximize the profit, we need to find the value of p that maximizes P(x,p). We can do this by taking the derivative of P(x,p) with respect to p and setting it equal to zero.
dP/dp = x – (3p – 6x)/p2 = 0
x = (3p – 6x)/p

d) Determine the maximum profit that can be made.
To determine the maximum profit that can be made, we substitute the value of p found in Part 3 back into the profit function P(x,p) and evaluate at x.
P(x,p) = px – (2000 + 2x2 + 3x) = √(6x/x + 3)x – (2000 + 2x2 + 3x)

e) Discuss the effect of changes in the cost function on the profit function.
Changes in the cost function will affect the profit function, as it is a part of the cost function which is subtracted from the revenue function to find the profit.

If the cost of production increases, the profit will decrease as the difference between revenue and cost decreases. Conversely, if the cost of production decreases, the profit will increase as the difference between revenue and cost increases.

However, the effect on the profit function will depend on the specifics of the change in the cost function.

8.) Consider the function f(x) = x3 – 6x2 + 9x + 2, defined for all real numbers x.

a)
i) Find the critical points of f(x)
Solution:
Step 1: Find the first derivative of f(x)
f'(x) = 3x2 – 12x + 9
Step 2: Set f'(x) = 0 and solve for x
3x2 – 12x + 9 = 0
Using the quadratic formula, we get:
x = (12 ± sqrt(122 – 4(3)(9))) / (2(3))
Simplifying further, we get:
x = (12 ± sqrt(144 – 108)) / 6
x = (12 ± sqrt(36)) / 6
x = (12 ± 6) / 6
So, the critical points of f(x) are x = 3 and x = 1.

ii) Determine whether they correspond to local maxima, local minima, or saddle points. Show all your work.
Solution:
Step 1: Find the second derivative of f(x)
f”(x) = 6x – 12
Step 2: Substitute critical points into f”(x) to determine the nature of critical points
Substituting x = 3 into f”(x), we get:
f”(3) = 6(3) – 12 = 6
Since f”(3) > 0, x = 3 corresponds to a local minimum.
Substituting x = 1 into f”(x), we get:
f”(1) = 6(1) – 12 = -6
Since f”(1) < 0, x = 1 corresponds to a local maximum.
So, the final answer for part (a) is:
The critical point x = 3 corresponds to a local minimum, and the critical point x = 1 corresponds to a local maximum.

b) Sketch the graph of f(x), clearly indicating the location of the critical points, any inflection points, and the concavity of the graph. Label any x-intercepts and y-intercepts, and provide a brief description of the overall behavior of the graph.
Based on the information obtained from part (a), we can sketch the graph of f(x) as follows:
The critical point x = 3 corresponds to a local minimum, so the graph of f(x) will have an upward concavity at x = 3.
The critical point x = 1 corresponds to a local maximum, so the graph of f(x) will have a downward concavity at x = 1.
The x-intercepts and y-intercepts of f(x) can be found by setting f(x) = 0 and x = 0, respectively, and solving for x and y.
The overall behavior of the graph can be described as a cubic function that approaches positive infinity as x goes to negative infinity and approaches negative infinity as x goes to positive infinity.
The final sketch of the graph of f(x) should clearly indicate the location of the critical points, any inflection points, the concavity of the graph, and label any x-intercepts and y-intercepts.

c) Find the points of symmetry
Step 1: Determine if f(x) has any points of symmetry
To determine if f(x) has any points of symmetry, we need to check if it is an even or odd function.
Step 2: Test for even function
A function is even if f(-x) = f(x) for all x in its domain. Substitute -x for x in f(x) and check if the equation holds true.
Step 3: Test for odd function
A function is odd if f(-x) = -f(x) for all x in its domain. Substitute -x for x in f(x) and check if the equation holds true.

d) Find the intervals of concavity
Step 1: Find the second derivative of f(x)
f”(x) = 6x – 2
Step 2: Set f”(x) = 0 and solve for x
6x – 2 = 0
x = 1/3
Choose a test point from each of the intervals (-∞, 1/3) and (1/3, ∞) and plug it into f”(x). Based on the sign of f”(x) at each test point, determine the intervals of concavity.

e) Find the points of inflection
Step 1: Set f”(x) = 0 and solve for x
From part (d), we found that f”(x) = 0 at x = 1/3.
Step 2: Use the second derivative test to determine if x = 1/3 is a point of inflection
Choose a test point within the interval (1/3 – ε, 1/3 + ε) where ε is a small positive value. Plug it into f”(x) and based on the sign of f”(x) at the test point, determine if x = 1/3 is a point of inflection.

9.) A factory produces two types of electronic devices, A and B. The factory has 10 machines that produce type A devices and 8 machines that produce type B devices. The probability of a type A device being defective is 0.04, while the probability of a type B device being defective is 0.02.

a) A device is randomly selected from the factory and found to be defective. What is the probability that it is a type A device?
To find the probability that a randomly selected defective device is of type A, we can use Bayes’ theorem. Let’s denote the events as follows: A = selecting a type A device, B = selecting a type B device, D = selecting a defective device.
Given: P(A) = probability of selecting a type A device = 10/18 (since there are 10 machines that produce type A devices out of a total of 18 machines producing both type A and type B devices)
P(B) = probability of selecting a type B device = 8/18 (since there are 8 machines that produce type B devices out of a total of 18 machines producing both type A and type B devices)
P(D|A) = probability of selecting a defective device given that it is a type A device = 0.04
P(D|B) = probability of selecting a defective device given that it is a type B device = 0.02
We need to find P(A|D), the probability of selecting a type A device given that the device is defective.
By Bayes’ theorem, we have:
P(A|D) = P(D|A) * P(A) / (P(D|A) * P(A) + P(D|B) * P(B))
Substituting the given values, we get:
P(A|D) = 0.04 * (10/18) / (0.04 * (10/18) + 0.02 * (8/18))
Simplifying the expression, we get:
P(A|D) = 20/43
So the probability that a randomly selected defective device is of type A is 20/43.

b) Two devices are randomly selected from the factory, one after the other without replacement. What is the probability that both devices are type B devices and both are defective?
To find the probability that both devices selected are type B devices and both are defective, we can use conditional probability.
The probability of selecting the first device as a type B device and it being defective is: P(B and D) = P(D|B) * P(B) = 0.02 * (8/18)
After selecting the first type B device and assuming it is defective, there are 7 type B devices left out of a total of 17 remaining devices. So the probability of selecting the second device as a type B device and it being defective is: P(B and D|B and D) = P(D|B and D) * P(B|B and D) = 0.02 * (7/17)
Thus, the probability of both devices being type B devices and both being defective is:
P(B and D and B and D) = P(B and D) * P(B and D|B and D) = (0.02 * (8/18)) * (0.02 * (7/17))
Simplifying the expression, we get:
P(B and D and B and D) = 0.0003847

c) Three devices are randomly selected from the factory without replacement. What is the probability that exactly two of them are type A devices?
To find the probability that exactly two out of three randomly selected devices are type A devices, we can use combinatorial methods.
There are three ways this can happen: either the first two devices are type A and the third is not, or the first and third devices are type A and the second is not, or the second and third devices are type A and the first is not.
The probability of the first two devices being type A and the third not being type A is: P(A and A and not A) = P(A)2 * (1 – P(A)) = (10/18)2 * (1 – 10/18)
The probability of the first and third devices being type A and the second device not being type A is: P(A and not A and A) = P(A) * (1 – P(A))2 = (10/18) * (1 – 10/18)2
The probability of the second and third devices being type A and the first device not being type A is: P(not A and A and A) = (1 – P(A)) * P(A)2 = (1 – 10/18) * (10/18)2
Adding these three probabilities together, we get the total probability of exactly two out of three devices being type A devices:
P(exactly 2 type A devices) = P(A and A and not A) + P(A and not A and A) + P(not A and A and A)
Simplifying the expression, we get:
P(exactly 2 type A devices) = (10/18)2 * (1 – 10/18) + (10/18) * (1 – 10/18)2 + (1 – 10/18) * (10/18)2
You can further simplify this expression and calculate the numerical value using a calculator.

d) The factory wants to select 4 devices at random for quality control testing. What is the probability that exactly 3 of them are type B devices?
P(exactly 2 type A devices) = (10/18)2 * (1 – 10/18) + (10/18) * (1 – 10/18)2 + (1 – 10/18) * (10/18)2
Plugging in the values and simplifying, we get:
P(exactly 2 type A devices) = (10/18)2 * (1 – 10/18) + (10/18) * (1 – 10/18)2 + (1 – 10/18) * (10/18)2
≈ 0.273 (rounded to three decimal places)
So, the probability of exactly two out of three devices being type A devices is approximately 0.273 or 27.3% (rounded to the nearest tenth of a percent).

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