IB Mathematics: Applications and Interpretation Quiz: Function Notation And Interpretation
20 questions · exam conditions
0:00
Function Notation And InterpretationQuestion 1 of 20

The perceived brightness, BB, of a light source is modelled by B(d)=1200d2B(d) = \frac{1200}{d^2}, where dd is the distance from the source in metres, for d>0d > 0. Which statement correctly interprets the function's behaviour as dd increases?

As distance increases, the brightness decreases and approaches zero.
As distance increases, the brightness increases towards a maximum value.
The brightness is constant regardless of the distance from the source.
The brightness decreases linearly as the distance increases.
← Back to quizzes

IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Function Notation And Interpretation

Practice Function Notation And Interpretation in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Function Notation And Interpretation, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The perceived brightness, BB, of a light source is modelled by B(d)=1200d2B(d) = \frac{1200}{d^2}, where dd is the distance from the source in metres, for d>0d > 0. Which statement correctly interprets the function's behaviour as dd increases?

  1. As distance increases, the brightness decreases and approaches zero. (correct answer)
  2. As distance increases, the brightness increases towards a maximum value.
  3. The brightness is constant regardless of the distance from the source.
  4. The brightness decreases linearly as the distance increases.
Explanation: When you encounter functions modeling real-world phenomena like light brightness, focus on analyzing the function's behavior by examining its mathematical form and considering what happens to the output as the input changes. The function B(d)=1200d2B(d) = \frac{1200}{d^2} is an inverse square relationship, which is fundamental in physics for phenomena like light intensity, gravitational force, and sound intensity. As distance dd increases, you're dividing 1200 by increasingly larger values of d2d^2, making B(d)B(d) smaller and smaller. To see this clearly, test some values: at d=1d = 1, B=1200B = 1200; at d=10d = 10, B=12B = 12; at d=100d = 100, B=0.12B = 0.12. As dd approaches infinity, BB approaches zero but never quite reaches it. This confirms that answer A is correct. Looking at the incorrect options: B suggests brightness increases, which contradicts the inverse relationship—larger denominators create smaller fractions. C claims brightness is constant, but this function clearly depends on distance; a constant function would have no variable in it. D describes linear decrease, but this function decreases much more rapidly than a linear relationship—it follows an inverse square pattern where doubling the distance quarters the brightness. Remember that inverse square relationships are everywhere in physics and always follow this pattern: as one quantity increases, the other decreases rapidly and approaches (but never reaches) zero. Recognizing constantx2\frac{\text{constant}}{x^2} immediately tells you the behavior.

Question 2

The cost, CC, in dollars, for parking in a garage for tt hours is given by the piecewise function:

Using the piecewise function for parking cost, what is the interpretation of C(3.5)C(3.5)?

  1. It is the cost for parking for 1 hour, which is $8.
  2. It is the cost for parking for more than 5 hours, which is $30.
  3. It is the cost for parking for 3.5 hours, calculated as 8+4(3.5−1)8 + 4(3.5-1), which is $18. (correct answer)
  4. It is the average cost per hour for parking for 3.5 hours.
Explanation: To find C(3.5)C(3.5), we must first determine which piece of the function to use. Since the input t=3.5t=3.5 satisfies the condition 1<t≤51 < t \le 5, we use the second formula: C(t)=8+4(t−1)C(t) = 8 + 4(t-1). Substituting t=3.5t=3.5 gives C(3.5)=8+4(3.5−1)=8+4(2.5)=8+10=18C(3.5) = 8 + 4(3.5-1) = 8 + 4(2.5) = 8 + 10 = 18. This value represents the total cost in dollars for parking for 3.5 hours.

Question 3

The depth of water, DD, in metres, at a harbour entrance is modelled by the function D(t)=12+4cos⁡(0.5t)D(t) = 12 + 4\cos(0.5t), where tt is the number of hours after midnight. What is the range of water depths predicted by this model?

  1. 8≤D≤168 \le D \le 16 (correct answer)
  2. 0≤t≤240 \le t \le 24
  3. 12≤D≤1612 \le D \le 16
  4. D≥0D \ge 0
Explanation: When you encounter a trigonometric function modeling a real-world situation, you need to identify the range by analyzing how the trigonometric component behaves. This function represents a sinusoidal model where the cosine function creates periodic variation around a central value. The function D(t)=12+4cos⁡(0.5t)D(t) = 12 + 4\cos(0.5t) has the form of a vertical shift plus an amplitude variation. Since cosine oscillates between -1 and 1, you can find the extreme values by substituting these bounds. When cos⁡(0.5t)=1\cos(0.5t) = 1, the depth reaches its maximum: D=12+4(1)=16D = 12 + 4(1) = 16 metres. When cos⁡(0.5t)=−1\cos(0.5t) = -1, the depth reaches its minimum: D=12+4(−1)=8D = 12 + 4(-1) = 8 metres. Therefore, the range is 8≤D≤168 \le D \le 16. Option A correctly identifies this range. Option B gives 0≤t≤240 \le t \le 24, which describes the domain (input values) for a 24-hour period, not the range of water depths. This is a common confusion between domain and range. Option C states 12≤D≤1612 \le D \le 16, which only captures the upper half of the oscillation—this would be correct if the cosine term could only be positive, but cosine takes both positive and negative values. Option D suggests D≥0D \ge 0, which is far too broad and doesn't reflect the specific constraints of this model. Remember: for sinusoidal functions in the form a+bcos⁡(ct)a + b\cos(ct), the range is always [a−∣b∣,a+∣b∣][a-|b|, a+|b|]. The coefficient aa is the midline, and ∣b∣|b| is the amplitude of oscillation.

Question 4

The temperature of a cup of coffee, TT, in degrees Celsius, is modelled by the function T(m)=20+70(0.95)mT(m) = 20 + 70(0.95)^m, where mm is the number of minutes after it is poured. What is the practical meaning of the expression T(5)−T(10)T(5) - T(10)?

  1. The average temperature of the coffee during the first 10 minutes.
  2. The decrease in the coffee's temperature between the 5th minute and the 10th minute. (correct answer)
  3. The time it takes for the temperature to drop from its value at 5 minutes to its value at 10 minutes.
  4. The average rate at which the coffee is cooling between the 5th and 10th minute.
Explanation: T(5)T(5) is the temperature at 5 minutes, and T(10)T(10) is the temperature at 10 minutes. The difference, T(5)−T(10)T(5) - T(10), represents the total change in temperature between these two points in time. Since the coffee is cooling, T(5)T(5) will be greater than T(10)T(10), so the result is a positive value representing the total temperature drop, or decrease, over that 5-minute interval.

Question 5

The concentration of a certain medication in a patient's bloodstream, CC, in milligrams per litre (mg/L), is modelled by the function C(t)=100tt2+4C(t) = \frac{100t}{t^2 + 4}, where tt is the number of hours after the medication was taken, for t≥0t \ge 0. What does the value C(2)C(2) represent?

  1. The time in hours it takes for the concentration to reach 2 mg/L.
  2. The concentration of the medication in the bloodstream after 2 hours. (correct answer)
  3. The maximum concentration of the medication in the bloodstream.
  4. The rate at which the medication's concentration is changing after 2 hours.
Explanation: In the function notation C(t)C(t), tt is the input (time in hours) and C(t)C(t) is the output (concentration in mg/L). Therefore, C(2)C(2) represents the output of the function when the input is t=2t=2. This means it is the concentration of the medication in the bloodstream after 2 hours. While t=2t=2 happens to be where the maximum concentration occurs for this function, the expression C(2)C(2) itself simply means the value of C at t=2.

Question 6

An open-top box is made by cutting congruent squares of side length xx cm from the corners of a rectangular piece of cardboard measuring 30 cm by 40 cm. The volume of the box, VV, in cm³, is modelled by V(x)=x(30−2x)(40−2x)V(x) = x(30-2x)(40-2x). Which of the following represents the practical domain of this function?

  1. x>0x > 0
  2. 0<x<150 < x < 15 (correct answer)
  3. 0<x<200 < x < 20
  4. x∈Rx \in \mathbb{R}
Explanation: The domain is the set of possible values for xx. For the volume to be physically possible, all dimensions must be positive. The side length of the cut square, xx, must be greater than 0. The side length of the box's base, 30−2x30-2x, must be greater than 0, which implies 30>2x30 > 2x, or x<15x < 15. The other side length, 40−2x40-2x, must also be greater than 0, which implies 40>2x40 > 2x, or x<20x < 20. For all three conditions to be met (x>0x>0, x<15x<15, and x<20x<20), xx must be between 0 and 15. Thus, the practical domain is 0<x<150 < x < 15.

Question 7

A rectangular garden has a fixed perimeter of 100 metres. The area of the garden, AA, can be modelled as a function of its width, ww, by A(w)=w(50−w)A(w) = w(50-w). What is the practical domain for the width, ww?

  1. w>0w > 0
  2. 0<w<250 < w < 25
  3. w∈Rw \in \mathbb{R}
  4. 0<w<500 < w < 50 (correct answer)
Explanation: When you encounter optimization problems involving real-world constraints, identifying the practical domain requires considering both mathematical and physical limitations. The function A(w)=w(50−w)A(w) = w(50-w) models area as a function of width, but not all mathematical values of ww make sense in reality. Since this is a rectangular garden, the width must be positive, so w>0w > 0. To find the upper bound, consider what the function tells us: if the area is A(w)=w(50−w)A(w) = w(50-w), then the length must be (50−w)(50-w) since area equals width times length. This makes sense because with a perimeter of 100 meters, we have 2w+2l=1002w + 2l = 100, which gives us l=50−wl = 50 - w. For a rectangle to exist, both dimensions must be positive. While w>0w > 0 is obvious, we also need the length to be positive: 50−w>050 - w > 0, which means w<50w < 50. Therefore, the practical domain is 0<w<500 < w < 50. Looking at the wrong answers: Choice A (w>0w > 0) ignores the upper constraint that length must also be positive. Choice B (0<w<250 < w < 25) incorrectly assumes the garden must be a square or that width cannot exceed length, but the problem doesn't specify which dimension is called "width." Choice C (w∈Rw \in \mathbb{R}) ignores all physical constraints entirely. Remember: In optimization problems with real-world contexts, always check that all physical quantities remain positive and meaningful throughout your domain.

Question 8

The velocity, vv, in metres per second, of a wave in shallow water is modelled by v(d)=9.8dv(d) = \sqrt{9.8d}, where dd is the depth of the water in metres. The model is used for a coastal area where the depth ranges from 0.5 metres to 10 metres. What is the domain of the function in this specific context?

  1. d≥0d \ge 0
  2. 0.5≤d≤100.5 \le d \le 10 (correct answer)
  3. 2.21≤v≤9.902.21 \le v \le 9.90
  4. d∈Rd \in \mathbb{R}
Explanation: The domain is the set of valid inputs, dd. Mathematically, the square root requires d≥0d \ge 0. However, the problem specifies that the model is being applied to a context where the depth, dd, ranges from 0.5 metres to 10 metres. The practical domain is therefore restricted by this context. The correct domain is 0.5≤d≤100.5 \le d \le 10.

Question 9

The profit, PP, in euros (EUR), from selling xx handmade scarves is modelled by the function P(x)=25x−300P(x) = 25x - 300. The model is considered valid for the first 100 scarves made each month. Which statement best describes the domain of this function in this specific context?

  1. {x∈R∣0≤x≤100}\{x \in \mathbb{R} \mid 0 \le x \le 100\}
  2. {x∈Z∣0≤x≤100}\{x \in \mathbb{Z} \mid 0 \le x \le 100\} (correct answer)
  3. {P∈R∣−300≤P≤2200}\{P \in \mathbb{R} \mid -300 \le P \le 2200\}
  4. {x∈Z∣x≥12}\{x \in \mathbb{Z} \mid x \ge 12\}
Explanation: The domain represents the set of all possible input values. In this context, the input xx is the number of handmade scarves. Since scarves are discrete, countable items, xx must be an integer (Z\mathbb{Z}). The context states the model is valid for the first 100 scarves, and the number of scarves cannot be negative, so the domain is all integers from 0 to 100, inclusive. This is represented by {x∈Z∣0≤x≤100}\{x \in \mathbb{Z} \mid 0 \le x \le 100\}.

Question 10

The concentration of a certain medication in a patient's bloodstream, CC, in milligrams per litre (mg/L), is modelled by the function C(t)=100tt2+4C(t) = \frac{100t}{t^2 + 4}, where tt is the number of hours after the medication was taken, for t≥0t \ge 0. What does the value C(2)C(2) represent?

  1. The time in hours it takes for the concentration to reach 2 mg/L.
  2. The concentration of the medication in the bloodstream after 2 hours. (correct answer)
  3. The maximum concentration of the medication in the bloodstream.
  4. The rate at which the medication's concentration is changing after 2 hours.
Explanation: In the function notation C(t)C(t), tt is the input (time in hours) and C(t)C(t) is the output (concentration in mg/L). Therefore, C(2)C(2) represents the output of the function when the input is t=2t=2. This means it is the concentration of the medication in the bloodstream after 2 hours. While t=2t=2 happens to be where the maximum concentration occurs for this function, the expression C(2)C(2) itself simply means the value of C at t=2.

Question 11

The height, hh, in metres, of a ball thrown upwards from the top of a 50-metre building is given by h(t)=−4.9t2+20t+50h(t) = -4.9t^2 + 20t + 50, where tt is the time in seconds. The ball hits the ground after approximately 5.88 seconds. What is the most appropriate range for this function in this context?

  1. {t∈R∣0≤t≤5.88}\{t \in \mathbb{R} \mid 0 \le t \le 5.88\}
  2. {h∈R∣h≥0}\{h \in \mathbb{R} \mid h \ge 0\}
  3. {h∈R∣0≤h≤70.4}\{h \in \mathbb{R} \mid 0 \le h \le 70.4\} (correct answer)
  4. {h∈R∣50≤h≤70.4}\{h \in \mathbb{R} \mid 50 \le h \le 70.4\}
Explanation: The range represents the set of all possible output values (heights). The minimum height is 0 metres, when the ball hits the ground. The maximum height occurs at the vertex of the parabola. The time at the vertex is t=−b/(2a)=−20/(2×−4.9)≈2.04t = -b/(2a) = -20/(2 \times -4.9) \approx 2.04 seconds. The maximum height is h(2.04)=−4.9(2.04)2+20(2.04)+50≈70.4h(2.04) = -4.9(2.04)^2 + 20(2.04) + 50 \approx 70.4 metres. Therefore, the practical range of heights is from 0 to 70.4 metres, inclusive.

Question 12

The value of an investment, VV, in thousands of dollars, is modelled by V(t)=10(1.04)tV(t) = 10(1.04)^t, where tt is the number of years since the initial investment. Let V−1(x)V^{-1}(x) be the inverse function of V(t)V(t). What is the practical interpretation of V−1(20)V^{-1}(20)?

  1. The value of the investment after 20 years, which would be V(20)V(20).
  2. The initial value of the investment required to reach a value of $20,000.
  3. The number of years it takes for the investment's value to reach $20,000. (correct answer)
  4. The rate of increase of the investment's value when it is worth $20,000.
Explanation: An inverse function reverses the input and output. If V(t)V(t) takes a time tt and gives a value VV, then V−1(x)V^{-1}(x) takes a value xx and gives the time tt at which that value was reached. The input to V−1V^{-1} is 20, which represents a value of $20,000 (since V is in thousands). The output will be the time tt. Therefore, V−1(20)V^{-1}(20) is the number of years it takes for the investment's value to reach $20,000.

Question 13

The cost, CC, in dollars, for parking in a garage for tt hours is given by the piecewise function:

Using the piecewise function for parking cost, what is the interpretation of C(3.5)C(3.5)?

  1. It is the cost for parking for 1 hour, which is $8.
  2. It is the cost for parking for more than 5 hours, which is $30.
  3. It is the cost for parking for 3.5 hours, calculated as 8+4(3.5−1)8 + 4(3.5-1), which is $18. (correct answer)
  4. It is the average cost per hour for parking for 3.5 hours.
Explanation: To find C(3.5)C(3.5), we must first determine which piece of the function to use. Since the input t=3.5t=3.5 satisfies the condition 1<t≤51 < t \le 5, we use the second formula: C(t)=8+4(t−1)C(t) = 8 + 4(t-1). Substituting t=3.5t=3.5 gives C(3.5)=8+4(3.5−1)=8+4(2.5)=8+10=18C(3.5) = 8 + 4(3.5-1) = 8 + 4(2.5) = 8 + 10 = 18. This value represents the total cost in dollars for parking for 3.5 hours.

Question 14

The number of users of a social media app, NN, in millions, is given by N(t)=2t2+10t+50N(t) = 2t^2 + 10t + 50, where tt is the number of months since its launch. What does the expression N(6)−N(3)6−3\frac{N(6) - N(3)}{6-3} represent?

  1. The total increase in users, in millions, from the 3rd month to the 6th month.
  2. The number of users exactly halfway between the 3rd and 6th month.
  3. The average rate of change in the number of users, in millions per month, from month 3 to month 6. (correct answer)
  4. The exact rate of growth of users, in millions per month, at the 3-month mark.
Explanation: The expression f(b)−f(a)b−a\frac{f(b) - f(a)}{b-a} is the formula for the average rate of change of a function ff over the interval [a,b][a, b]. In this context, N(6)−N(3)N(6) - N(3) is the total change in the number of users (in millions) from month 3 to month 6. Dividing by the time interval, 6−36-3 months, gives the average rate of change over that period, with units of millions of users per month.

Question 15

The velocity, vv, in metres per second, of a wave in shallow water is modelled by v(d)=9.8dv(d) = \sqrt{9.8d}, where dd is the depth of the water in metres. The model is used for a coastal area where the depth ranges from 0.5 metres to 10 metres. What is the domain of the function in this specific context?

  1. d≥0d \ge 0
  2. 0.5≤d≤100.5 \le d \le 10 (correct answer)
  3. 2.21≤v≤9.902.21 \le v \le 9.90
  4. d∈Rd \in \mathbb{R}
Explanation: The domain is the set of valid inputs, dd. Mathematically, the square root requires d≥0d \ge 0. However, the problem specifies that the model is being applied to a context where the depth, dd, ranges from 0.5 metres to 10 metres. The practical domain is therefore restricted by this context. The correct domain is 0.5≤d≤100.5 \le d \le 10.

Question 16

The depth of water, DD, in metres, at a harbour entrance is modelled by the function D(t)=12+4cos⁡(0.5t)D(t) = 12 + 4\cos(0.5t), where tt is the number of hours after midnight. What is the range of water depths predicted by this model?

  1. 8≤D≤168 \le D \le 16 (correct answer)
  2. 0≤t≤240 \le t \le 24
  3. 12≤D≤1612 \le D \le 16
  4. D≥0D \ge 0
Explanation: When you encounter a trigonometric function modeling a real-world situation, you need to identify the range by analyzing how the trigonometric component behaves. This function represents a sinusoidal model where the cosine function creates periodic variation around a central value. The function D(t)=12+4cos⁡(0.5t)D(t) = 12 + 4\cos(0.5t) has the form of a vertical shift plus an amplitude variation. Since cosine oscillates between -1 and 1, you can find the extreme values by substituting these bounds. When cos⁡(0.5t)=1\cos(0.5t) = 1, the depth reaches its maximum: D=12+4(1)=16D = 12 + 4(1) = 16 metres. When cos⁡(0.5t)=−1\cos(0.5t) = -1, the depth reaches its minimum: D=12+4(−1)=8D = 12 + 4(-1) = 8 metres. Therefore, the range is 8≤D≤168 \le D \le 16. Option A correctly identifies this range. Option B gives 0≤t≤240 \le t \le 24, which describes the domain (input values) for a 24-hour period, not the range of water depths. This is a common confusion between domain and range. Option C states 12≤D≤1612 \le D \le 16, which only captures the upper half of the oscillation—this would be correct if the cosine term could only be positive, but cosine takes both positive and negative values. Option D suggests D≥0D \ge 0, which is far too broad and doesn't reflect the specific constraints of this model. Remember: for sinusoidal functions in the form a+bcos⁡(ct)a + b\cos(ct), the range is always [a−∣b∣,a+∣b∣][a-|b|, a+|b|]. The coefficient aa is the midline, and ∣b∣|b| is the amplitude of oscillation.

Question 17

A rectangular garden has a fixed perimeter of 100 metres. The area of the garden, AA, can be modelled as a function of its width, ww, by A(w)=w(50−w)A(w) = w(50-w). What is the practical domain for the width, ww?

  1. w>0w > 0
  2. 0<w<250 < w < 25
  3. w∈Rw \in \mathbb{R}
  4. 0<w<500 < w < 50 (correct answer)
Explanation: When you encounter optimization problems involving real-world constraints, identifying the practical domain requires considering both mathematical and physical limitations. The function A(w)=w(50−w)A(w) = w(50-w) models area as a function of width, but not all mathematical values of ww make sense in reality. Since this is a rectangular garden, the width must be positive, so w>0w > 0. To find the upper bound, consider what the function tells us: if the area is A(w)=w(50−w)A(w) = w(50-w), then the length must be (50−w)(50-w) since area equals width times length. This makes sense because with a perimeter of 100 meters, we have 2w+2l=1002w + 2l = 100, which gives us l=50−wl = 50 - w. For a rectangle to exist, both dimensions must be positive. While w>0w > 0 is obvious, we also need the length to be positive: 50−w>050 - w > 0, which means w<50w < 50. Therefore, the practical domain is 0<w<500 < w < 50. Looking at the wrong answers: Choice A (w>0w > 0) ignores the upper constraint that length must also be positive. Choice B (0<w<250 < w < 25) incorrectly assumes the garden must be a square or that width cannot exceed length, but the problem doesn't specify which dimension is called "width." Choice C (w∈Rw \in \mathbb{R}) ignores all physical constraints entirely. Remember: In optimization problems with real-world contexts, always check that all physical quantities remain positive and meaningful throughout your domain.

Question 18

The sound intensity level, LL, in decibels (dB), is given by the function L(I)=10log⁡10(II0)L(I) = 10 \log_{10}\left(\frac{I}{I_0}\right), where II is the sound intensity and I0I_0 is a reference intensity. What is the correct interpretation of L(100I0)L(100I_0)?

  1. The sound intensity when the decibel level is 100.
  2. The decibel level of a sound that is 100 times less intense than the reference.
  3. The increase in decibels when the sound intensity is multiplied by 100.
  4. The decibel level of a sound that is 100 times more intense than the reference. (correct answer)
Explanation: When you encounter logarithmic functions like this sound intensity formula, focus on understanding what the input variable represents and how the function transforms it. The function L(I)=10log⁡10(II0)L(I) = 10 \log_{10}\left(\frac{I}{I_0}\right) converts sound intensity ratios into decibel levels. To find L(100I0)L(100I_0), substitute I=100I0I = 100I_0 into the formula: L(100I0)=10log⁡10(100I0I0)=10log⁡10(100)=10×2=20 dBL(100I_0) = 10 \log_{10}\left(\frac{100I_0}{I_0}\right) = 10 \log_{10}(100) = 10 \times 2 = 20 \text{ dB} This represents the decibel level when the sound intensity is 100 times the reference intensity I0I_0. The key insight is that L(100I0)L(100I_0) is asking for the output (decibel level) when the input intensity is 100I0100I_0. Option A incorrectly reverses the relationship—it describes finding intensity from a given decibel level, which would require solving L(I)=100L(I) = 100 for II. Option B contains two errors: it suggests the sound is less intense (when 100I0>I0100I_0 > I_0) and misinterprets what the function calculates. Option C describes the change in decibels rather than the absolute decibel level, which isn't what L(100I0)L(100I_0) represents. Option D correctly identifies that we're finding the decibel level (the function's output) for a sound with intensity 100I0100I_0, which is indeed 100 times more intense than the reference I0I_0. Study tip: In function notation f(x)f(x), always identify what goes in (the input) and what comes out (the output). Don't confuse finding f(a)f(a) with solving f(x)=af(x) = a.

Question 19

The profit, PP, in euros (EUR), from selling xx handmade scarves is modelled by the function P(x)=25x−300P(x) = 25x - 300. The model is considered valid for the first 100 scarves made each month. Which statement best describes the domain of this function in this specific context?

  1. {x∈R∣0≤x≤100}\{x \in \mathbb{R} \mid 0 \le x \le 100\}
  2. {x∈Z∣0≤x≤100}\{x \in \mathbb{Z} \mid 0 \le x \le 100\} (correct answer)
  3. {P∈R∣−300≤P≤2200}\{P \in \mathbb{R} \mid -300 \le P \le 2200\}
  4. {x∈Z∣x≥12}\{x \in \mathbb{Z} \mid x \ge 12\}
Explanation: The domain represents the set of all possible input values. In this context, the input xx is the number of handmade scarves. Since scarves are discrete, countable items, xx must be an integer (Z\mathbb{Z}). The context states the model is valid for the first 100 scarves, and the number of scarves cannot be negative, so the domain is all integers from 0 to 100, inclusive. This is represented by {x∈Z∣0≤x≤100}\{x \in \mathbb{Z} \mid 0 \le x \le 100\}.

Question 20

The population of a species of owl in a forest is modelled by P(t)=120+20t−0.5t2P(t) = 120 + 20t - 0.5t^2, where tt is the number of years since 2020. The model is considered valid for 0≤t≤300 \le t \le 30. What is the practical meaning of solving the inequality P(t)>250P(t) > 250?

  1. Finding the number of owls in the forest after 250 years.
  2. Finding the time period during which the owl population is above 250. (correct answer)
  3. Finding the maximum population of owls and the year in which it occurs.
  4. Finding the number of years it takes for the owl population to reach 250 for the first time.
Explanation: The function P(t)P(t) gives the population at a given time tt. The inequality P(t)>250P(t) > 250 means 'the population is greater than 250'. Solving for tt in this inequality means finding the values of tt (the time in years) for which this condition is true. Therefore, the solution represents the time period during which the owl population exceeds 250.