Math 2 Quiz: Domain Restrictions In Context

9 questions · exam conditions

0:00

Domain Restrictions In ContextQuestion 1 of 9

A population of bacteria grows according to $P(t) = \frac{5000}{1 + 24e^{-0.3t}}$ where $t$ is time in hours. Lab safety requires the culture to be destroyed when population exceeds 4000, but the experiment needs at least 1000 bacteria to produce meaningful data. The lab can only monitor cultures during business hours (8 AM to 6 PM). If the experiment starts at 9 AM, what is the valid time domain?

0 ≤ t ≤ 9

representing business hours from start to 6 PM

t_1 ≤ t ≤ \min(9, t_2)

where

P(t_1) = 1000

and

P(t_2) = 4000

0 ≤ t ≤ t_2

where

P(t_2) = 4000

and monitoring ends at 6 PM

1 ≤ t ≤ 8

ensuring sufficient population throughout business hours

← Back to quizzes

Math 2 Quiz

Math 2 Quiz: Domain Restrictions In Context

Practice Domain Restrictions In Context in Math 2 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 Domain Restrictions In Context, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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

$0 ≤ t ≤ 9$ representing business hours from start to 6 PM
$t_1 ≤ t ≤ \min(9, t_2)$ where $P(t_1) = 1000$ and $P(t_2) = 4000$ (correct answer)
$0 ≤ t ≤ t_2$ where $P(t_2) = 4000$ and monitoring ends at 6 PM
$1 ≤ t ≤ 8$ ensuring sufficient population throughout business hours

Explanation: The domain needs: (1)

P(t) ≥ 1000

, (2)

P(t) ≤ 4000

, and (3)

t ≤ 9

(business hours). At

t = 0

P(0) = \frac{5000}{25} = 200 < 1000

. The population grows to 1000 at some

t_1 > 0

and reaches 4000 at some

t_2 > t_1

. The domain is

t_1 ≤ t ≤ \min(9, t_2)

. Choice A ignores population constraints. Choice C doesn't account for the minimum population requirement. Choice D arbitrarily assumes specific times without calculation.

Question 2

A roller coaster's height above ground is modeled by $h(x) = 100 + 80\cos\left(\frac{\pi x}{200}\right)$ feet, where $x$ is horizontal distance in feet from the start. Safety regulations require the track to be at least 50 feet high, and the ride experience needs height variations of at least 60 feet between peaks and valleys. The coaster track extends 800 feet horizontally. What domain satisfies safety while preserving ride quality?

No valid domain exists since minimum height violates safety requirements (correct answer)
$x_1 ≤ x ≤ x_2$ where $h(x) ≥ 50$ and height variation equals 160 feet
$0 ≤ x ≤ 800$ since the minimum height is $100 - 80 = 20$ feet
$0 ≤ x ≤ 600$ restricting track length to ensure safety compliance

Explanation: When analyzing cosine functions for real-world constraints, you need to examine both the function's range and how domain restrictions affect the output values. The height function

h(x) = 100 + 80\cos\left(\frac{\pi x}{200}\right)

has a range determined by the cosine component. Since cosine oscillates between -1 and 1, this function varies between

100 + 80(-1) = 20

feet and

100 + 80(1) = 180

feet. The height variation between peaks and valleys is

180 - 20 = 160

feet, which exceeds the required 60 feet for ride quality. However, the minimum height of 20 feet falls well below the 50-foot safety requirement. This creates an insurmountable problem because restricting the domain cannot eliminate the low points of a cosine function—it can only potentially cut off some cycles. Answer A correctly identifies that no valid domain exists since the minimum height violates safety requirements. The function inherently drops to 20 feet, making safety compliance impossible regardless of domain restrictions. Answer B incorrectly suggests finding a domain where

h(x) ≥ 50

while maintaining 160 feet of variation—mathematically impossible since achieving the full variation requires including the minimum value of 20 feet. Answer C ignores the safety constraint entirely, accepting the dangerous 20-foot minimum. Answer D arbitrarily restricts the domain length without addressing the fundamental safety violation that occurs throughout the function's range. Study tip: When real-world constraints conflict with a function's inherent properties (like range), check whether the constraints can actually be satisfied before exploring domain restrictions.

Question 3

A satellite's signal strength follows $S(d) = \frac{100}{d^2 + 1}$ where $d$ is distance in thousands of kilometers from the transmission point. Communication requires signal strength ≥ 4, but interference occurs when strength > 50. The satellite orbit restricts $d$ to the interval $[0.5, 8]$ . What domain ensures reliable communication without interference?

$0.5 ≤ d ≤ 8$ since orbital constraints are most restrictive
$d_1 ≤ d ≤ \min(8, d_2)$ where $S(d_1) = 50$ and $S(d_2) = 4$
$\max(0.5, d_1) ≤ d ≤ \min(8, d_2)$ where signal bounds are satisfied (correct answer)
$1 ≤ d ≤ 5$ avoiding interference while maintaining communication strength

Explanation: We need: (1)

0.5 ≤ d ≤ 8

(orbital), (2)

S(d) ≥ 4

(communication), and (3)

S(d) ≤ 50

(interference). For

S(d) = 4

\frac{100}{d^2 + 1} = 4

, so

d^2 + 1 = 25

, giving

d = \sqrt{24} ≈ 4.9

. For

S(d) = 50

\frac{100}{d^2 + 1} = 50

, so

d^2 + 1 = 2

, giving

d = 1

. Since

S(d)

decreases as

d

increases, we need

1 ≤ d ≤ 4.9

. Combined with orbital constraints:

\max(0.5, 1) ≤ d ≤ \min(8, 4.9)

, which gives

1 ≤ d ≤ 4.9

. Choice A ignores signal constraints. Choice B doesn't account for orbital limits. Choice D uses approximate values without justification.

Question 4

An investment account balance follows $B(t) = 10000(1.05)^t - 500t$ dollars after $t$ years, where the second term represents annual fees. The account must maintain a minimum balance of $8000 to avoid penalties, and the investor plans to withdraw funds when the balance first reaches $15000. However, tax regulations require the account to remain open for at least 3 years. What domain represents the viable investment period?

$3 ≤ t ≤ t_{15000}$ where $B(t_{15000}) = 15000$
$0 ≤ t ≤ t_{15000}$ where $B(t) ≥ 8000$ throughout
$\max(3, t_{min}) ≤ t ≤ t_{15000}$ where $t_{min}$ ensures $B(t) ≥ 8000$ (correct answer)
$3 ≤ t ≤ 10$ assuming reasonable investment horizon with penalty avoidance

Explanation: The domain must satisfy: (1)

t ≥ 3

(tax rules), (2)

B(t) ≥ 8000

(penalty avoidance), and (3)

t ≤ t_{15000}

(withdrawal plan). Since

B(0) = 10000 > 8000

but the balance may dip below $8000 due to fees, we need to check if there's a period where $

B(t) < 8000

$. The domain starts at the later of year 3 or when the balance recovers to $8000, whichever is more restrictive. Choice A ignores the minimum balance constraint. Choice B ignores the 3-year tax requirement. Choice D arbitrarily assumes a 10-year endpoint.

Question 5

A geothermal power plant's efficiency follows $E(T) = \frac{40(T-20)}{T+50}$ percent, where $T$ is ground temperature in Celsius. Environmental regulations require efficiency ≥ 25%, and the plant becomes economically viable only when efficiency ≥ 30%. Geological surveys show ground temperatures between 80°C and 300°C are accessible. However, drilling costs become prohibitive when efficiency gains per degree drop below 0.2% per °C. What temperature range satisfies all operational criteria?

$80 ≤ T ≤ 300$ since geological access defines the operational range
$T_1 ≤ T ≤ 300$ where $E(T_1) = 30$ ensures economic viability
$T_2 ≤ T ≤ T_3$ where $E(T_2) = 30$ and $E'(T_3) = 0.002$
$\max(80, T_{econ}) ≤ T ≤ \min(300, T_{drill})$ accounting for all constraints (correct answer)

Explanation: We need: (1)

80 ≤ T ≤ 300

(geological), (2)

E(T) ≥ 30

(economic), (3)

E'(T) ≥ 0.002

(drilling cost). For

E(T) = 30

\frac{40(T-20)}{T+50} = 30

, so

40(T-20) = 30(T+50)

, giving

T = 230

°C. For the drilling constraint:

E'(T) = \frac{2800}{(T+50)^2}

. Setting

E'(T) = 0.002

gives

T ≈ 1133

°C. Since this exceeds 300°C, the drilling constraint is never restrictive within the geological range. Therefore, the domain is

\max(80, 230) ≤ T ≤ 300

, which simplifies to

230 ≤ T ≤ 300

. Choice D represents the correct approach of taking the intersection of all constraints.

Question 6

An online retailer's shipping cost function is S(w,z) = 5 + 0.8w + 0.3z, where w is weight in pounds, z is the zip code zone number (1-8), and S is shipping cost in dollars.

The company only ships items weighing 0.1 to 50 pounds to zones 1-8, and company policy requires all shipping costs to be rounded up to the nearest dollar. For an item weighing 12.7 pounds shipped to zone 5, what constraint most affects the final charged amount?

The weight restriction since 12.7 pounds is within the 0.1-50 pound range
The zone restriction since zone 5 is within the allowable 1-8 range
The rounding policy since S(12.7,5) = 16.66, which rounds up to $17 (correct answer)
No constraints affect this shipment since all conditions are naturally satisfied

Explanation: Calculate S(12.7,5) = 5 + 0.8(12.7) + 0.3(5) = 5 + 10.16 + 1.5 = 16.66. The weight (12.7 lbs) is within 0.1-50 range, and zone 5 is within 1-8 range, so these don't restrict the domain. However, the rounding policy changes the actual charged amount from $16.66 to $17, representing a $0.34 increase (about 2% more). This policy constraint has the most practical impact on the final cost. Choices A and B identify satisfied conditions that don't restrict this case. Choice D ignores that rounding affects the final amount charged.

Question 7

A rental car company charges according to C(m,d) = 45d + 0.25m, where m is miles driven, d is days rented, and C is total cost in dollars. Company policy limits rentals to 1-30 days and 3000 miles total. A customer planning a 12-day trip estimates driving 280 miles per day. How do the company's restrictions affect this rental?

Only the day limit applies since 12 ≤ 30, creating no additional restrictions
Only the mileage limit applies since 280 × 12 = 3360 > 3000 miles
Both restrictions apply equally since the customer exceeds mileage by exactly 12%
The mileage restriction forces a reduction to 250 miles per day maximum (correct answer)

Explanation: The customer plans 12 days (satisfies 1 ≤ d ≤ 30) and 280 × 12 = 3360 miles (exceeds 3000-mile limit). To stay within the mileage limit over 12 days, the maximum daily mileage is 3000/12 = 250 miles per day. The mileage restriction is the binding constraint, forcing the customer to reduce daily driving from 280 to 250 miles. Choice A ignores the mileage constraint violation. Choice B correctly identifies the violation but doesn't specify the solution. Choice C focuses on the percentage excess rather than the practical restriction needed.

Question 8

A projectile is launched from a platform. Its height function is $h(t) = -16t^2 + 64t + 80$ , where $h$ is height in feet and $t$ is time in seconds. The projectile lands when it hits the ground, and the measuring equipment only records data for the first 6 seconds after launch. What is the appropriate domain for this function in this context?

$0 \leq t \leq 6$ since the equipment stops recording after 6 seconds
$0 \leq t \leq 5$ since the projectile hits the ground at $t = 5$ seconds (correct answer)
$0 \leq t \leq 5$ since this represents the complete flight time
$0 \leq t \leq 4$ since the projectile reaches maximum height at $t = 2$ seconds

Explanation: To find when the projectile hits the ground, set h(t) = 0: -16t² + 64t + 80 = 0. Dividing by -16: t² - 4t - 5 = 0, which factors as (t-5)(t+1) = 0. So t = 5 or t = -1. Since t ≥ 0, the projectile hits the ground at t = 5 seconds. The domain is limited by whichever comes first: the equipment stopping (t = 6) or the projectile landing (t = 5). Since the projectile lands first, the domain is 0 ≤ t ≤ 5. Choice A ignores that the projectile lands before 6 seconds. Choice C gives the right interval but wrong reasoning. Choice D confuses maximum height time with landing time.

Question 9

A ride-sharing company uses surge pricing during peak hours. The fare function is F(d,s) = 2.50 + 1.20d⋅s, where d is distance in miles, s is the surge multiplier, and F is the fare in dollars.

Company policy limits surge multipliers to between 1.0 and 4.0, rides are limited to 25 miles maximum, and the payment system requires fares to be multiples of $0.25. If a customer travels exactly 8.5 miles, which surge multipliers are permissible?

All surge multipliers s where 1.0 ≤ s ≤ 4.0 and 1.20(8.5)s results in a multiple of 0.25
All surge multipliers s where 1.0 ≤ s ≤ 4.0 and the total fare is a multiple of 0.25 (correct answer)
Only surge multipliers s = 1.0, 2.0, 3.0, 4.0 since these are whole number multiples
All surge multipliers s where 1.0 ≤ s ≤ 4.0 since 8.5 < 25 mile limit

Explanation: For d = 8.5 miles, F = 2.50 + 1.20(8.5)s = 2.50 + 10.2s. For the fare to be a multiple of $0.25, we need 2.50 + 10.2s = 0.25k for some integer k. Since 2.50 = 10(0.25), we need 10.2s to also be a multiple of 0.25. This means 10.2s = 0.25m for integer m, so s = 0.25m/10.2 = 25m/102. The constraint affects the total fare, not just the distance component. Choice A incorrectly focuses only on the distance component. Choice C incorrectly assumes only integer surge values work. Choice D ignores the payment system constraint entirely.