UG Applied Mathematics Booster Test 2 - Mathematical Modeling -

🚀

UG Applied Mathematics Booster Test 2 - Mathematical Modeling

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

UG Applied Mathematics Booster Test 2 - Mathematical Modeling

1 / 20

QUESTION 1 OF 20

Differential equations play a pivotal role in describing real-world problems. Which of the following is NOT listed in the text as a diverse range application for mathematical models?

Answer Review

1 Differential equations play a pivotal role in describing real-world problems. Which of the following is NOT listed in the text as a diverse range application for mathematical models?

A Astronomy

B Social Science

C Quantum Mechanics

D Financial Mathematics

💡 Short Explanation

Applications list in modeling Common fields included Identify excluded domain

✅ Correct Answer Explanation

Mathematical modeling typically includes astronomy, social sciences, and finance, but quantum mechanics is not explicitly listed in the given context.

❓ Why Other Options Are Incorrect

A → standard application area
B → included in modeling scope
D → financial modeling domain

🎯 Strategy

Extreme word filter

Final Logic: identify non-listed field

🧠 Memory Trick

"QM is extra here"

📚 Class XII – NCERT Mathematics – Chapter: Introduction to Differential Equations – Topic: Applications of Modeling

2 Match the scenario to the corresponding mathematical formulation step based on rate of change principles:

List I	List II
1. A quantity y grows at a rate proportional to its size	a. dx/dt = -k₁ x
2. A quantity x decays at a rate proportional to its size	b. dA/dt = rA
3. Temperature T changes proportional to difference from surrounding A	c. dy/dt = ky
4. Money A compounds continuously at rate r	d. dT/dt = -k(T - A)

A (1)-(a), (2)-(c), (3)-(d), (4)-(b)

B (1)-(c), (2)-(a), (3)-(d), (4)-(b)

C (1)-(c), (2)-(a), (3)-(b), (4)-(d)

D (1)-(b), (2)-(a), (3)-(d), (4)-(c)

💡 Short Explanation

Growth → positive k Decay → negative k Match standard models

✅ Correct Answer Explanation

1: growth → dy/dt = ky 2: decay → dx/dt = −kx 3: Newton cooling 4: compound interest model dA/dt = rA 3Why Other Options Are Incorrect • A → mismatches decay/growth pairing • C → swaps compound interest mapping • D → incorrect assignment of models

🎯 Strategy

Model identification

Final Logic: standard rate equations mapping

🧠 Memory Trick

"Growth positive, decay negative"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Mathematical Modelling

3 Which steps describe the transition from a real-world scenario to a mathematical conclusion?
1. Formulation in terms of Mathematical Model
2. Convert into differential equation
3. Solution of the problem
4. Interpretation of the solution

A 1,2,4

B 1,2

C 1,2,3

D 1,2,3,4

💡 Short Explanation

Full modeling cycle Formulation → solution → interpretation Complete process

✅ Correct Answer Explanation

All steps are part of modeling: formulation, equation setup, solving, and interpreting results.

❓ Why Other Options Are Incorrect

1 → missing key steps
2 → incomplete process
3 → missing interpretation

🎯 Strategy

Process completeness check

Final Logic: modeling is full 4-step cycle

🧠 Memory Trick

"Model = Form–Solve–Interpret"

📚 Class XII – NCERT Mathematics – Chapter: Introduction to Differential Equations – Topic: Modeling Process

4 Identify the incorrect statement concerning mathematical interpretation of physical problems.

A Real-life applications often involve quantities changing at a rate proportional to the amount present.

B The amount present at time t is mathematically modeled as a constant independent of t.

C The physical law must be translated into a differential equation.

D The interpretation of the result brings the mathematical solution back to the original problem context.

💡 Short Explanation

Real systems vary with time Modeling uses functions Constants are invalid assumption

✅ Correct Answer Explanation

In differential equations, quantities depend on time t; assuming constant independence is incorrect.

❓ Why Other Options Are Incorrect

A → correct modeling idea
C → correct modeling step
D → correct interpretation stage

🎯 Strategy

Extreme statement filter

Final Logic: time dependence is essential

🧠 Memory Trick

"Time means change"

📚 Class XII – NCERT Mathematics – Chapter: Introduction to Differential Equations – Topic: Modeling Assumptions

5 When observing dynamic systems, what justifies the necessity of differential equations over standard algebra?

A Algebra cannot relate variables to one another.

B Natural phenomena involving change require equations relating changing quantities.

C Dynamic systems only contain integer values.

D Algebra is only used for medicine and social sciences.

💡 Short Explanation

Change is central Rates of change needed Algebra insufficient

✅ Correct Answer Explanation

Differential equations model how quantities change continuously, unlike algebraic equations which are static.

❓ Why Other Options Are Incorrect

A → incorrect generalization
C → irrelevant restriction
D → false statement

🎯 Strategy

Conceptual reasoning

Final Logic: change → differential equations

🧠 Memory Trick

"Change needs derivatives"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Need for Differential Equations

6 If dP/dt = (α - β)P represents population, what is the constraint for strict decrease?

A α > β

B α = β

C α < β

D α ≥ β

💡 Short Explanation

Net growth rate Negative condition Decay occurs

✅ Correct Answer Explanation

For decrease, (α − β) must be negative → α < β.

❓ Why Other Options Are Incorrect

A → growth condition
B → constant population
D → includes growth cases

🎯 Strategy

Sign analysis

Final Logic: negative rate condition

🧠 Memory Trick

"Death > Birth → decay"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Population Model Conditions

7 For a positive growth constant k > 0, what does the parameter A represent in the exponential growth formula
\(y=Ae^{kt},\)

A Growth rate

B Time elasped

C Original amount

D Maximum capacity

💡 Short Explanation

Initial condition t = 0 value Exponential model

✅ Correct Answer Explanation

At t = 0, y = A e⁰ = A → initial amount.

❓ Why Other Options Are Incorrect

A → k is growth rate
B → independent variable
D → unrelated

🎯 Strategy

Initial condition substitution

Final Logic: t = 0 gives A

🧠 Memory Trick

"A = At start"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Exponential Growth Model

8 The growth function f(t) = A e^(kt) with k > 0 has what key graphical property as time t increases?

A It approaches zero asymptotically.

B It demonstrates a linear moving trend.

C It continues to increase exponentially.

D It fluctuates around the original amount A

💡 Short Explanation

Positive exponent Increasing function Rapid growth

✅ Correct Answer Explanation

For k > 0, exponential function grows rapidly as t increases.

❓ Why Other Options Are Incorrect

A → decay case
B → not linear
D → no oscillation

🎯 Strategy

Graph behavior recognition

Final Logic: positive k → growth

🧠 Memory Trick

"Positive k → up curve"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Exponential Growth Graph

9 What is the explicit general solution to the decay differential equation
\(\frac{dy}{dt}=ky,k<0?\)
\(y=Ae^{-kt}\)
\(y=Ae^{kt}\)
\(y=Aln⁡(kt)\)
\(y=A+e^{kt}\)

💡 Short Explanation

The differential equation \(\frac{dy}{dt}=ky\) represents exponential growth or decay. Its general solution is \(y=Ae^{kt}.\) Since \(k<0\), the exponential term decreases with time, representing exponential decay.

✅ Correct Answer Explanation

Given, \(\frac{dy}{dt}=ky,k<0.\) Separate the variables: \(\frac{dy}{y}=k dt.\) Integrate both sides: \(\int \frac{1}{y} dy=\int k dt,\) which gives \(log⁡∣y∣=kt+C.\) Exponentiating, \(y=e^{kt+C}.\) Since \(e^{C}=A,\) the general solution becomes \(y=Ae^{kt}.\) Because \(k<0\), \(e^{kt}=e^{-∣k∣t},\) which decreases as \(t\) increases, representing exponential decay. Hence, Option B is correct

❓ Why Other Options Are Incorrect

Option A)
\(y=Ae^{-kt}\)
Incorrect because if \(k<0\),
\(-k>0,\)
so
\(e^{-kt}=e^{∣k∣t},\)
which represents exponential growth, not decay.
Option B)
\(y=Ae^{kt}\)
Correct.
This is the standard general solution of
\(\frac{dy}{dt}=ky.\)
When \(k<0\), it automatically represents exponential decay.
Option C)
\(y=Alog⁡(kt)\)
Incorrect because integrating
\(\frac{dy}{dt}=ky\)
produces an exponential function, not a logarithmic function.
Option D)
\(y=A+e^{kt}\)
Incorrect because this expression does not satisfy
\(\frac{dy}{dt}=ky.\)
The arbitrary constant appears as a multiplicative constant, not an additive one.

🎯 Strategy

Used

Separation of Variables

Application:

1. Separate the variables:
\(\frac{dy}{y}=k dt.\)
1. Integrate both sides.
2. Exponentiate to remove the logarithm.
3. Replace \(e^{C}\)by the arbitrary constant \(A\).

🧠 Memory Trick

"Sign of \(k\) decides Growth or Decay; the solution always stays \(Ae^{kt}\).

📚 Chapter: Differential Equations

10 In the decreasing function graph for
\(y=Ae^{kt},k<0,\)
as the variable \(t\) moves towards positive infinity, the value of \(y\) approaches:

A Negative infinity

B Original amount \(A\)

C Zero

D Positive infinity

💡 Short Explanation

When \(k<0\), the function \(y=Ae^{kt}\) represents exponential decay. As time \(t\) increases indefinitely, the exponential term approaches zero, so the value of \(y\) also approaches zero.

✅ Correct Answer Explanation

The given function is \(y=Ae^{kt},\) where \(A\) is the initial amount. \(k<0\), indicating exponential decay. As \(t\rightarrow \infty ,\) the exponent \(kt\rightarrow -\infty .\) Since \(e^{-\infty }=0,\) we have \(Ae^{kt}\rightarrow A\times 0=0.\) Thus, \({lim⁡}_{t\rightarrow \infty }Ae^{kt}=0.\) Therefore, the function approaches the x-axis (horizontal asymptote) but never becomes negative. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A) Negative infinity
Incorrect because an exponential decay function remains non-negative (for \(A>0\)) and approaches 0, not negative infinity.
Option B) Original amount \(A\)
Incorrect because \(A\) is the initial value at
\(t=0,\)
since
\(y(0)=Ae^{0}=A.\)
As time increases, the value decreases from \(A\) toward zero.
Option C) Zero
Correct.
Since
\(e^{kt}\rightarrow 0\)
for \(k<0\),
\(y=Ae^{kt}\rightarrow 0.\)
Option D) Positive infinity
Incorrect because exponential growth occurs only when
\(k>0.\)
Here, \(k<0\), so the function decreases instead of increasing.

🎯 Strategy

Used

Analyse the Long-Term Behaviour of an Exponential Function

Application:

1. Identify the sign of \(k\).
2. If \(k<0\), the function represents exponential decay.
3. Evaluate the limit as
\(t\rightarrow \infty .\)
1. Use the fact that
\(e^{-\infty }=0.\)

🧠 Memory Trick

"Negative \(k\), Zero is the Key."

📚 Chapter: Differential Equations

11 In the birth rate model
\(\frac{dP}{dt}=(\alpha -\beta )P,\)
if the death rate \(\beta\) is completely ignored (i.e., \(\beta =0\)), the differential equation simplifies to:
\(\frac{dP}{dt}=\alpha\)
\(\frac{dP}{dt}=\alpha P\)
\(\frac{dP}{dt}=-\beta P\)
\(\frac{dP}{dt}=P\)

💡 Short Explanation

When the death rate is ignored (\(\beta =0\)), the net population growth depends only on the birth rate. Substituting \(\beta =0\) into the equation gives \(\frac{dP}{dt}=\alpha P.\)

✅ Correct Answer Explanation

The population growth model is \(\frac{dP}{dt}=(\alpha -\beta )P,\) where \(\alpha\)= birth rate, \(\beta\)= death rate. If \(\beta =0,\) then \(\frac{dP}{dt}=(\alpha -0)P=\alpha P.\) Thus, the population changes only due to births, resulting in exponential growth. Hence, \(\frac{dP}{dt}=\alpha P.\) Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A)
\(\frac{dP}{dt}=\alpha\)
Incorrect because the rate of change is proportional to the population \(P\), not just the constant birth rate.
Option B)
\(\frac{dP}{dt}=\alpha P\)
Correct.
Setting \(\beta =0\) directly simplifies the equation to
\(\frac{dP}{dt}=\alpha P.\)
Option C)
\(\frac{dP}{dt}=-\beta P\)
Incorrect because this represents only the effect of deaths, whereas births are being considered.
Option D)
\(\frac{dP}{dt}=P\)
Incorrect because the birth-rate constant \(\alpha\) is omitted.
This equation would be valid only if \(\alpha =1\).

🎯 Strategy

Used

Substitution into the Differential Equation

Application:

1. Write the given model:
\(\frac{dP}{dt}=(\alpha -\beta )P.\)
1. Substitute the given condition:
\(\beta =0.\)
1. Simplify the equation.

🧠 Memory Trick

Mnemonic: "Remove Death → Only Birth Remains."

📚 Chapter: Differential Equations

12 To solve the death rate model, the integration
\(\int \frac{1}{P} dP=\int k dt\)
is performed. What is the resulting integrated equation before removing the logarithm?
\(P=kt+c\)
\(log⁡P=k+t+c\)
\(log⁡P=kt+c\)
\(\frac{P^{2}}{2}=kt+c\)

💡 Short Explanation

After separating the variables, integrating both sides gives \(log⁡P=kt+c.\) The logarithm is removed only after exponentiating both sides.

✅ Correct Answer Explanation

Given the differential equation \(\frac{dP}{dt}=kP,\) separate the variables: \(\frac{1}{P} dP=k dt.\) Integrate both sides: \(\int \frac{1}{P} dP=\int k dt.\) Using the standard integration formulas, \(log⁡∣P∣=kt+c.\) Since population \(P>0\), \(log⁡P=kt+c.\) This is the required equation before removing the logarithm. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A)
\(P=kt+c\)
Incorrect because integrating
\(\frac{1}{P}\)
produces a logarithmic function, not a linear function.
Option B)
\(log⁡P=k+t+c\)
Incorrect because
\(\int k dt=kt+c,\)
not \(k+t+c\).
Option C)
\(log⁡P=kt+c\)
Correct.
This is the standard result obtained after integrating both sides.
Option D)
\(\frac{P^{2}}{2}=kt+c\)
Incorrect because
\(\int \frac{1}{P} dP\)
does not equal
\(\frac{P^{2}}{2}.\)
That result comes from integrating \(P dP\).

🎯 Strategy

Used

Separation of Variables

Application:

1. Separate the variables:
\(\frac{1}{P} dP=k dt.\)
1. Integrate both sides.
2. Apply the standard integral:
\(\int \frac{1}{P} dP=log⁡∣P∣.\)
1. Obtain the integrated equation before exponentiating.

🧠 Memory Trick

Mnemonic: "Log First, Exponential Next."

📚 Chapter: Differential Equations

13 The bacteria grew from 10,000 to 40,000 at 5 hours. What was the exact mathematical value found for \(e^{2k}\) ?

A 2

B 4

C 1250

D 10000

💡 Short Explanation

Ratio method Exponential growth Solve using division

✅ Correct Answer Explanation

40,000 / 10,000 = 4 = e^(2k), so e^(2k) = 4.

❓ Why Other Options Are Incorrect

A → incorrect ratio
C → unrelated scale
D → initial value confusion

🎯 Strategy

Ratio comparison

Final Logic: final/initial = e^(kt)

🧠 Memory Trick

"4× growth → e²k = 4"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Bacterial Growth Model

14 Based on the initial population calculation in Example 13, if
\(\lambda e^{3k}=10,000\)
and
\(e^{k}=2,\)
what is the calculated value of the initial population \(\lambda\)?
\(2500\)
\(5000\)
\(1250\)
\(1000\)

💡 Short Explanation

Use the law of exponents to evaluate \(e^{3k}\), then substitute its value into the given equation and solve for \(\lambda\).

✅ Correct Answer Explanation

Given, \(\lambda e^{3k}=10,000\) and \(e^{k}=2.\) Step 1: Evaluate \(e^{3k}\) Using the law of exponents, \(e^{3k}=(e^{k})^{3}=2^{3}=8.\) Step 2: Substitute into the given equation \(\lambda \times 8=10,000.\) Step 3: Solve for \(\lambda\) \(\lambda =\frac{10,000}{8}=1250.\) Therefore, the initial population is \(\lambda =1250.\) Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A) 2500
Incorrect because
\(2500\times 8=20,000\neq 10,000.\)
Option B) 5000
Incorrect because
\(5000\times 8=40,000\neq 10,000.\)
Option C) 1250
Correct.
Since
\(1250\times 8=10,000,\)
it satisfies the given equation.
Option D) 1000
Incorrect because
\(1000\times 8=8000\neq 10,000.\)

🎯 Strategy

Used

Substitution and Exponent Law

Application:

1. Use the exponent rule:
\(e^{3k}=(e^{k})^{3}.\)
1. Substitute the given value of \(e^{k}\).
2. Solve the resulting linear equation for \(\lambda\).

🧠 Memory Trick

Mnemonic: "Cube the exponential, then divide."

📚 Chapter: Differential Equations

15 Ms. Rajni deposits Rs. 10,000 at 4% continuous compound interest. The formula evaluated to find the amount after 10 years is:
\(10000\times e^{0.04}\)
\(10000\times e^{0.4}\)
\(10000\times e^{4}\)
\(10000\times e^{40}\)

💡 Short Explanation

Substituting \(P=10000\), \(r=0.04\), and \(t=10\), \(A=10000e^{0.04\times 10}=10000e^{0.4}.\)Correct Answer Explanation The continuous compounding formula is \(A=Pe^{rt},\) where: \(P=10000\)(Principal) \(r=0.04\)(4% per annum) \(t=10\) years Substituting, \(A=10000e^{0.04\times 10}=10000e^{0.4}.\) Hence, the amount after 10 years is \(10000e^{0.4}.\) Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A)
\(10000e^{0.04}\)
Incorrect because it considers only 1 year instead of 10 years.
Option B)
\(10000e^{0.4}\)
Correct.
Since
\(rt=0.04\times 10=0.4,\)
this is the correct expression.
Option C)
\(10000e^{4}\)
Incorrect because the exponent should be 0.4, not 4.
Option D)
\(10000e^{40}\)
Incorrect because
\(0.04\times 10=0.4,\)
not 40.

🎯 Strategy

Used

Apply the Continuous Compound Interest Formula

Application:

1. Identify the values of \(P\), \(r\), and \(t\).
2. Use the formula
\(A=Pe^{rt}.\)
1. Calculate the exponent \(rt\).
2. Substitute the values to obtain the required expression.

🧠 Memory Trick

Simple Interest: \(P(1+rt)\)

📚 which is used for continuous compound interest applications.

16 To find the doubling time of the money, the equation simplifies to
\(0.04t=ln⁡2.\)
What is the approximate numerical value of
\(\ln\,2\)
used to solve for \(t\)?

A 0.3010

B 0.6931

C 2.718

D 1.414

💡 Short Explanation

The natural logarithm of 2 is approximately \(ln⁡2\approx 0.6931.\) This value is commonly used to calculate the doubling time in exponential growth and continuous compound interest problems.

✅ Correct Answer Explanation

To find the doubling time, we use \(0.04t=ln⁡2.\) The standard approximation is \(ln⁡2\approx 0.6931.\) Therefore, \(t=\frac{0.6931}{0.04}\approx 17.33 years.\) Hence, the numerical value used is \(0.6931.\) Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A) 0.3010
Incorrect because 0.3010 is approximately
\({log⁡}_{10}2,\)
the common (base-10) logarithm, not the natural logarithm.
Option B) 0.6931
Correct.
This is the approximate value of
\(ln⁡2,\)
which is used in exponential growth and continuous compounding.
Option C) 2.718
Incorrect because 2.718 is the approximate value of Euler's number,
\(e,\)
not \(\ln\,2\).
Option D) 1.414
Incorrect because 1.414 is approximately
\(\sqrt{2},\)
not \(\ln\,2\).

🎯 Strategy

Used

Use Standard Mathematical Constants

Application:

1. Recognize the equation
\(0.04t=ln⁡2.\)
1. Recall the standard approximation
\(ln⁡2\approx 0.6931.\)
1. Substitute this value to compute the doubling time.

🧠 Memory Trick

"Natural Log of 2 → 0.693."

📚 Chapter: Differential Equations

17 The solution to the cooling process differential equation
\(\frac{dT}{dt}=-k(T-A)\)
integrates to
\(log⁡(T-A)=-kt+c.\)
What does \(c\) represent when converted to the exponential form
\(T-A=\lambda e^{-kt}?\)
\(c=\lambda\)
\(e^{c}=\lambda\)
\(ln⁡c=\lambda\)
\(-k=\lambda\)

💡 Short Explanation

After integrating, exponentiating both sides converts the additive constant \(c\) into a multiplicative constant. Thus, \(e^{c}\) is replaced by a new arbitrary constant \(\lambda\).

✅ Correct Answer Explanation

Given, \(log⁡(T-A)=-kt+c.\) Step 1: Exponentiate both sides \(e^{log⁡(T-A)}=e^{-kt+c}.\) Using the exponential law, \(e^{a+b}=e^{a}⋅e^{b},\) we get \(T-A=e^{-kt}e^{c}.\) Step 2: Replace the constant Since \(e^{c}\)is also an arbitrary positive constant, let \(\lambda =e^{c}.\) Therefore, \(T-A=\lambda e^{-kt}.\) Hence, \(\lambda =e^{c}.\) Thus, Option B is correct

❓ Why Other Options Are Incorrect

Option A)
\(c=\lambda\)
Incorrect because the arbitrary constant is transformed after exponentiation.
The correct relation is
\(\lambda =e^{c},\)
not \(\lambda =c\).
Option B)
\(e^{c}=\lambda\)
Correct.
Exponentiating the integration constant converts it into the multiplicative constant \(\lambda\).
Option C)
\(ln⁡c=\lambda\)
Incorrect because no logarithm of the constant is taken during the derivation.
Option D)
\(-k=\lambda\)
Incorrect because \(k\) is the cooling constant, whereas \(\lambda\) is the arbitrary constant determined by the initial condition.

🎯 Strategy

Used

Exponentiation After Integration

Application:

1. Integrate the separated differential equation.
2. Exponentiate both sides.
3. Use the identity
\(e^{a+b}=e^{a}e^{b}.\)
1. Replace \(e^{c}\)with a new arbitrary constant.

🧠 Memory Trick

Mnemonic: "After taking \(e\), \(c\) becomes \(e^{c}\)."

📚 Chapter: Differential Equations

18 A cake cools from 185°F to 150°F in 30 minutes in a 75°F room. What is the value of λ derived from the initial conditions at t=0? :

A 185

B 75

C 110

D 150

💡 Short Explanation

λ = initial excess temperature Subtract surrounding temp Use initial condition

✅ Correct Answer Explanation

λ = T(0) − A = 185 − 75 = 110

❓ Why Other Options Are Incorrect

A → raw temperature
B → ambient temp
D → final temp

🎯 Strategy

Initial condition substitution

Final Logic: difference at t = 0

🧠 Memory Trick

"Start minus surround"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Cooling Applications

19 (Passage-Based)
Utilizing the half-life concept from the passage, substituting
\(t=5700\)
and
\(x=\frac{x_{0}}{2}\)
into
\(x=x_{0}e^{-k_{1}t}\)
results in which exact equation?
\(e^{-5700k_{1}}=2\)
\(e^{-5700k_{1}}=\frac{1}{2}\)
\(e^{5700k_{1}}=\frac{1}{2}\)
\(e^{-k_{1}}=\frac{2}{5700}\)

💡 Short Explanation

Substituting the half-life values into the exponential decay equation and cancelling the initial amount \(x_{0}\)gives \(e^{-5700k_{1}}=\frac{1}{2}.\)

✅ Correct Answer Explanation

Given, \(x=x_{0}e^{-k_{1}t}.\) At half-life, \(t=5700,x=\frac{x_{0}}{2}.\) Substitute these values: \(\frac{x_{0}}{2}=x_{0}e^{-5700k_{1}}.\) Divide both sides by \(x_{0}\): \(e^{-5700k_{1}}=\frac{1}{2}.\) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A)
\(e^{-5700k_{1}}=2\)
Incorrect because radioactive decay decreases the quantity, so the exponential term must be less than 1.
Option B)
\(e^{-5700k_{1}}=\frac{1}{2}\)
Correct.
This follows directly from the half-life condition.
Option C)
\(e^{5700k_{1}}=\frac{1}{2}\)
Incorrect because the exponent should be negative in the decay model.
Option D)
\(e^{-k_{1}}=\frac{2}{5700}\)
Incorrect because this equation is not obtained by substituting the half-life values.

🎯 Strategy

Used

Substitution into the Exponential Decay Model

Application:

1. Write the decay equation.
2. Substitute the half-life values.
3. Cancel the common factor \(x_{0}\).
4. Simplify to obtain the required equation.

🧠 Memory Trick

\(e^{-kt}=\frac{1}{2}.\)

📚 Chapter: Differential Equations

20 (Passage-Based)
The mathematical model for exponential growth or decay is given by
\(\frac{df}{dt}=kf(t)\)
or
\(\frac{dy}{dt}=ky.\)
For carbon dating, Carbon-14 decays exponentially with a half-life of 5700 years. If \(x\) is the mass of Carbon-14 remaining after \(t\) years, the differential equation is
\(\frac{dx}{dt}=-k_{1}x,\)
whose general solution is
\(x=x_{0}e^{-k_{1}t},\)
where \(x_{0}\)is the initial amount of Carbon-14. The half-life principle states that after 5700 years, only half of the original Carbon-14 remains. Thus,
\(t=5700,x=\frac{x_{0}}{2}.\)
Substituting these values into the exponential solution gives
\(\frac{x_{0}}{2}=x_{0}e^{-5700k_{1}},\)
which simplifies to
\(e^{-5700k_{1}}=\frac{1}{2}.\)
This equation is used to determine the decay constant \(k_{1}\). Once the decay constant is known, scientists can estimate the age of archaeological samples by measuring the remaining Carbon-14
If charcoal from an ancient pit contained only one-fourth \(\left(1/4\right)\)of the Carbon-14 found in a living sample, what is its estimated age?

A 5700 years

B 11400 years

C 17100 years

D 2850 years

💡 Short Explanation

One-fourth of the original Carbon-14 means two half-lives have passed. Since each half-life is 5700 years, the estimated age is \(2\times 5700=11400 years.\)

✅ Correct Answer Explanation

Carbon-14 has a half-life of \(5700 years.\) After one half-life: \(x=\frac{x_{0}}{2}.\) After two half-lives: \(x=\frac{x_{0}}{4}.\) Since the charcoal contains \(\frac{1}{4}\) of the original Carbon-14, \(Age=2\times 5700=11400 years.\) Thus, \(11400 years.\) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A) 5700 years
Incorrect because one half-life leaves
\(\frac{1}{2}\)
of the original Carbon-14, not
\(\frac{1}{4}.\)
Option B) 11400 years
Correct.
Two half-lives produce
\(\frac{1}{4}\)
of the original Carbon-14.
Option C) 17100 years
Incorrect because three half-lives leave
\(\frac{1}{8}\)
of the original Carbon-14.
Option D) 2850 years
Incorrect because 2850 years is only half of one half-life.

🎯 Strategy

Used

Half-Life Counting

Application:

1. Determine the remaining fraction.
2. Count the number of half-lives.
3. Multiply by 5700 years.

🧠 Memory Trick

Mnemonic: "Every Half-Life Halves Again."

📚 Chapter: Differential Equations

UG Applied Mathematics Booster Test 2 - Mathematical Modeling

Test Complete!

Answer Review