UG Applied Mathematics Booster Test 3 - Mathematical Modeling -

🚀

UG Applied Mathematics Booster Test 3 - Mathematical Modeling

⏱️ 30 Minutes

❓ 20 Questions

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

UG Applied Mathematics Booster Test 3 - Mathematical Modeling

1 / 20

QUESTION 1 OF 20

Why is an in-depth study of differential equations of great importance in modern scientific investigations, rather than relying solely on algebra?

Answer Review

1 Why is an in-depth study of differential equations of great importance in modern scientific investigations, rather than relying solely on algebra?

A Algebra only computes probabilities, whereas differential equations solve geometric shapes.

B Many interesting natural phenomena involve change and require equations that relate changing quantities.

C Differential equations are the only mathematical tool accepted in social sciences.

D Algebra cannot be used for any physical problems.

💡 Short Explanation

Real-world systems involve change Differential equations model rates of change Algebra alone is static

✅ Correct Answer Explanation

Natural phenomena such as motion, growth, decay, and heat transfer involve continuously changing quantities. Differential equations are essential because they relate variables with their rates of change (derivatives), which algebra cannot describe dynamically.

❓ Why Other Options Are Incorrect

A → Incorrect role of algebra and DE
C → Not true; many tools exist in social sciences
D → Algebra does apply to physical problems

🎯 Strategy

Used: Contextual/Tonal Matching

Final Logic: Identify true statement about dynamic systems

🧠 Memory Trick

"Change → Differential Equations"

📚 Class XII – NCERT Mathematics – Differential Equations – Introduction to Modeling

2 Match the modeled processes to their mathematical interpretation equations given in the text:

List I	List II
1. Rate of change of drug in GI-tract	a. (I) dA/dt = rA
2. Change of population	b. dT/dt = -k(T - A)
3. Rate of change of temperature	c. dP/dt = (α - β)P
4. Rate of change of money continuously compounded	d. dx/dt = -k₁ x

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

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

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

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

💡 Short Explanation

Match each physical process to DE Use standard growth/decay laws Temperature uses Newton's law

✅ Correct Answer Explanation

(1) Drug decay → dx/dt = -k₁x (d) (2) Population → dP/dt = (α − β)P (c) (3) Temperature → dT/dt = -k(T − A) (b) (4) Continuous money growth → dA/dt = rA (a)

❓ Why Other Options Are Incorrect

A → incorrect pairing of all processes
B → mismatches population and decay
C → incorrect assignment of equations

🎯 Strategy

Used: Option Grouping

Final Logic: Direct mapping of standard models

🧠 Memory Trick

"Decay–Population–Heat–Money order"

📚 Class XII – NCERT Mathematics – Differential Equations – Mathematical Modelling

3 During the problem formulation phase, establishing the rate of change of population P with respect to time involves which constants?
1. Constant birth rate α
2. Constant death rate β
3. Surrounding temperature A
4. Annual interest rate r

A 1,2,4

B 1,2

C 1,2,3

D 1,2,3,4

💡 Short Explanation

Population depends on birth and death rates Only α and β are relevant Other parameters belong to different models

✅ Correct Answer Explanation

Population growth is modeled by dP/dt = (α − β)P, where only α (birth rate) and β (death rate) affect population change.

❓ Why Other Options Are Incorrect

C → belongs to cooling model
D → belongs to interest model
C and D irrelevant to population equation

🎯 Strategy

Used: Elimination

Final Logic: Identify relevant parameters in model

🧠 Memory Trick

"Population = Birth − Death only"

📚 Class XII – NCERT Mathematics – Differential Equations – Population Model

4 Select the incorrect statement regarding the mathematical interpretation of models like the decay model.

A If k < 0, the formula represents exponential decay.

B The physical law must be successfully translated into a differential equation.

C The parameter t universally represents the mass of the quantity.

D The mathematical result must be interpreted in the context of the original problem.

💡 Short Explanation

t represents time, not mass Only modeling steps are valid Misinterpretation in option C

✅ Correct Answer Explanation

In differential equation models, t always represents time, not physical mass. Hence C is incorrect, while other options correctly describe modeling steps.

❓ Why Other Options Are Incorrect

A → correct property of decay
B → correct modeling step
D → correct interpretation principle

🎯 Strategy

Used: Extreme Word Filter

Final Logic: Identify incorrect universal claim

🧠 Memory Trick

"t = time, never mass"

📚 Class XII – NCERT Mathematics – Differential Equations – Modeling Concepts

5 When representing dynamic systems of growth or mixture, the general mathematical model f'(t) = k f(t) uses k as a constant that specifically depends on:

A The initial value only

B The rate of growth or decay

C The total time passed

D The surrounding temperature

💡 Short Explanation

k controls growth speed Determines increase or decrease Independent of time/initial value

✅ Correct Answer Explanation

In f'(t) = kf(t), k represents proportionality constant defining growth or decay rate, not initial conditions or external variables.

❓ Why Other Options Are Incorrect

A → initial value is separate constant
C → time is independent variable
D → temperature not part of general model

🎯 Strategy

Used: Conceptual Matching

Final Logic: Interpret meaning of k in DE

🧠 Memory Trick

"k = growth speed"

📚 Class XII – NCERT Mathematics – Differential Equations – Exponential Model

6 What constraint explicitly differentiates the exponential growth model from the exponential decay model in changing quantities modeled by
\(y=Ae^{kt}?\)

A Growth requires \(t>0\), decay requires \(t<0\)

B Growth requires \(A>0\), decay requires \(A<0\)

C Growth requires \(k>0\), decay requires \(k<0\)

D Growth requires \(k<1\), decay requires \(k>1\)

💡 Short Explanation

The sign of the constant \(k\) determines whether the quantity grows or decays. A positive \(k\) gives exponential growth, while a negative \(k\) gives exponential decay.

✅ Correct Answer Explanation

The exponential model is \(y=Ae^{kt}.\) The constant \(k\) is called the growth/decay constant. If \(k>0,\) then \(e^{kt}\) increases as \(t\) increases, so the quantity grows exponentially. If \(k<0,\) then \(e^{kt}=e^{-∣k∣t},\) which decreases as \(t\) increases, so the quantity decays exponentially. Thus, Growth: \(k>0\) Decay: \(k<0\) Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A)
Incorrect because growth and decay are not determined by whether \(t\) is positive or negative.
The determining factor is the sign of \(k\).
Option B)
Incorrect because \(A\) represents the initial amount, not whether the function grows or decays.
The sign of \(k\) determines the behavior.
Option C)
✅ Correct.
The sign of the growth constant \(k\) distinguishes exponential growth from exponential decay.
Option D)
Incorrect because the comparison is not between \(k<1\) and \(k>1\).
Only the sign of \(k\) matters.

🎯 Strategy

Used

Analyse the Sign of the Growth Constant

Application:

1. Identify the exponential model
\(y=Ae^{kt}.\)
1. Examine the sign of \(k\).
2. Conclude:
\(k>0\)⇒ Growth
\(k<0\)⇒ Decay

🧠 Memory Trick

Mnemonic: "The Sign of \(k\) Controls the Curve."

📚 Chapter: Differential Equations

7 In exponential growth with a positive constant, if a quantity doubles from its initial amount \(A\) to \(2A\), the doubling time \(t\) can be mathematically derived as:
\(t=\frac{k}{\ln\,2}\)
\(t=\frac{\ln\,2}{k}\)
\(t=ln⁡(2k)\)
\(t=\frac{2}{\ln\,k}\)

💡 Short Explanation

For exponential growth, \(y=Ae^{kt},k>0.\) When the quantity doubles, \(y=2A\). Solving the resulting equation gives the doubling time \(t=\frac{\ln\,2}{k}.\)

✅ Correct Answer Explanation

The exponential growth model is \(y=Ae^{kt},\) where \(A\)= initial amount, \(k>0\)= growth constant. For doubling, \(y=2A.\) Substitute into the model: \(2A=Ae^{kt}.\) Divide both sides by \(A\): \(2=e^{kt}.\) Take the natural logarithm on both sides: \(ln⁡2=kt.\) Hence, \(t=\frac{\ln\,2}{k}.\) Therefore, Option B is correct

❓ Why Other Options Are Incorrect

Option A)
\(t=\frac{k}{\ln\,2}\)
Incorrect because the numerator and denominator are interchanged.
The correct formula is
\(t=\frac{\ln\,2}{k}.\)
Option B)
\(t=\frac{\ln\,2}{k}\)
✅ Correct.
It is obtained directly from
\(2=e^{kt}.\)
Option C)
\(t=ln⁡(2k)\)
Incorrect because the logarithm is not applied to \(2k\).
The correct derivation gives
\(kt=ln⁡2.\)
Option D)
\(t=\frac{2}{\ln\,k}\)
Incorrect because this expression does not follow from the exponential growth equation.

🎯 Strategy

Used

Substitution and Natural Logarithm

Application:

1. Use the growth model
\(y=Ae^{kt}.\)
1. Substitute \(y=2A\).
2. Divide by \(A\).
3. Take the natural logarithm.
4. Solve for \(t\).

🧠 Memory Trick

Mnemonic: "Double → \(\ln\,2\); Half → also \(\ln\,2\)."

📚 Chapter: Differential Equations

8 In the graph depicting
\(y=Ae^{kt}\)
for an increasing function (Exponential Growth), the y-intercept is denoted by the point:

A \(\left(A,\ 0\right)\)

B \(\left(0,\ A\right)\)

C \(\left(k,\ A\right)\)

D \(\left(0,\ k\right)\)

💡 Short Explanation

The y-intercept is obtained by substituting \(t=0.\) Since \(y=Ae^{k(0)}=Ae^{0}=A,\) the graph passes through \((0,A).\)

✅ Correct Answer Explanation

The exponential growth model is \(y=Ae^{kt},\) where \(A\) is the initial amount, \(k>0\) is the growth constant. To find the y-intercept, substitute \(t=0.\) Then, \(y=Ae^{0}=A.\) Therefore, the graph intersects the y-axis at \((0,A).\) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A) \(\left(A,\ 0\right)\)
Incorrect because this point lies on the x-axis.
The y-intercept must have an x-coordinate (time) of 0.
Option B) \(\left(0,\ A\right)\)
Correct.
At
\(t=0,\)
the function value is
\(A.\)
Option C) \(\left(k,\ A\right)\)
Incorrect because \(k\) is the growth constant, not the x-coordinate of the y-intercept.
Option D) \(\left(0,\ k\right)\)
Incorrect because the y-coordinate at
\(t=0\)
is \(A\), not \(k\).

🎯 Strategy

Used

Evaluate the Function at \(t=0\)

Application:

1. Write the function:
\(y=Ae^{kt}.\)
1. Substitute
\(t=0.\)
1. Use
\(e^{0}=1.\)
1. Obtain the y-intercept.

🧠 Memory Trick

Mnemonic: "Time Zero → Initial Amount."

📚 Chapter: Differential Equations

9 In a negative growth constant scenario, what happens to the value of the exponential term
\(e^{kt}\)
as time \(t\) becomes very large?

A It approaches 1

B It approaches -1

C It approaches \(k\)

D It approaches 0

💡 Short Explanation

When \(k<0\), the exponent \(kt\) becomes a very large negative number as \(t\) increases. Therefore, y = \(e^{-kt}\) and hence, \(e^{kt}\rightarrow 0.\)

✅ Correct Answer Explanation

The exponential function is \(e^{kt}.\) If \(k<0,\) then \(kt\rightarrow -\infty\) as \(t\rightarrow \infty .\) Since \({lim⁡}_{x\rightarrow -\infty }e^{x}=0,\) we obtain \({lim⁡}_{t\rightarrow \infty }e^{kt}=0.\) Thus, the exponential term continuously decreases and approaches zero, which is the basis of exponential decay. Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A) It approaches 1
Incorrect because \(e^{kt}=1\) only when \(kt=0\)(for example, at \(t=0\)), not as \(t\) becomes very large.
Option B) It approaches -1
Incorrect because the exponential function is always positive and can never become negative.
Option C) It approaches \(k\)
Incorrect because \(e^{kt}\)approaches 0, not the constant \(k\).
Option D) It approaches 0
Correct.
For \(k<0\),
\(e^{kt}\rightarrow 0\)
as
\(t\rightarrow \infty .\)

🎯 Strategy

Used

Analyse the Limit of the Exponential Function

Application:

1. Identify that \(k<0\).
2. Observe that as \(t\rightarrow \infty\),
\(kt\rightarrow -\infty .\)
1. Use the standard limit
\(e^{-\infty }=0.\)

🧠 Memory Trick

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

📚 Chapter: Differential Equations

10 Observing the exponential decay of a decreasing function, the initial amount \(A\) changes to \(A/2\) after a specific period known as the half-life. Using
\(x=x_{0}e^{-kt},\)
what is the exact expression for the half-life \(t\)?
\(t=\frac{\ln\,2}{k}\)
\(t=-\frac{\ln\,2}{k}\)
\(t=ln⁡\left(\frac{1}{2}\right)k\)
\(t=\frac{k}{\ln\,2}\)

💡 Short Explanation

At half-life, \(x=\frac{x_{0}}{2}.\) Substituting this into the exponential decay equation and solving gives \(t=\frac{\ln\,2}{k}.\)

✅ Correct Answer Explanation

The exponential decay model is \(x=x_{0}e^{-kt},\) where \(k>0\) is the decay constant. At half-life, \(x=\frac{x_{0}}{2}.\) Substitute into the equation: \(\frac{x_{0}}{2}=x_{0}e^{-kt}.\) Divide both sides by \(x_{0}\): \(\frac{1}{2}=e^{-kt}.\) Take the natural logarithm: \(ln⁡\left(\frac{1}{2}\right)=-kt.\) Since \(ln⁡\left(\frac{1}{2}\right)=-ln⁡2,\) we get \(-ln⁡2=-kt.\) Hence, \(t=\frac{\ln\,2}{k}.\) Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
Correct.
It is the standard half-life formula for exponential decay.
Option B)
Incorrect because the negative signs cancel during simplification, making the time positive.
Option C)
Incorrect because \(\ln\,\left(\frac{1}{2}\right)\)should be divided by \(k\), not multiplied by it.
Option D)
Incorrect because the numerator and denominator are interchanged.

🎯 Strategy

Used

Half-Life Formula Using Exponential Decay
1. Write the decay model:
\(x=x_{0}e^{-kt}.\)
1. Substitute the half-life condition:
\(x=\frac{x_{0}}{2}.\)
1. Cancel \(x_{0}\).
2. Apply the natural logarithm.
3. Solve for \(t\).

🧠 Memory Trick

Mnemonic: "Half or Double → Think \(\ln\,2\).

📚 I will follow this exact format (without any graphs or interactive widgets) for all your upcoming CUET questions

11 In the differential equation for population growth dP/dt = kP, the constant k is derived from the birth rate model (α) and death rate model (β) as:

A k = α + β

B k = α β

C k = α - β

D k = α / β

💡 Short Explanation

Population change depends on birth minus death rate Net growth rate is difference of α and β Standard population model form

✅ Correct Answer Explanation

The population model is given by dP/dt = (α - β)P, so the coefficient of P is k. Thus, k = α - β, where α is birth rate and β is death rate. • A is incorrect since addition does not represent net change • B is unrelated to rate modeling • D has no biological interpretation in this model

❓ Why Other Options Are Incorrect

Option A → Would imply both birth and death increase population
Option B → Product has no rate interpretation
Option D → Ratio is not used in standard population models

🎯 Strategy

Used: Elimination

Application: Standard form comparison with known model

Final Logic: Compare coefficient of P in given DE

🧠 Memory Trick

"Population = Birth − Death"

📚 Class XII – NCERT Mathematics – Differential Equations – Population Growth Model

12 If a population model is integrated to yield
\(P=Ae^{kt},\)
where
\(k=\alpha -\beta ,\)
under what condition will the population size remain perfectly static over time?

A \(\alpha >\beta\)

B \(\alpha <\beta\)

C \(\alpha =\beta\)

D \(A=0\) only

💡 Short Explanation

The population remains constant when the birth rate equals the death rate. In this case, \(k=\alpha -\beta =0,\) so \(P=Ae^{0}=A,\) which is constant for all time.

✅ Correct Answer Explanation

The population model is \(P=Ae^{kt},\) where \(k=\alpha -\beta ,\) with \(\alpha\)= birth rate, \(\beta\)= death rate. For the population to remain unchanged, \(k=0.\) Therefore, \(\alpha -\beta =0,\) which implies \(\alpha =\beta .\) Substituting \(k=0\) into the model, \(P=Ae^{0}=A.\) Since \(A\) is constant, \(P\) remains constant over time. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A) \(\alpha >\beta\)
Incorrect because
\(k=\alpha -\beta >0,\)
leading to exponential growth, not a constant population.
Option B) \(\alpha <\beta\)
Incorrect because
\(k=\alpha -\beta <0,\)
leading to exponential decay.
Option C) \(\alpha =\beta\)
✅ Correct.
Here,
\(k=0,\)
so
\(P=A,\)
which is constant.
Option D) \(A=0\) only
Incorrect because \(A=0\) merely represents a zero initial population.
A static population occurs whenever
\(\alpha =\beta ,\)
regardless of the value of \(A\).

🎯 Strategy

Use

Analyse the Growth Constant

Application:

1. Identify
\(k=\alpha -\beta .\)
1. For a constant population, set
\(k=0.\)
1. Solve for the relationship between the birth and death rates.

🧠 Memory Trick

Mnemonic: "Equal Rates, No Change."

📚 Chapter: Differential Equations

13 A proportional growth model
\(P(t)=\lambda e^{kt}\)
is used for bacteria. If
\(P(3)=10,000\)
and
\(P(5)=40,000,\)
the division of the two equations
\(\frac{\lambda e^{5k}}{\lambda e^{3k}}\)
yields:
\(e^{2k}=4\)
\(e^{k}=4\)
\(e^{8k}=4\)
\(e^{15k}=4\)

💡 Short Explanation

Divide the equation at \(t=5\) by the equation at \(t=3\). The constant \(\lambda\) cancels, and using the law of exponents, \(\frac{e^{5k}}{e^{3k}}=e^{2k}.\) Hence, \(e^{2k}=4.\)

✅ Correct Answer Explanation

The population model is \(P(t)=\lambda e^{kt}.\) Given, \(P(3)=10,000,\) so \(\lambda e^{3k}=10,000.\) Also, \(P(5)=40,000,\) so \(\lambda e^{5k}=40,000.\) Divide the second equation by the first: \(\frac{\lambda e^{5k}}{\lambda e^{3k}}=\frac{40,000}{10,000}.\) The constant \(\lambda\) cancels: \(e^{5k-3k}=4.\) Therefore, \(e^{2k}=4.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(e^{2k}=4\)
Correct.
Applying the exponent law,
\(\frac{e^{5k}}{e^{3k}}=e^{2k}.\)
Option B)
\(e^{k}=4\)
Incorrect because the exponent difference is
\(5k-3k=2k,\)
not \(k\).
Option C)
\(e^{8k}=4\)
Incorrect because exponents are subtracted during division, not added.
Option D)
\(e^{15k}=4\)
Incorrect because exponents are not multiplied when dividing exponential terms.

🎯 Strategy

Used

Division of Exponential Equations

Application:

1. Write the two population equations.
2. Divide one equation by the other.
3. Cancel the common constant \(\lambda\).
4. Apply the exponent law
\(\frac{e^{a}}{e^{b}}=e^{a-b}.\)
1. Simplify the resulting equation.

🧠 Memory Trick

Mnemonic: "Divide Exponentials → Exponents Subtract."

📚 Chapter: Differential Equations

14 After finding
\(e^{k}=2,\)
in the bacterial growth problem, substituting this back to find the initial population \(\lambda\) through
\(\lambda (2)^{3}=10,000\)
evaluates to:
\(\frac{10,000}{6}\)
\(\frac{10,000}{8}\)
\(\frac{10,000}{4}\)
\(\frac{10,000}{2}\)

💡 Short Explanation

Since \(e^{k}=2,\) we have \(e^{3k}=(e^{k})^{3}=2^{3}=8.\) Therefore, \(\lambda =\frac{10,000}{8}=1250.\)

✅ Correct Answer Explanation

The population model is \(P(t)=\lambda e^{kt}.\) Given, \(P(3)=10,000,\) so \(\lambda e^{3k}=10,000.\) Since \(e^{k}=2,\) using the exponent law, \(e^{3k}=(e^{k})^{3}=2^{3}=8.\) Substitute into the equation: \(\lambda (8)=10,000.\) Hence, \(\lambda =\frac{10,000}{8}=1250.\) Thus, the required expression is \(\frac{10,000}{8}.\) Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A)
\(\frac{10,000}{6}\)
Incorrect because
\(2^{3}=8,\)
not 6.
Option B)
\(\frac{10,000}{8}\)
Correct.
Since
\(e^{3k}=(e^{k})^{3}=2^{3}=8,\)
this gives the correct initial population.
Option C)
\(\frac{10,000}{4}\)
Incorrect because
\(2^{3}\neq 4.\)
Option D)
\(\frac{10,000}{2}\)
Incorrect because the exponent is 3, so the exponential factor is 8, not 2.

🎯 Strategy

Used

Substitution Using Exponent Laws

Application:

1. Use the known value
\(e^{k}=2.\)
1. Apply the exponent law
\(e^{3k}=(e^{k})^{3}.\)
1. Substitute into the population equation.
2. Solve for the unknown constant \(\lambda\).

🧠 Memory Trick

Mnemonic: "Power Before Division."

📚 Chapter: Differential Equations

15 During a short time interval \(\Delta t\), the amount of continuous compounding interest added to an account \(A\) with annual rate \(r\) is approximately given by which expression before taking the limit?
\(\Delta A=r⋅A(t)⋅\Delta t\)
\(\Delta A=\frac{r}{A(t)}⋅\Delta t\)
\(\Delta A=\frac{A(t)}{r\Delta t}\)
\(\Delta A=r^{A(t)}\Delta t\)

💡 Short Explanation

Rate proportional to amount Linear approximation over small interval Standard growth model

✅ Correct Answer Explanation

Continuous compounding assumes rate of change proportional to amount: dA/dt = rA So over small Δt: ΔA ≈ rA(t)Δt

❓ Why Other Options Are Incorrect

B → inverse relation incorrect
C → dimensionally invalid
D → exponential of power incorrect

🎯 Strategy

Used: Dimensional/Unit Analysis

Final Logic: Only rAΔt has correct growth units

🧠 Memory Trick

"Rate × Amount × Time"

📚 Class XII – NCERT Mathematics – Differential Equations – Continuous Compounding

16 A bank pays 4% interest compounded continuously. What is the precise analytical solution for the time it takes the money to double, given that
\(e^{0.04t}=2?\)
\(t=\frac{\ln\,2}{0.4}\)
\(t=\frac{\ln\,2}{0.04}\)
\(t=\frac{2}{0.04}\)
\(t=\frac{\ln\,0.04}{2}\)

💡 Short Explanation

Take natural log Solve exponential equation Isolate time variable

✅ Correct Answer Explanation

\(e^{0.04t}=2?\)Taking ln: 0.04t = ln2 So t = ln2 / 0.04

❓ Why Other Options Are Incorrect

A → incorrect decimal shift
C → ignores logarithm
D → wrong log placement

🎯 Strategy

Used: Substitution

Final Logic: Apply logarithm to exponential equation

🧠 Memory Trick

"Log kills exponential"

📚 Class XII – NCERT Mathematics – Differential Equations – Continuous Growth

17 Evaluating the cake cooling process from
\(t=0 (T={185}^{\circ })\)
to
\(t=30 (T={150}^{\circ }),\)
with
\(A={75}^{\circ },\)
the constant \(k\) is embedded in the equation
\(e^{-30k}=\frac{75}{110}.\)
What is the fraction
\(\frac{75}{110}\)
reduced to decimal form as given in the text?

A 0.7500

B 0.6818

C 0.5000

D 0.8500

💡 Short Explanation

Simplify fraction Convert to decimal Approximate value

✅ Correct Answer Explanation

75/110 = 15/22 ≈ 0.6818

❓ Why Other Options Are Incorrect

A → too large
C → incorrect halving
D → incorrect approximation

🎯 Strategy

Used: Elimination

Final Logic: Direct fraction simplification

🧠 Memory Trick

"15/22 ≈ 0.68"

📚 Class XII – NCERT Mathematics – Differential Equations – Newton's Law of Cooling

18 For the heating process under Newton's Law of Cooling, the temperature of the body T is less than the surrounding medium A. Consequently, dT/dt is positive, indicating that the temperature T is an:

A Oscillating function of time

B Inverse function of time

C Increasing function of time

D Exponential decay function of time

💡 Short Explanation

Heat flows into body Temperature rises Positive rate of change

✅ Correct Answer Explanation

Since T < A, heat flows from environment to body, so dT/dt > 0, meaning T increases over time until equilibrium.

❓ Why Other Options Are Incorrect

A → no oscillation in cooling law
B → unrelated concept
D → contradicts heating case

🎯 Strategy

Used: Contextual/Tonal Matching

Final Logic: Physical interpretation of sign of derivative

🧠 Memory Trick

"Cold → heats up"

📚 Class XII – NCERT Mathematics – Differential Equations – Newton's Law of Cooling

📄 Passage

To find the differential equation governing the variation in the amount of Carbon-14, scientists use the model

\(\frac{dx}{dt}=-k_{1}x,\)
whose general solution is
\(x(t)=x_{0}e^{-k_{1}t},\)
where \(x_{0}\)is the initial amount of Carbon-14 in the living organism. Carbon-14 has a half-life of 5700 years, which means that after 5700 years only half of the original Carbon-14 remains. Thus,
\(\frac{x_{0}}{2}=x_{0}e^{-5700k_{1}}.\)
Cancelling \(x_{0}\)and solving for the decay constant gives
\(k_{1}=\frac{\ln\,2}{5700}.\)
Substituting this value into the exponential decay equation, we obtain
\(x(t)=x_{0}e^{-\left(\frac{\ln\,2}{5700}\right)t}.\)
Using this model, the age of an archaeological sample can be estimated by comparing its remaining Carbon-14 with the original amount \(x_{0}\).
Question : Utilizing the half-life concept and substituting
\(k_{1}=\frac{\ln\,2}{5700}\)
into the general equation, what is the exact formula for
\(x(t)?\)
\(x(t)=x_{0}e^{\frac{(ln⁡2)t}{5700}}\)
\(x(t)=x_{0}e^{-\left(\frac{\ln\,2}{5700}\right)t}\)
\(x(t)=x_{0}e^{-5700(ln⁡2)t}\)
\(x(t)=x_{0}ln⁡\left(\frac{2t}{5700}\right)\)

💡 Short Explanation

Substitute \(k_{1}=\frac{\ln\,2}{5700}\) into \(x(t)=x_{0}e^{-k_{1}t},\) to obtain \(x(t)=x_{0}e^{-\left(\frac{\ln\,2}{5700}\right)t}.\)

✅ Correct Answer Explanation

The exponential decay model is \(x(t)=x_{0}e^{-k_{1}t}.\) Using \(k_{1}=\frac{\ln\,2}{5700},\) we get \(x(t)=x_{0}e^{-\left(\frac{\ln\,2}{5700}\right)t}.\) Hence, \(x(t)=x_{0}e^{-\left(\frac{\ln\,2}{5700}\right)t}.\) Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A)
Incorrect because the exponent is positive, representing exponential growth rather than decay.
Option B)
Correct.
This is obtained by direct substitution into the decay model.
Option C)
Incorrect because \(5700\) should divide \(\ln\,2\), not multiply it.
Option D)
Incorrect because the solution of the differential equation is exponential, not logarithmic.

🎯 Strategy

Use

Substitution into the Exponential Decay Formula
1. Write the general solution.
2. Substitute the value of the decay constant.
3. Simplify the exponent.

🧠 Memory Trick

Simply replace \(k\) in the formula.

📚 Chapter: Differential Equations

20 (Passage-Based)
To find the differential equation governing the variation in the amount of Carbon-14, scientists use the model
\(\frac{dx}{dt}=-k_{1}x,\)
whose general solution is
\(x(t)=x_{0}e^{-k_{1}t},\)
where \(x_{0}\)is the initial amount of Carbon-14 in the living organism. Carbon-14 has a half-life of 5700 years, which means that after 5700 years only half of the original Carbon-14 remains. Thus,
\(\frac{x_{0}}{2}=x_{0}e^{-5700k_{1}}.\)
Cancelling \(x_{0}\)and solving for the decay constant gives
\(k_{1}=\frac{\ln\,2}{5700}.\)
Substituting this value into the exponential decay equation, we obtain
\(x(t)=x_{0}e^{-\left(\frac{\ln\,2}{5700}\right)t}.\)
Using this model, the age of an archaeological sample can be estimated by comparing its remaining Carbon-14 with the original amount \(x_{0}\).
Question: In the age estimation process for a sample containing
\(\frac{1}{4}\)
of the original Carbon-14, substituting
\(x(t)=\frac{x_{0}}{4}\)
into the decay equation
\(\frac{x_{0}}{4}=x_{0}e^{-\left(\frac{\ln\,2}{5700}\right)t}\)
yields which analytical step?
\(ln⁡\left(\frac{1}{4}\right)=\left(,\ \frac{\ln\,2}{5700}\right)t\)
\(ln⁡4=\left(\frac{\ln\,2}{5700}\right)t\)
\(-ln⁡4=\left(\frac{\ln\,2}{5700}\right)t\)
\(\frac{1}{4}=\left(,\ \frac{\ln\,2}{5700}\right)t\)

💡 Short Explanation

After cancelling \(x_{0}\), \(\frac{1}{4}=e^{-\left(\frac{\ln\,2}{5700}\right)t}.\) Taking the natural logarithm of both sides gives \(ln⁡\left(\frac{1}{4}\right)=\left(,\ \frac{\ln\,2}{5700}\right)t.\)

✅ Correct Answer Explanation

Given, \(\frac{x_{0}}{4}=x_{0}e^{-\left(\frac{\ln\,2}{5700}\right)t}.\) Divide both sides by \(x_{0}\): \(\frac{1}{4}=e^{-\left(\frac{\ln\,2}{5700}\right)t}.\) Taking the natural logarithm, \(ln⁡\left(\frac{1}{4}\right)=ln⁡\left(e^{-\left(\frac{\ln\,2}{5700}\right)t}\right).\) Therefore, \(ln⁡\left(\frac{1}{4}\right)=\left(,\ \frac{\ln\,2}{5700}\right)t.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
Correct.
This is obtained directly after taking the natural logarithm.
Option B)
Incorrect because the negative sign from the exponential exponent is missing.
Option C)
Incorrect because although \(ln⁡\left(\frac{1}{4}\right)=-ln⁡4\), the immediate analytical step after taking logarithms is
\(ln⁡\left(\frac{1}{4}\right)=\left(,\ \frac{\ln\,2}{5700}\right)t.\)
Option D)
Incorrect because the natural logarithm must be applied before solving for \(t\).

🎯 Strategy

Used

Apply Natural Logarithms
1. Substitute the remaining fraction into the decay equation.
2. Cancel the initial amount.
3. Take the natural logarithm of both sides.
4. Simplify and solve for time.

🧠 Memory Trick

Mnemonic: "Substitute → Log → Solve."

📚 Chapter: Differential Equations

UG Applied Mathematics Booster Test 3 - Mathematical Modeling

Test Complete!

Answer Review