CUET UG Mathematics Booster Test 1 - Rate of Change Applications -

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CUET UG Mathematics Booster Test 1 - Rate of Change Applications

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 1 - Rate of Change Applications

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QUESTION 1 OF 20

If the derivative dy/dx represents instantaneous rate, then ∫ₐᵇ (dy/dx) dx provides:

Answer Review

1 If the derivative dy/dx represents instantaneous rate, then ∫ₐᵇ (dy/dx) dx provides:

A The rate of change at b

B The total net change of y from x = a to x = b

C The absolute maximum of y

D The continuous rate of y

💡 Short Explanation

Integration reverses differentiation. Definite integral accumulates change. Net change equals final value minus initial value.

✅ Correct Answer Explanation

By the Fundamental Theorem of Calculus, integrating a derivative over an interval gives the net change in the original function. Thus, \(\int_{a}^{b}\,\frac{dy}{dx}dx=y(b)-y(a)\) Option B correctly represents total net change. Option A gives only a point rate, Option C concerns extrema, and Option D misinterprets the role of integration.

❓ Why Other Options Are Incorrect

Option A → Gives the derivative at a single point, not accumulated change.
Option C → Integration does not directly determine maximum values.
Option D → The integral gives accumulated change, not a continuous rate.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Recognize the Fundamental Theorem of Calculus connecting derivatives and integrals.

Final Logic: Integral of rate of change equals total change.

🧠 Memory Trick

Derivative → Rate, Integral → Change

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Fundamental Theorem of Calculus

2 If a position vector maps a curve y = x², what is the interpretation of dy/dx at x = 0?

A Maximum rate of motion

B Constant velocity vector

C Zero instantaneous rate of change (horizontal tangent)

D Negative displacement

💡 Short Explanation

Differentiate y = x². Slope at x = 0 equals zero. Zero slope means horizontal tangent.

✅ Correct Answer Explanation

For y = x², \(\frac{dy}{dx}=2x\) At x = 0, dy/dx = 0. A zero derivative indicates a horizontal tangent and zero instantaneous rate of change of y with respect to x at that point. Therefore Option C is correct. The other options have no relation to the derivative value here.

❓ Why Other Options Are Incorrect

Option A → No maximum rate exists at x = 0.
Option B → Derivative is a scalar slope, not a constant velocity vector.
Option D → Negative displacement is unrelated to the derivative.

🎯 Strategy

Used: Substitution

Application: Differentiate and substitute x = 0.

Final Logic: dy/dx = 0 gives a horizontal tangent.

🧠 Memory Trick

x² → 2x → 0 at origin

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Geometrical Meaning of Derivative

3 Assertion (A): The formula dy/dx = (dy/dt)/(dx/dt) is valid only if dx/dt ≠ 0.
Reason (R): Division by zero makes the expression undefined.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Chain Rule connects related rates. Division by zero is undefined. The reason correctly explains the assertion.

✅ Correct Answer Explanation

The formula \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\) requires dx/dt ≠ 0 because division by zero is undefined. Therefore the assertion is true. The reason is also true and directly explains why the condition is necessary. Hence Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A → Both statements are mathematically true.
Option B → The reason is not false.
Option D → The assertion is also true.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Verify the mathematical condition required for the Chain Rule form.

Final Logic: Undefined division justifies the restriction.

🧠 Memory Trick

No Divide by Zero

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Chain Rule

4 To find how fast the enclosed area of a circular wave increases when a stone drops:
1. Calculate dr/dt.
2. Form area A = πr².
3. Determine dA/dt using the chain rule.
4. Evaluate at the specific radius.

A 1, 2, 3, 4

B 2, 4, 1, 3

C 3, 2, 1, 4

D 2, 1, 3, 4

💡 Short Explanation

Start with the formula. Obtain radius rate. Apply Chain Rule and evaluate.

✅ Correct Answer Explanation

The logical procedure is: write the area formula A = πr², obtain the radius rate dr/dt, use the Chain Rule to find dA/dt, and finally substitute the required radius. This corresponds to the sequence 2, 1, 3, 4. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Formula should precede differentiation.
Option B → Evaluation occurs too early.
Option C → Chain Rule cannot be applied before forming the equation.

🎯 Strategy

Used: Elimination

Application: Follow the natural sequence of solving related-rate problems.

Final Logic: Formula → Rate → Chain Rule → Evaluation.

🧠 Memory Trick

Formula → Rate → Differentiate → Evaluate

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

5 Which statements are correct when the radius of a circle increases at 0.7 cm/s?
I. The circumference increases at 1.4π cm/s.
II. The rate depends on the current radius.
III. The area increases at a constant rate.

A I only

B I and II only

C II and III only

D I, II, and III

💡 Short Explanation

Circumference rate is constant. Area rate depends on radius. Area growth is not constant.

✅ Correct Answer Explanation

For circumference C = 2πr, \(\frac{dC}{dt}=2\pi \frac{dr}{dt}=1.4\pi\) Thus Statement I is true. For area, \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\) which depends on r, so Statement II is true. Since dA/dt varies with r, Statement III is false. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Omits Statement II.
Option C → Includes false Statement III.
Option D → Statement III is incorrect.

🎯 Strategy

Used: Option Grouping

Application: Evaluate each statement separately.

Final Logic: Only Statements I and II are true.

🧠 Memory Trick

Circumference Constant, Area Depends on r

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

6 A cube has edge increasing at 3 cm/s. At edges x = 5, 10, 15 cm, the values of dV/dt are:

A 225, 900, 2025 cm³/s

B 75, 300, 675 cm³/s

C 15, 30, 45 cm³/s

D 45, 90, 135 cm³/s

💡 Short Explanation

Volume V = x³. Differentiate using Chain Rule. Substitute x values.

✅ Correct Answer Explanation

For a cube, \(V=x^{3}\) Hence, \(\frac{dV}{dt}=3x^{2}\frac{dx}{dt}\) Given dx/dt = 3, \(\frac{dV}{dt}=9x^{2}\) At x = 5, 10, 15, the values are 225, 900, and 2025 cm³/s respectively. Thus Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Uses an incorrect coefficient.
Option C → Ignores the x² factor.
Option D → Underestimates the derivative values.

🎯 Strategy

Used: Substitution

Application: Substitute edge lengths into dV/dt = 9x².

Final Logic: Numerical evaluation matches Option A.

🧠 Memory Trick

Cube Rate = 3x²(dx/dt)

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

7 Match the wave expansion attributes when wave speed is 5 cm/s and radius is 8 cm:

List I	List II
1. Wave speed (dr/dt)	a. 80π cm²/s
2. Radius r	b. 8 cm
3. Area formula	c. 5 cm/s
4. Rate of increase of area	d. A = πr²

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

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

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

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

💡 Short Explanation

Identify given quantities. Use circle area formula. Compute area growth rate.

✅ Correct Answer Explanation

Wave speed corresponds to 5 cm/s (c). Radius corresponds to 8 cm (b). Area formula is A = πr² (d). Using \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\) gives \(2\pi (8)(5)=80\pi\) cm²/s (a). Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Mismatches all key quantities.
Option B → Assigns incorrect values to wave speed and radius.
Option D → Interchanges formulas and numerical values.

🎯 Strategy

Used: Matching / Substitution

Application: Identify known quantities and compute dA/dt.

Final Logic: Correct matching yields Option C.

🧠 Memory Trick

Area Rate = 2πrv

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

8 A particle moves along 6y = x³ + 2. Which statement is incorrect?

A dy/dt = x²/2 dx/dt

B 6 dy/dt = 3x² dx/dt

C If dy/dt = 8 dx/dt, then x² = 16

D The points satisfying the condition lie only in the first quadrant

💡 Short Explanation

Differentiate implicitly. Apply the rate condition. Solutions occur in more than one quadrant.

✅ Correct Answer Explanation

Differentiating gives \(6\frac{dy}{dt}=3x^{2}\frac{dx}{dt}\) or \(\frac{dy}{dt}=\frac{x^{2}}{2}\frac{dx}{dt}\) If dy/dt = 8 dx/dt, then x² = 16, giving x = ±4. Hence valid points occur in more than one quadrant. Therefore Option D is incorrect.

❓ Why Other Options Are Incorrect

Option A → Correct result after simplification.
Option B → Correct implicit differentiation form.
Option C → Correct consequence of the given rate condition.

🎯 Strategy

Used: Elimination

Application: Differentiate and test each statement.

Final Logic: Solutions exist for both x = 4 and x = −4.

🧠 Memory Trick

x² = 16 ⇒ ±4

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

9 For a rectangle with length decreasing at 5 cm/min and width increasing at 4 cm/min, the rate of change of perimeter is:

A Positive slope

B Negative slope of 2 cm/min

C Exponential decline

D Zero slope

💡 Short Explanation

Perimeter depends on both dimensions. Differentiate P = 2(x + y). Substitute rates.

✅ Correct Answer Explanation

For P = 2(x + y), \(\frac{dP}{dt}=2\left(\frac{dx}{dt}\ +\ \frac{dy}{dt}\right)\) Substituting dx/dt = −5 and dy/dt = 4 gives \(dP/dt=2(-1)=-2\) cm/min. Therefore the perimeter decreases at 2 cm/min and Option B is correct.

❓ Why Other Options Are Incorrect

Option A → The rate is negative.
Option C → No exponential behaviour is involved.
Option D → The derivative is not zero.

🎯 Strategy

Used: Substitution

Application: Insert rates into the perimeter derivative formula.

Final Logic: Net rate equals −2 cm/min.

🧠 Memory Trick

Perimeter = Twice the Sum

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

10 When length x = 8 cm (decreasing at 5 cm/min) and width y = 6 cm (increasing at 4 cm/min), the rate of change of area is:

A 2 cm²/min

B -2 cm²/min

C 32 cm²/min

D -30 cm²/min

💡 Short Explanation

Area = xy. Apply product rule. Substitute given values.

✅ Correct Answer Explanation

For A = xy, \(\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}\) Substituting x = 8, y = 6, dx/dt = −5, dy/dt = 4: \(dA/dt=8(4)+6(-5)=32-30=2\) cm²/min. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Wrong sign.
Option C → Ignores the second product-rule term.
Option D → Uses only one contribution.

🎯 Strategy

Used: Substitution

Application: Apply product rule directly.

Final Logic: Net area change equals +2 cm²/min.

🧠 Memory Trick

Area Rate = xdy/dt + ydx/dt

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

11 Given \(C(x)=0.005x^{3}-0.02x^{2}+30x+5000\). Calculate the marginal cost when \(x=3\).

A 30.00

B 30.50

C 30.02

D 31.00

💡 Short Explanation

Marginal cost is the derivative of cost. Differentiate the cost function. Substitute x = 3.

✅ Correct Answer Explanation

Marginal cost is \(C^{'}(x)\). \(C^{'}(x)=0.015x^{2}-0.04x+30\) At \(x=3\), \(C^{'}(3)=0.015(9)-0.04(3)+30=0.135-0.12+30=30.015\) Rounded to two decimal places, the marginal cost is 30.02. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores the derivative contributions from the polynomial terms.
Option B → Overestimates the calculated marginal cost.
Option D → Much larger than the actual derivative value.

🎯 Strategy

Used: Substitution

Application: Differentiate first and then substitute x = 3.

Final Logic: \(C^{'}(3)=30.015\approx 30.02\).

🧠 Memory Trick

MC = dC/dx

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Marginal Cost

12 If total revenue \(R(x)=13x^{2}+26x+15\), what is the marginal revenue at \(x=7\)?

A 182

B 196

C 130

D 208

💡 Short Explanation

Marginal revenue is derivative of revenue. Differentiate R(x). Evaluate at x = 7.

✅ Correct Answer Explanation

Marginal revenue is \(R^{'}(x)=26x+26\) At \(x=7\), \(R^{'}(7)=26(7)+26=182+26=208\) Thus Option D is correct. The remaining options result from incomplete or incorrect evaluation of the derivative.

❓ Why Other Options Are Incorrect

Option A → Omits the constant derivative term 26.
Option B → Incorrect arithmetic.
Option C → Does not correspond to the derivative value.

🎯 Strategy

Used: Substitution

Application: Differentiate and substitute x = 7.

Final Logic: \(26\times 7+26=208\).

🧠 Memory Trick

MR = dR/dx

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Marginal Revenue

13 If a circular boundary grows uniformly with \(dr/dt=a\), and probability depends on area, the rate of change of area is:

A Linear over time

B Proportional to the radius

C Inversely proportional to the radius

D Constant

💡 Short Explanation

Area depends on r². Differentiate with respect to time. Area rate contains the radius factor.

✅ Correct Answer Explanation

For a circle, \(A=\pi r^{2}\) Differentiating, \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\) Since \(dr/dt=a\), \(\frac{dA}{dt}=2\pi ar\) The rate of area change is directly proportional to the radius. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Area itself may not grow linearly with time.
Option C → The formula contains r in the numerator, not denominator.
Option D → dA/dt changes as r changes.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Observe the dependence of dA/dt on r.

Final Logic: \(dA/dt\propto r\).

🧠 Memory Trick

Area Rate ∝ Radius

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

📄 Passage

Sand pours from a pipe at 12 cm³/s, forming a cone such that the height is always one-sixth of the radius of the base.

Using the passage, the volume of the cone in terms of height h is:

A \(V=\frac{1}{3}\pi h^{3}\)

B \(V=12\pi h^{2}\)

C \(V=36\pi h^{3}\)

D \(V=72\pi h^{3}\)

💡 Short Explanation

Cone volume depends on radius and height. Express radius using the given relation. Substitute into the volume formula.

✅ Correct Answer Explanation

Given height is one-sixth of radius: \(h=\frac{r}{6}\Rightarrow r=6h\) For a cone, \(V=\frac{1}{3}\pi r^{2}h\) Substituting \(r=6h\), \(V=\frac{1}{3}\pi (36h^{2})h=12\pi h^{3}\) Therefore Option B is correct. The provided answer is incorrect and has been corrected.

❓ Why Other Options Are Incorrect

Option A → Uses r = h incorrectly.
Option C → Coefficient is three times larger than the correct value.
Option D → Results from an incorrect substitution and is not valid.

🎯 Strategy

Used: Substitution

Application: Replace r with 6h in the cone volume formula.

Final Logic: \(V=12\pi h^{3}\).

🧠 Memory Trick

h = r/6 ⇒ r = 6h

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

📄 Passage

Sand pours from a pipe at 12 cm³/s, forming a cone such that the height is always one-sixth of the radius of the base.

From the passage, how fast is the height increasing when h = 4 cm?

A \(1/(12\pi )\)cm/s

B \(1/(24\pi )\)cm/s

C \(1/(36\pi )\)cm/s

D \(1/(48\pi )\)cm/s

💡 Short Explanation

Use corrected cone-volume relation. Differentiate with respect to time. Substitute h = 4.

✅ Correct Answer Explanation

From Question 14, \(V=12\pi h^{3}\) Differentiating, \(\frac{dV}{dt}=36\pi h^{2}\frac{dh}{dt}\) Given \(dV/dt=12\) and \(h=4\), \(12=36\pi (16)\frac{dh}{dt}\frac{dh}{dt}=\frac{1}{48\pi }\) Therefore Option D is mathematically correct. The provided answer is incorrect and must be corrected.

❓ Why Other Options Are Incorrect

Option A → Uses an incorrect coefficient.
Option B → Does not satisfy the differentiated equation.
Option C → Derived from an incorrect volume expression.

🎯 Strategy

Used: Substitution

Application: Differentiate V and substitute known values.

Final Logic: \(dh/dt=1/(48\pi )\).

🧠 Memory Trick

Differentiate Before Substituting

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

16 The area of a circle increases at \(10\pi\) cm²/s. At \(r=5\) cm, find \(dr/dt\).

A 2 cm/s

B 1 cm/s

C 0.5 cm/s

D 5 cm/s

💡 Short Explanation

Use the area-rate formula. Substitute the given values. Solve for dr/dt.

✅ Correct Answer Explanation

For a circle, \(A=\pi r^{2}\frac{dA}{dt}=2\pi r\frac{dr}{dt}\) Given \(dA/dt=10\pi\) and \(r=5\), \(10\pi =2\pi (5)\frac{dr}{dt}10\pi =10\pi \frac{dr}{dt}\frac{dr}{dt}=1\) Thus Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Twice the correct value.
Option C → Half the correct value.
Option D → Much larger than required.

🎯 Strategy

Used: Substitution

Application: Insert known values into dA/dt.

Final Logic: dr/dt = 1 cm/s.

🧠 Memory Trick

10π = 10π(dr/dt)

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

17 A spherical balloon has volume increasing at 900 cm³/s. The rate of increase of radius when r = 15 cm is:

A \(1/\pi\) cm/s

B \(3/\pi\) cm/s

C \(4/\pi\) cm/s

D \(9/\pi\) cm/s

💡 Short Explanation

Differentiate sphere volume. Use related rates. Solve for dr/dt.

✅ Correct Answer Explanation

For a sphere, \(V=\frac{4}{3}\pi r^{3}\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\) Given \(dV/dt=900\) and \(r=15\), \(900=4\pi (225)\frac{dr}{dt}900=900\pi \frac{dr}{dt}\frac{dr}{dt}=\frac{1}{\pi }\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Triple the actual rate.
Option C → Four times the correct rate.
Option D → Nine times the correct rate.

🎯 Strategy

Used: Substitution

Application: Use sphere related-rate formula.

Final Logic: dr/dt = 1/π.

🧠 Memory Trick

900 Cancels 900π

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

18 A ladder 5 m long has its base moving away from the wall at 2 cm/s. When the base is 4 m from the wall, the rate of change of height is:

A 2 cm/s

B 8/3 cm/s

C 3 cm/s

D 4/3 cm/s

💡 Short Explanation

Apply Pythagoras theorem. Differentiate implicitly. Compute dy/dt.

✅ Correct Answer Explanation

For the ladder, \(x^{2}+y^{2}=25\) When \(x=4\), \(y=3\). Differentiating: \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=-\frac{4}{3}(2)=-\frac{8}{3}\) The height decreases at \(8/3\) cm/s. Hence Option B is selected as the magnitude.

❓ Why Other Options Are Incorrect

Option A → Does not satisfy the differentiated equation.
Option C → Incorrect numerical value.
Option D → Half the required magnitude.

🎯 Strategy

Used: Substitution

Application: Use the 3–4–5 triangle and related rates.

Final Logic: \(∣dy/dt∣=8/3\) cm/s.

🧠 Memory Trick

3–4–5 Triangle

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Related Rates

19 A positive rate of change \(dV/dt>0\) indicates:
I. The object is expanding.
II. The graph has positive slope.
III. The radius is decreasing.

A I only

B II and III only

C I and II only

D I, II, and III

💡 Short Explanation

Positive rate means increase. Positive slope corresponds to growth. Decreasing radius contradicts positive growth.

✅ Correct Answer Explanation

A positive derivative indicates the quantity increases with the independent variable. Therefore Statement I is true. A positive derivative corresponds to a positive graph slope, making Statement II true. Statement III is false because a decreasing radius implies a negative rate. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Omits Statement II.
Option B → Includes false Statement III.
Option D → Statement III is incorrect.

🎯 Strategy

Used: Option Grouping

Application: Test each statement individually.

Final Logic: Only I and II are true.

🧠 Memory Trick

Positive = Growing

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Sign of Derivative

20 Assertion (A): A negative rate \(dx/dt\) implies the variable x is increasing.
Reason (R): A negative sign indicates increase in magnitude.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Negative rate means decrease. Increasing quantity requires positive rate. Reason is also incorrect.

✅ Correct Answer Explanation

A negative rate \(dx/dt\) indicates that x decreases as time increases. Therefore the assertion is false. The reason is also false because a negative sign represents reduction rather than increase in magnitude. Hence both statements are false and Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Assertion is false.
Option C → Both statements are not true.
Option D → Reason is also false.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret the meaning of a negative derivative.

Final Logic: Negative rate implies decrease, not increase.

🧠 Memory Trick

Negative = Decreasing

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Interpretation of Derivative Sign

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