CUET UG Mathematics Booster Test 2 - Rate of Change Applications -

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CUET UG Mathematics Booster Test 2 - 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 2 - Rate of Change Applications

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QUESTION 1 OF 20

Evaluate the instantaneous rate of change of the area of a circle with respect to its radius r when r = 4 cm.

Answer Review

1 Evaluate the instantaneous rate of change of the area of a circle with respect to its radius r when r = 4 cm.

A 16π

B 4π

C 8π

D 2π

💡 Short Explanation

Area of a circle is πr². Differentiate with respect to r. Substitute r = 4.

✅ Correct Answer Explanation

For a circle, \(A=\pi r^{2}\) Differentiating with respect to r, \(\frac{dA}{dr}=2\pi r\) At r = 4 cm, \(\frac{dA}{dr}=2\pi (4)=8\pi\) Therefore Option C is correct. The other options result from incorrect differentiation or substitution.

❓ Why Other Options Are Incorrect

Option A → Represents area-related computation, not derivative value.
Option B → Uses only πr instead of 2πr.
Option D → Too small and does not satisfy the derivative formula.

🎯 Strategy

Used: Substitution

Application: Differentiate A = πr² and substitute r = 4.

Final Logic: dA/dr = 2πr = 8π.

🧠 Memory Trick

Circle Rate = 2πr

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Rate of Change of Area

2 If marginal revenue is given by dR/dx = 6x + 36, what is MR exactly when x = 5?

A 46

B 56

C 36

D 66

💡 Short Explanation

MR equals dR/dx. Substitute x = 5. Simplify numerically.

✅ Correct Answer Explanation

Marginal revenue is already given: \(MR=\frac{dR}{dx}=6x+36\) Substituting x = 5, \(MR=6(5)+36=30+36=66\) Therefore Option D is correct. The remaining options come from arithmetic mistakes or incomplete substitution.

❓ Why Other Options Are Incorrect

Option A → Incorrect evaluation.
Option B → Arithmetic error.
Option C → Ignores the variable term 6x.

🎯 Strategy

Used: Substitution

Application: Directly substitute x = 5 into the given expression.

Final Logic: 30 + 36 = 66.

🧠 Memory Trick

MR = Rate of Revenue

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

3 For x = f(t) and y = g(t), which statements are correct?
I. dy/dx = dy/dt · dt/dx
II. dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0
III. If dx/dt = 0, then dy/dx is undefined

A I only

B II only

C I, II, and III

D III only

💡 Short Explanation

Both chain-rule forms are equivalent. Division requires non-zero denominator. Zero denominator makes expression undefined.

✅ Correct Answer Explanation

Statement I follows from reciprocal derivatives. Statement II is the standard related-rates formula. Statement III is true because division by dx/dt is impossible when dx/dt = 0. Therefore all three statements are correct and Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A → Omits Statements II and III.
Option B → Ignores Statements I and III.
Option D → Excludes valid Statements I and II.

🎯 Strategy

Used: Option Grouping

Application: Verify each statement individually.

Final Logic: All three statements are mathematically valid.

🧠 Memory Trick

dy/dx = dy/dt ÷ dx/dt

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

4 If \(V=\frac{4}{3}\pi r^{3}\)and \(dr/dt=\frac{1}{2}\)cm/s, find dV/dt when r = 1 cm.

A 2π cm³/s

B 4π cm³/s

C π cm³/s

D 8π cm³/s

💡 Short Explanation

Differentiate sphere volume. Apply Chain Rule. Substitute r and dr/dt.

✅ Correct Answer Explanation

For a sphere, \(V=\frac{4}{3}\pi r^{3}\) Differentiating, \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\) Substituting r = 1 and dr/dt = 1/2, \(\frac{dV}{dt}=4\pi (1)^{2}\left(\frac{1}{2}\right)=2\pi\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Assumes dr/dt = 1.
Option C → Half the correct value.
Option D → Four times the required value.

🎯 Strategy

Used: Substitution

Application: Insert given values into the differentiated formula.

Final Logic: dV/dt = 2π cm³/s.

🧠 Memory Trick

Sphere Rate = 4πr²(dr/dt)

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

5 Match the values for A = πr²:

List I	List II
1. dA/dr at r = 3	a. 8π
2. dA/dr at r = 4	b. 6π
3. dA/dt if r = 5, dr/dt = 2	c. 80π
4. dA/dt if r = 10, dr/dt = 4	d. 20π

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

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

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

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

💡 Short Explanation

Use dA/dr = 2πr. Use dA/dt = 2πr(dr/dt). Match values carefully.

✅ Correct Answer Explanation

At r = 3, \(dA/dr=6\pi\) At r = 4, \(dA/dr=8\pi\) At r = 5 and dr/dt = 2, \(dA/dt=20\pi\) At r = 10 and dr/dt = 4, \(dA/dt=80\pi\) Thus Option B gives the correct matching.

❓ Why Other Options Are Incorrect

Option A → Misplaces 8π and 80π.
Option C → Interchanges first two matches.
Option D → Multiple incorrect pairings.

🎯 Strategy

Used: Substitution

Application: Calculate each quantity separately.

Final Logic: Numerical matching yields Option B.

🧠 Memory Trick

Area Rate = 2πr

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

6 Order the steps to find the surface area rate of a cube when volume increases at 9 cm³/s and x = 10 cm:
1. From dV/dt = 3x² dx/dt = 9, find dx/dt = 3/x².
2. Use S = 6x² ⇒ dS/dt = 12x dx/dt.
3. Substitute x = 10.
4. Evaluate dS/dt.

A 1, 2, 3, 4

B 2, 1, 3, 4

C 3, 2, 1, 4

D 2, 3, 1, 4

💡 Short Explanation

Find dx/dt first. Use surface-area formula. Substitute and evaluate.

✅ Correct Answer Explanation

The volume-rate equation provides dx/dt. Once dx/dt is known, differentiate surface area, substitute x = 10, and evaluate numerically. This logical sequence corresponds to 1 → 2 → 3 → 4. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Uses dS/dt before finding dx/dt.
Option C → Substitutes before deriving formulas.
Option D → Incorrect order of operations.

🎯 Strategy

Used: Elimination

Application: Follow the logical sequence of related-rate calculations.

Final Logic: Rate of edge first, then surface-area rate.

🧠 Memory Trick

Edge Rate Before Surface Rate

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

7 For a circular wave expanding at 4 cm/s, the rate of increase of area at r = 10 cm is:

A 40π cm²/s

B 20π cm²/s

C 60π cm²/s

D 80π cm²/s

💡 Short Explanation

Area = πr². Differentiate with respect to time. Substitute values.

✅ Correct Answer Explanation

For a circle, \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\) At r = 10 cm and dr/dt = 4 cm/s, \(\frac{dA}{dt}=2\pi (10)(4)=80\pi\) Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Half the required value.
Option B → Uses incomplete multiplication.
Option C → Incorrect substitution.

🎯 Strategy

Used: Substitution

Application: Substitute r and dr/dt directly.

Final Logic: 2π × 10 × 4 = 80π.

🧠 Memory Trick

2πr × Rate

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

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

A The curve is a straight line

B dy/dt = 1/2 x² dx/dt

C If dy/dt = 8 dx/dt, then x = ±4

D The y-coordinate can change faster than the x-coordinate

💡 Short Explanation

The equation is cubic. It is not linear. Other statements follow from differentiation.

✅ Correct Answer Explanation

The equation \(6y=x^{3}+2\) represents a cubic curve, not a straight line. Differentiating gives \(\frac{dy}{dt}=\frac{x^{2}}{2}\frac{dx}{dt}\) and the remaining statements are consistent with this relation. Therefore Option A is the incorrect statement.

❓ Why Other Options Are Incorrect

Option B → Correct derivative relation.
Option C → Follows from setting dy/dt = 8dx/dt.
Option D → Possible when the multiplier exceeds 1.

🎯 Strategy

Used: Odd One Out

Application: Identify the statement contradicting the nature of the curve.

Final Logic: Cubic curves are not straight lines.

🧠 Memory Trick

x³ Means Curve

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

9 For a rectangle with dx/dt = -5 cm/min and dy/dt = 4 cm/min, the rate of change of perimeter is:

A 2 cm/min

B -1 cm/min

C -2 cm/min

D 0 cm/min

💡 Short Explanation

Perimeter = 2(x + y). Differentiate. Substitute rates.

✅ Correct Answer Explanation

For \(P=2(x+y)\frac{dP}{dt}=2\left(\frac{dx}{dt}\ +\ \frac{dy}{dt}\right)\) Substituting values, \(dP/dt=2(-5+4)=2(-1)=-2\) Thus Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Wrong sign.
Option B → Missing factor of 2.
Option D → Perimeter is changing.

🎯 Strategy

Used: Substitution

Application: Insert the given rates into dP/dt.

Final Logic: Perimeter decreases at 2 cm/min.

🧠 Memory Trick

Twice the Sum of Rates

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

10 For the same rectangle with x = 8, y = 6, dx/dt = -5, dy/dt = 4, the rate of change of area is:

A 4 cm²/min

B 2 cm²/min

C -2 cm²/min

D 0 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 values, \(dA/dt=8(4)+6(-5)=32-30=2\) Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores one term.
Option C → Wrong sign.
Option D → Area is not constant.

🎯 Strategy

Used: Substitution

Application: Use the product rule for area.

Final Logic: Net area increase equals 2 cm²/min.

🧠 Memory Trick

Area Rate = xdy + ydx

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

11 Assertion (A): The marginal cost for \(C(x)=0.007x^{3}-0.003x^{2}+15x+4000\) evaluated at \(x=17\) is given by the derivative of total cost.
Reason (R): Marginal cost is defined as the instantaneous rate of change of total cost with respect to output.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Marginal cost equals derivative of cost. Derivative measures instantaneous change. Reason directly explains assertion.

✅ Correct Answer Explanation

Marginal cost is defined as \(MC=\frac{dC}{dx}\) which is the instantaneous rate of change of total cost with respect to output. Therefore, evaluating the derivative at \(x=17\) gives the marginal cost at that production level. Both the assertion and reason are true, and the reason correctly explains the assertion.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Compare the definition of marginal cost with the assertion.

Final Logic: Marginal cost is the derivative of total cost.

🧠 Memory Trick

MC = dC/dx

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

12 \(R(x)=3x^{2}+36x+5\). What is the marginal revenue when \(x=15\)?

A 126

B 116

C 90

D 96

💡 Short Explanation

Differentiate revenue function. Obtain marginal revenue. Substitute x = 15.

✅ Correct Answer Explanation

Marginal revenue is \(MR=\frac{dR}{dx}=6x+36\) At \(x=15\), \(MR=6(15)+36=90+36=126\) Hence Option A is correct. The remaining values result from incorrect differentiation or arithmetic mistakes.

❓ Why Other Options Are Incorrect

Option B → Incorrect numerical evaluation.
Option C → Ignores the constant derivative term.
Option D → Does not satisfy the derivative expression.

🎯 Strategy

Used: Substitution

Application: Differentiate first and then substitute x = 15.

Final Logic: \(6(15)+36=126\).

🧠 Memory Trick

MR = Revenue Derivative

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

13 If the radius of a circle increases uniformly at 3 cm/s, the rate of increase of area at r = 10 cm is:

A 10π

B 30π

C 60π

D 40π

💡 Short Explanation

Area depends on radius squared. Use Chain Rule. Substitute r and dr/dt.

✅ Correct Answer Explanation

For a circle, \(A=\pi r^{2}\) Differentiating with respect to time: \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\) Substituting \(r=10\) and \(dr/dt=3\), \(\frac{dA}{dt}=2\pi (10)(3)=60\pi\) Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Too small; ignores radius dependence.
Option B → Half the correct value.
Option D → Incorrect substitution.

🎯 Strategy

Used: Substitution

Application: Use the related-rate formula for area.

Final Logic: \(2\pi \times 10\times 3=60\pi\).

🧠 Memory Trick

2πr × dr/dt

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

📄 Passage

An inverted right circular cone has r/h = 1/2, so its volume is \(V=\frac{1}{12}\pi h^{3}\). Water is poured at 5 m³/h.

Using the passage, the rate equation is:

A \(dV/dt=\pi h dh/dt\)

B \(dV/dt=\frac{1}{3}\pi h^{2} dh/dt\)

C \(dV/dt=\pi h^{3} dh/dt\)

D \(dV/dt=\frac{1}{4}\pi h^{2} dh/dt\)

💡 Short Explanation

Differentiate the given volume formula. Apply Chain Rule. Obtain the rate equation.

✅ Correct Answer Explanation

Given \(V=\frac{1}{12}\pi h^{3}\) Differentiating with respect to time: \(\frac{dV}{dt}=\frac{1}{12}\pi (3h^{2})\frac{dh}{dt}=\frac{1}{4}\pi h^{2}\frac{dh}{dt}\) This matches Option D exactly. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect differentiation of \(h^{3}\).
Option B → Corresponds to a different volume expression.
Option C → Contains the wrong power of h after differentiation.

🎯 Strategy

Used: Substitution

Application: Differentiate the given volume formula directly.

Final Logic: \(dV/dt=(1/4)\pi h^{2}(dh/dt)\).

🧠 Memory Trick

\(h^{3}\rightarrow 3h^{2}\)

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

📄 Passage

An inverted right circular cone has r/h = 1/2, so its volume is \(V=\frac{1}{12}\pi h^{3}\). Water is poured at 5 m³/h.

When h = 4 m, the rate dh/dt is:

A 35/88 m/h

B 45/88 m/h

C 25/88 m/h

D 15/88 m/h

💡 Short Explanation

Use the rate equation from Question 14. Substitute h = 4 and dV/dt = 5. Solve for dh/dt.

✅ Correct Answer Explanation

From Question 14, \(\frac{dV}{dt}=\frac{1}{4}\pi h^{2}\frac{dh}{dt}\) Substituting \(dV/dt=5\) and \(h=4\), \(5=\frac{1}{4}\pi (16)\frac{dh}{dt}5=4\pi \frac{dh}{dt}\frac{dh}{dt}=\frac{5}{4\pi }\) This value is not among the options. Therefore the provided answer A is incorrect, and none of the options exactly matches the mathematically correct result.

❓ Why Other Options Are Incorrect

Option A → Does not satisfy the derived equation.
Option B → Incorrect numerical value.
Option C → Also inconsistent with the rate equation.

🎯 Strategy

Used: Substitution

Application: Substitute the given values into the differentiated relation.

Final Logic: \(dh/dt=5/(4\pi )\), which is not listed.

🧠 Memory Trick

Substitute After Differentiating

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

16 A man 2 m tall walks away from a 6 m lamp post at 5 km/h. The rate of increase of the length of his shadow is:

A 5/2 km/h

B 2/5 km/h

C 5/3 km/h

D 3/5 km/h

💡 Short Explanation

Use similar triangles. Relate shadow length and walking speed. Differentiate with respect to time.

✅ Correct Answer Explanation

Let x be the man's distance from the lamp and y the shadow length. Using similar triangles: \(\frac{6}{x+y}=\frac{2}{y}\) This simplifies to \(6y=2x+2y4y=2xy=\frac{x}{2}\) Differentiating, \(\frac{dy}{dt}=\frac{1}{2}\frac{dx}{dt}\) Since \(dx/dt=5\), \(dy/dt=5/2\) Thus Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Inverse relationship error.
Option C → Incorrect triangle ratio.
Option D → Not obtained from similar triangles.

🎯 Strategy

Used: Substitution

Application: Form similar-triangle equation and differentiate.

Final Logic: Shadow grows at half the walking speed.

🧠 Memory Trick

6–2 Triangle Ratio

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

17 For \(V=\frac{4}{3}\pi r^{3}\)and \(dV/dt=900\), the rate \(dr/dt\) at \(r=15\) is:

A 2/π

B 1/π

C 3/π

D 4/π

💡 Short Explanation

Differentiate sphere volume. Substitute known values. Solve for radius rate.

✅ Correct Answer Explanation

Differentiating, \(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\) Substituting \(r=15\) and \(dV/dt=900\), \(900=4\pi (225)\frac{dr}{dt}900=900\pi \frac{dr}{dt}\frac{dr}{dt}=\frac{1}{\pi }\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Twice the correct value.
Option C → Triple the correct value.
Option D → Four times the correct value.

🎯 Strategy

Used: Substitution

Application: Use the sphere volume-rate relation.

Final Logic: \(900/(900\pi )=1/\pi\).

🧠 Memory Trick

900 Cancels 900π

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

18 A 5 m ladder slides with base moving at 2 cm/s. When the base is 4 m away, the rate of change of height is:

A 4/3 cm/s

B 2/3 cm/s

C 8/3 cm/s

D 5/3 cm/s

💡 Short Explanation

Use Pythagoras theorem. Differentiate implicitly. Calculate dy/dt.

✅ Correct Answer Explanation

For the ladder, \(x^{2}+y^{2}=25\) At \(x=4\), \(y=3\). 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. Thus Option C represents the magnitude.

❓ Why Other Options Are Incorrect

Option A → Half the correct magnitude.
Option B → Too small.
Option D → Does not satisfy the relation.

🎯 Strategy

Used: Substitution

Application: Use x = 4, y = 3 in the differentiated equation.

Final Logic: \(∣dy/dt∣=8/3\).

🧠 Memory Trick

3–4–5 Triangle

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

19 For \(f(x)=x^{3}-3x^{2}+4x\), if \(f^{'}(x)>0\) for all x, then:

A The function is strictly decreasing on R

B The function has a local minimum

C The function is constant

D The function is strictly increasing on R

💡 Short Explanation

Positive derivative implies increasing function. Slope remains positive. Function rises everywhere.

✅ Correct Answer Explanation

A function with \(f^{'}(x)>0\) throughout its domain is strictly increasing. This is a standard result from applications of derivatives. Therefore Option D is correct. The function cannot be decreasing, constant, or possess a local minimum under the stated condition.

❓ Why Other Options Are Incorrect

Option A → Requires negative derivative.
Option B → A local minimum requires derivative sign change.
Option C → Constant functions have zero derivative.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret the meaning of a positive derivative.

Final Logic: Positive derivative ⇒ increasing function.

🧠 Memory Trick

Positive Slope = Increasing

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Increasing Functions

20 For \(f(x)=cos⁡x\), since \(f^{'}(x)=-sin⁡x\), in \(\left(0\ ,\ \pi \right)\)the function is:

A Neither increasing nor decreasing

B Strictly decreasing

C Strictly increasing

D Having a local maximum

💡 Short Explanation

sin x is positive in (0, π). Therefore −sin x is negative. Negative derivative means decreasing.

✅ Correct Answer Explanation

For \(0<x<\pi\), \(sin⁡x>0\) Hence, \(f^{'}(x)=-sin⁡x<0\) throughout the interval. Since the derivative remains negative, the function decreases continuously on \(\left(0\ ,\ \pi \right)\). Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → The derivative has a definite sign.
Option C → Increasing functions require positive derivative.
Option D → No local maximum exists inside the interval.

🎯 Strategy

Used: Elimination

Application: Determine the sign of the derivative on the interval.

Final Logic: Negative derivative throughout implies strict decrease.

🧠 Memory Trick

−sin x < 0 on (0, π)

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Monotonicity of Functions

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