CUET UG Applied Mathematics Booster Test 1 - Applications of Derivatives -

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CUET UG Applied Mathematics Booster Test 1 - Applications of Derivatives

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 1 - Applications of Derivatives

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QUESTION 1 OF 20

Using the first derivative test, identify the critical points where local extrema occur for the function, f(x) = 4x³ − 24x² + 44x − 24

Answer Review

1 Using the first derivative test, identify the critical points where local extrema occur for the function, f(x) = 4x³ − 24x² + 44x − 24

A x = 2 − (√3)/3, 2 + (√3)/3

B x = 1, 2, 3

C x = −2, 2

D x = 0, 4

💡 Short Explanation

Differentiate the function. Set f′(x) = 0. Solve the quadratic equation to obtain the critical points where local extrema may occur.

✅ Correct Answer Explanation

Differentiate the function: f′(x) = 12x² − 48x + 44 Setting f′(x) = 0, 12x² − 48x + 44 = 0 Divide by 4: 3x² − 12x + 11 = 0 Using the quadratic formula, x = [12 ± √(144 − 132)] / 6 x = (12 ± √12) / 6 x = 2 ± (√3)/3 Thus, the critical points are x = 2 − (√3)/3 and x = 2 + (√3)/3. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because the derivative is a quadratic equation and cannot have three distinct critical points.
Option C → Incorrect because neither x = −2 nor x = 2 satisfies f′(x) = 0.
Option D → Incorrect because substituting x = 0 and x = 4 into f′(x) does not give zero.

🎯 Strategy

Used: Substitution

Application: Differentiate the function, solve f′(x) = 0, and compare the obtained critical points with the given options.

Final Logic: The derivative has two real roots: x = 2 − (√3)/3 and x = 2 + (√3)/3.

🧠 Memory Trick

Differentiate → Set Equal to Zero → Solve Quadratic

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Local Maxima and Local Minima (First Derivative Test).

2 Match the evaluated x-coordinates to their respective local extremum behaviours for the function
\(f(x)=\frac{x^{3}}{3}-\frac{x^{2}}{2}-6x\). (Given \(f^{'}(x)=x^{2}-x-6\))

List I	List II
1. x = 3	a. Negative to Positive
2. x = -2	b. Positive to Negative
3. Sign change of \(f^{'}(x)\)at x = 3	c. Point of local minimum
4. Sign change of \(f^{'}(x)\)at x = -2	d. Point of local maximum

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

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

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

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

💡 Short Explanation

Factorize \(f^{'}(x)=x^{2}-x-6=(x-3)(x+2)\). Critical points are \(x=-2\) and \(x=3\). Apply the first derivative test to determine local maximum and local minimum.

✅ Correct Answer Explanation

The derivative is \(f^{'}(x)=x^{2}-x-6=(x-3)(x+2).\) Hence, the critical points are \(x=-2\) and \(x=3\). Testing the sign of \(f^{'}(x)\): \(x<-2: f^{'}(x)>0\) \(-2<x<3: f^{'}(x)<0\) \(x>3: f^{'}(x)>0\) Therefore, at \(x=-2\), the derivative changes from Positive to Negative, indicating a local maximum. At \(x=3\), it changes from Negative to Positive, indicating a local minimum. Thus, the correct matching is 1-c, 2-d, 3-a, 4-b , making Option B correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect because it assigns x = 3 to Negative→Positive instead of local minimum and x = -2 to Positive→Negative instead of local maximum. The coordinate and behaviour are mismatched.
Option C → Incorrect because it reverses the nature of both extrema. It treats x = 3 as a local maximum and x = -2 as a local minimum, contradicting the first derivative test.
Option D → Incorrect because both the sign changes and extremum classifications are incorrectly paired. None of the matches agree with the derivative sign analysis.

🎯 Strategy

Used: Option Grouping

Application: First determine the derivative's sign in each interval. Then group the corresponding extremum with its sign change and compare the complete set with the given options.

Final Logic: Positive→Negative indicates a local maximum, Negative→Positive indicates a local minimum. Only Option B correctly groups all four matches.

🧠 Memory Trick

"PN = Peak, NP = Pit."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: First Derivative Test for Local Maxima and Minima

3 For the function \(f(x)=x^{3}-1.5x^{2}-18x+1\) evaluated strictly on the closed interval \(\left[-4,6\right]\), which of the following statements are correct?
1) The critical points in the open interval are \(x=-2\) and \(x=3\).
2) The absolute maximum value occurs at \(x=6\) and evaluates to 55.
3) The absolute minimum value occurs at \(x=3\) and evaluates to \(-39.5\).
4) The absolute minimum value occurs at \(x=-4\).

A 1, 2, 3

B 1, 2

C 2, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

Differentiate to find critical points. Evaluate the function at critical points and interval endpoints. Compare all values to identify the absolute maximum and minimum on the closed interval.

✅ Correct Answer Explanation

Differentiate: \(f^{'}(x)=3x^{2}-3x-18=3(x+2)(x-3).\) Hence, the critical points are \(x=-2\) and \(x=3\). Evaluate: \(f(-4)=-7\) \(f(-2)=23\) \(f(3)=-39.5\) \(f(6)=55\) Therefore, the absolute maximum is 55 at \(x=6\), while the absolute minimum is −39.5 at \(x=3\). Thus, statements 1, 2, and 3 are true, whereas 4 is false. Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B (1, 2) → Incorrect because it omits statement 3, which is true. The value \(f(3)=-39.5\) is the smallest among all evaluated points.
Option C (2, 3, 4) → Incorrect because statement 1 is true, while statement 4 is false. The minimum does not occur at \(x=-4\).
Option D (1, 2, 3, 4) → Incorrect because statement 4 is false. Although \(f(-4)=-7\), this is greater than \(-39.5\), so it is not the absolute minimum.

🎯 Strategy

Used: Substitution

Application: Find the critical points using the derivative, substitute both the critical points and interval endpoints into the function, then compare the resulting values.

Final Logic: The largest function value is 55 at \(x=6\), and the smallest is −39.5 at \(x=3\), making 1, 2, and 3 the correct statements.

🧠 Memory Trick

"Closed Interval = Endpoints + Critical Points = Compare All."

📚 Bottom of Form

4 Identify the incorrect statement regarding the absolute extrema of the quadratic function
\(f(x)=9x^{2}+12x+2\).

A The function can be rewritten as \(f(x)=(3x+2)^{2}-2\).

B The absolute minimum value of the function is \(-2\).

C The function possesses an absolute maximum value at infinity.

D Since \(\left(3x+2)^{2}\geq 0\right.\), \(f(x)\geq -2\) for all real \(x\).

💡 Short Explanation

Complete the square. The parabola opens upward since the coefficient of \(x^{2}\)is positive. Therefore, the function has an absolute minimum but no absolute maximum.

✅ Correct Answer Explanation

Completing the square, \(9x^{2}+12x+2=(3x+2)^{2}-2.\) Since \(\left(3x+2)^{2}\geq 0\right.\), the minimum value is \(-2\), attained at \(x=-\frac{2}{3}\). Thus, A, B, and D are correct. An upward-opening parabola increases without bound, so it has no absolute maximum. Therefore, Option C is the incorrect statement.

❓ Why Other Options Are Incorrect

Option A → Correct because completing the square gives \(\left(3x+2)^{2}-2\right.\), which is algebraically equivalent to the original quadratic function.
Option B → Correct because \({\left(3x+2\right)}^{2}\)is minimum when it equals zero, giving the least function value as \(-2\).
Option D → Correct because \({\left(3x+2\right)}^{2}\)is always non-negative, implying \(f(x)\geq -2\) for every real value of \(x\).

🎯 Strategy

Used: Substitution

Application: Rewrite the quadratic by completing the square. This immediately reveals the vertex, minimum value, and whether the graph has an absolute maximum or minimum.

Final Logic: Since the parabola opens upward, it has a minimum of −2 but no absolute maximum, making Option C the incorrect statement.

🧠 Memory Trick

"Positive \(x^{2}\)→ Opens Up → Minimum Only."

📚 Class XI – NCERT Mathematics – Chapter: Quadratic Equations – Topic: Completing the Square and Nature of a Quadratic Function

5 Find the critical points of the function
\(f(x)=\frac{x^{4}}{4}-2x^{3}+\frac{11x^{2}}{2}-6x\)
by examining its critical point behaviour.

A 1, 2, 3

B 0, 1, 2

C -1, 1, 2

D 2, 3, 4

💡 Short Explanation

Differentiate the function. Set the first derivative equal to zero. Factor the derivative to obtain all critical points where the function's slope becomes zero.

✅ Correct Answer Explanation

Differentiate the function: \(f^{'}(x)=x^{3}-6x^{2}+11x-6.\) Factoring, \(f^{'}(x)=(x-1)(x-2)(x-3).\) Hence, the critical points are \(x=1, 2, 3\). These are the only points where \(f^{'}(x)=0\), making them the function's critical points. Therefore, Option A is correct. Options B, C, and D contain values that do not satisfy the derivative equation.

❓ Why Other Options Are Incorrect

Option B → Incorrect because \(x=0\) is not a critical point. Substituting into \(f^{'}(x)\)gives \(-6\), not zero, while omitting the valid critical point \(x=3\).
Option C → Incorrect because \(x=-1\) is not a root of the derivative. Although \(x=1\) and \(x=2\) are correct, the complete set must also include \(x=3\).
Option D → Incorrect because \(x=4\) is not a critical point, and the valid critical point \(x=1\) is missing from the list.

🎯 Strategy

Used: Substitution

Application: Differentiate the function, solve \(f^{'}(x)=0\), and compare the resulting critical points with the given options to identify the correct set.

Final Logic: Since \(f^{'}(x)=(x-1)(x-2)(x-3)\), the only critical points are 1, 2, and 3, so Option A is correct.

🧠 Memory Trick

"Derivative Zero → Critical Hero."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Critical Points, Local Maxima and Minima.

6 For the cubic function h(x) = x³, the derivative h'(0) = 0. However, 0 is neither a point of local maximum nor local minimum because the curve maintains an upward smooth turning trend. This specific critical point constraint is known as a:

A Point of absolute maximum

B Point of inflexion

C Vertical asymptote

D Cusp

💡 Short Explanation

\(h^{'}(0)=0\) gives a critical point. Concavity changes at \(x=0\). Hence, \(x=0\) is a stationary point of inflexion, not a maximum or minimum.

✅ Correct Answer Explanation

For \(h(x)=x^{3}\), \(h^{'}(x)=3x^{2},h^{''}(x)=6x.\) At \(x=0\), \(h^{'}(0)=0\), but the second derivative changes sign from negative to positive across zero. Therefore, the curve changes concavity without attaining a local maximum or minimum. Such a stationary critical point is called a point of inflexion. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect because an absolute maximum is the greatest function value on a domain. The function \(x^{3}\)is unbounded above and has no absolute maximum.
Option C → Incorrect because a vertical asymptote is a line approached by a curve where the function becomes unbounded. The graph of \(x^{3}\)is continuous and has no asymptotes.
Option D → Incorrect because a cusp occurs where the curve has a sharp point and is generally not differentiable. Since \(h^{'}(0)\)exists, \(x=0\) is not a cusp.

🎯 Strategy

Used: Elimination

Application: Eliminate options inconsistent with the properties of \(x^{3}\). A continuous, differentiable cubic cannot have a cusp or vertical asymptote, leaving the point of inflexion.

Final Logic: A zero derivative without a local extremum indicates a stationary point of inflexion, so Option B is correct.

🧠 Memory Trick

"Zero slope + Concavity change = Inflexion."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Stationary Point of Inflexion

7 Apply the second derivative test to \(f(x)=x^{3}-3x^{2}+3x+5\). At the critical point \(x=1\), \(f^{''}(1)=0\). This strictly implies:

A \(x=1\) is definitively a local minimum.

B \(x=1\) is definitively a local maximum.

C The second derivative test fails, requiring the first derivative test.

D The function is not differentiable at \(x=1\).

💡 Short Explanation

The second derivative equals zero. The second derivative test becomes inconclusive. Apply the first derivative test or higher derivative test to classify the critical point.

✅ Correct Answer Explanation

For \(f(x)=x^{3}-3x^{2}+3x+5,f^{'}(x)=3(x-1)^{2},f^{''}(x)=6(x-1).\) At \(x=1\), both derivatives satisfy \(f^{'}(1)=0\) and \(f^{''}(1)=0\). Since the second derivative is zero, the second derivative test cannot determine whether the point is a maximum or minimum. Therefore, the first derivative test (or higher derivative test) must be used. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect because \(f^{''}(1)=0\) does not prove a local minimum. The second derivative test is inconclusive when the second derivative is zero.
Option B → Incorrect because there is no evidence of a local maximum. The second derivative test cannot classify the critical point in this case.
Option D → Incorrect because the polynomial is differentiable for every real number. The existence of \(f^{''}(1)=0\) does not imply non-differentiability.

🎯 Strategy

Used: Elimination

Application: Recall the conditions of the second derivative test. When \(f^{''}(c)=0\), eliminate conclusions about maxima and minima because the test is inconclusive.

Final Logic: Since \(f^{''}(1)=0\), the second derivative test fails, making Option C the correct choice.

🧠 Memory Trick

"\(f^{''}=0\) → Don't Decide, Test Again!"

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Second Derivative Test

8 In objective function analysis, an open tank with a square base (side 'x') and depth 'y' holds exactly 4000 cubic cm of liquid. To minimize the surface area \(S=x^{2}+4xy\), substituting the volume constraint creates the single-variable function \(S(x)\)equal to:

A \(x^{2}+\frac{4000}{x}\)

B \(x^{2}+\frac{16000}{x}\)

C \(x^{2}+\frac{16000}{x^{2}}\)

D \(2x+\frac{16000}{x}\)

💡 Short Explanation

Use the volume constraint \(x^{2}y=4000\). Express \(y\) in terms of \(x\). Substitute into the surface area equation to obtain a single-variable function.

✅ Correct Answer Explanation

The volume of the open tank is \(x^{2}y=4000.\) Hence, \(y=\frac{4000}{x^{2}}.\) Substituting into \(S=x^{2}+4xy,\) gives \(S=x^{2}+4x\left(\frac{4000}{x^{2}}\right)=x^{2}+\frac{16000}{x}.\) Thus, the optimization problem is reduced to one variable. Therefore, Option B is correct. The remaining options arise from incorrect substitution or simplification.

❓ Why Other Options Are Incorrect

Option A → Incorrect because the factor of 4 in the lateral surface area \(4xy\) is ignored, giving \(\frac{4000}{x}\)instead of \(\frac{16000}{x}\).
Option C → Incorrect because after substitution, one power of \(x\) cancels. The denominator should be \(x\), not \(x^{2}\).
Option D → Incorrect because the first term should remain \(x^{2}\), representing the square base area, not \(2x\).

🎯 Strategy

Used: Substitution

Application: Use the given constraint to eliminate one variable. Substituting \(y=\frac{4000}{x^{2}}\)into the objective function simplifies the optimization problem.

Final Logic: Correct substitution directly yields

\(S=x^{2}+\frac{16000}{x},\)
so Option B is correct.

🧠 Memory Trick

"Constraint first, substitute next."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Optimization Problems (Maximum and Minimum Values)

9 A manufacturer's profit function is evaluated as
\(P(x)=-9x^{2}+522x-2500.\)
Finding the optimal production output requiring \(P^{'}(x)=0\) yields \(x\) equal to:

A 25

B 29

C 30

D 18

💡 Short Explanation

Differentiate the profit function. Set the derivative equal to zero. Solve for the production level where profit is maximized.

✅ Correct Answer Explanation

Differentiate: \(P^{'}(x)=-18x+522.\) Setting \(P^{'}(x)=0\) gives \(-18x+522=0\Rightarrow x=\frac{522}{18}=29.\) Also, \(P^{''}(x)=-18<0,\) confirming a maximum profit at \(x=29\). Hence, Option B is the correct answer. The remaining options do not satisfy \(P^{'}(x)=0\).

❓ Why Other Options Are Incorrect

Option A → Incorrect because substituting \(x=25\) into \(P^{'}(x)\)gives \(72\), not zero. Therefore, profit is still increasing.
Option C → Incorrect because \(P^{'}(30)=-18\), indicating the maximum has already been crossed.
Option D → Incorrect because \(P^{'}(18)=198\), so the profit function is increasing and has not yet reached its optimum.

🎯 Strategy

Used: Substitution

Application: Differentiate the profit function, equate the derivative to zero, and solve for the stationary point. Then verify with the second derivative if required.

Final Logic: Since \(P^{'}(29)=0\) and \(P^{''}<0\), profit is maximized at 29, making Option B correct.

🧠 Memory Trick

"Profit Peak ⇒ \(P^{'}(x)=0\)."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Optimization in Economics (Maximum Profit)

10 For the cost function
\(C(x)=300x-10x^{2}+\frac{1}{3}x^{3},\)
determining the output at which the Marginal Cost (MC) is minimum (utilizing the marginal equality principle mathematically) yields \(x\) equal to:

A 20

B 10

C 300

D 5

💡 Short Explanation

Marginal Cost is the derivative of the cost function. Differentiate MC again. Set the second derivative of the cost function equal to zero to obtain the minimum MC.

✅ Correct Answer Explanation

The cost function is \(C(x)=300x-10x^{2}+\frac{x^{3}}{3}.\) Marginal Cost: \(MC=C^{'}(x)=300-20x+x^{2}.\) To minimize MC, \(\frac{d(MC)}{dx}=2x-20=0,\) giving \(x=10.\) Since \(\frac{d^{2}(MC)}{dx^{2}}=2>0,\) the MC is minimum at \(x=10\). Hence, Option B is correct. The remaining options do not satisfy the minimization condition.

❓ Why Other Options Are Incorrect

Option A → Incorrect because at \(x=20\), \(\frac{d(MC)}{dx}=20\neq 0\). Thus, MC is increasing and is not minimum.
Option C → Incorrect because \(x=300\) is unrelated to the stationary condition. It does not satisfy \(2x-20=0\).
Option D → Incorrect because at \(x=5\), \(\frac{d(MC)}{dx}=-10\), indicating MC is still decreasing and has not reached its minimum.

🎯 Strategy

Used: Substitution

Application: Differentiate the cost function to obtain MC, differentiate again, and solve for the stationary point of the marginal cost.

Final Logic: Since \(MC^{'}(10)=0\) and \(MC^{''}>0\), the minimum marginal cost occurs at \(x=10\). Therefore, Option B is correct.

🧠 Memory Trick

"Cost → MC → Differentiate Again."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Marginal Cost and Optimization

11 A printed page must contain 180 sq cm of text with margins. The total area function is
\(A(x)=xy+5x+4y+20.\)
By substituting
\(y=\frac{180}{x},\)
solving \(A^{'}(x)=0\) for the most economical dimension 'x' yields:

A 10

B 12

C 15

D 18

💡 Short Explanation

Use the text-area constraint \(xy=180\). Express \(y\) in terms of \(x\). Differentiate the resulting single-variable function and solve for the stationary point.

✅ Correct Answer Explanation

Since \(xy=180,y=\frac{180}{x},\) the area function becomes \(A(x)=180+5x+\frac{720}{x}+20=200+5x+\frac{720}{x}.\) Differentiate: \(A^{'}(x)=5-\frac{720}{x^{2}}.\) Setting \(A^{'}(x)=0\), \(5=\frac{720}{x^{2}}\Rightarrow x^{2}=144\Rightarrow x=12\) (positive value only). Thus, the optimum page width is 12 cm, making Option B correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect because substituting \(x=10\) into \(A^{'}(x)\)does not give zero. Hence, it is not the optimum dimension.
Option C → Incorrect because \(x=15\) does not satisfy the stationary condition obtained from differentiation.
Option D → Incorrect because \(x=18\) also fails to satisfy \(A^{'}(x)=0\); therefore, it cannot minimize the total page area.

🎯 Strategy

Used: Substitution

Application: Replace one variable using the given constraint, differentiate the resulting expression, and solve the stationary equation to determine the optimal dimension.

Final Logic: Substituting \(y=\frac{180}{x}\)and solving \(A^{'}(x)=0\) gives \(x=12\), so Option B is correct.

🧠 Memory Trick

"Constraint → One Variable → Differentiate → Optimize."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Optimization Problems (Minimum Area)

12 For the minimum distance/wire problem where a 40 m wire is cut to form a square and circle, the total area
\(A(x)=\frac{x^{2}}{16}+\frac{{\left(40-x\right)}^{2}}{4\pi }.\)
Solving \(A^{'}(x)=0\) yields the length of the square piece 'x' as:

A \(\frac{160}{\pi +4}\)

B \(\frac{40}{\pi +4}\)

C \(\frac{160}{\pi }\)

D \(\frac{40\pi }{\pi +4}\)

💡 Short Explanation

Differentiate the total area function. Equate the derivative to zero. Solve the resulting equation to obtain the optimal wire length for the square.

✅ Correct Answer Explanation

Differentiate \(A(x)=\frac{x^{2}}{16}+\frac{{\left(40-x\right)}^{2}}{4\pi }.\) Then, \(A^{'}(x)=\frac{x}{8}-\frac{40-x}{2\pi }.\) Setting \(A^{'}(x)=0\), \(\frac{x}{8}=\frac{40-x}{2\pi }.\) Cross-multiplying, \(2\pi x=320-8x,x(\pi +4)=160.\) Hence, \(x=\frac{160}{\pi +4}.\) Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because it misses the factor of 4 obtained during cross-multiplication and algebraic simplification.
Option C → Incorrect because it ignores the contribution of the square term and omits the \(\left(\pi \ +\ 4\right)\)denominator.
Option D → Incorrect because the numerator should be 160, not 40π. It results from incorrect rearrangement of the derivative equation.

🎯 Strategy

Used: Substitution

Application: Differentiate the given objective function, set the derivative equal to zero, and solve the resulting linear equation carefully.

Final Logic: Correct differentiation and algebra give

\(x=\frac{160}{\pi +4},\)
so Option A is correct.

🧠 Memory Trick

"Differentiate → Equal Zero → Simplify Carefully."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Optimization Problems

📄 Passage

A farmer has a piece of land and observes that planting up to 25 trees yields 600 fruits per tree. For each additional tree planted in excess of 25, the yield per tree decreases by 15 fruits. Let 'x' be the number of trees planted in excess of 25. The total number of trees is (25 + x), and the yield per tree becomes (600 - 15x). The total production function of the entire garden is the product of the number of trees and the yield per tree.

Based on the passage, the total production function \(P(x)\)representing the behaviour on closed intervals for 'x' excess trees is exactly:

A \(P(x)=(25+x)(600+15x)\)

B \(P(x)=(25-x)(600-15x)\)

C \(P(x)=(25+x)(600-15x)\)

D \(P(x)=(25+x)(15x-600)\)

💡 Short Explanation

Total production equals number of trees × fruits per tree. Additional trees increase tree count. Each extra tree reduces the yield per tree by 15 fruits.

✅ Correct Answer Explanation

From the passage: Number of trees = \(25+x\). Yield per tree = \(600-15x\). Hence, the total production is \(P(x)=(25+x)(600-15x).\) The first factor represents the increased number of trees, while the second reflects the reduced yield per tree. Their product gives the total production. Therefore, Option C correctly models the situation.

❓ Why Other Options Are Incorrect

Option A → Incorrect because it assumes the yield per tree increases by 15 fruits instead of decreasing.
Option B → Incorrect because the total number of trees should increase to \(25+x\), not decrease to \(25-x\).
Option D → Incorrect because the yield expression has the wrong sign, producing negative values even when no extra trees are planted.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Translate each statement in the passage directly into an algebraic expression before multiplying them to form the production function.

Final Logic: Production = Number of Trees × Yield per Tree, giving

\((25+x)(600-15x),\)
which is Option C.

🧠 Memory Trick

"More Trees × Less Yield = Total Production."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Formulation of Optimization Functions

📄 Passage

To optimize the production via endpoint evaluation and critical point analysis, the marginal production \(P^{'}(x)\)equates to 225 − 30x. Setting this to zero yields the critical point \(x\) as:

A 7

B 8

C 7.5

D 8.5

💡 Short Explanation

Differentiate the production function. Set the derivative equal to zero. Solve the resulting linear equation to obtain the critical point for maximum production.

✅ Correct Answer Explanation

From the production function, \(P(x)=(25+x)(600-15x),\) the derivative is \(P^{'}(x)=225-30x.\) Setting \(225-30x=0,\) gives \(30x=225x=\frac{225}{30}=7.5.\) Thus, production is optimized when 7.5 additional trees are considered in the mathematical model. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect because substituting \(x=7\) gives \(P^{'}(7)=15\), not zero. Therefore, it is not the critical point.
Option B → Incorrect because \(P^{'}(8)=-15\), indicating the derivative has already changed sign after the optimum.
Option D → Incorrect because \(P^{'}(8.5)=-30\), which is not zero and therefore cannot represent the stationary point.

🎯 Strategy

Used: Substitution

Application: Differentiate the function, equate the derivative to zero, and solve the resulting equation to determine the production level where optimization occurs.

Final Logic: Since

\(225-30x=0\)
gives
\(x=7.5,\)
Option C is the correct answer.

🧠 Memory Trick

"Derivative Zero → Optimum Hero."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Optimization Problems

15 A cylindrical vessel of radius 0.5 m is being filled with oil at a volume rate
\(\frac{dV}{dt}=0.25\pi m^{3}/min.\)
Using real-life change models (where \(V=\pi r^{2}h\)), the rate at which the oil surface is rising \(\left(\frac{dh}{dt}\right)\)is:

A 0.5 m/min

B 1 m/min

C 0.25 m/min

D 2 m/min

💡 Short Explanation

Differentiate the cylinder volume formula with respect to time. Radius remains constant. Substitute the given values to determine the rate of increase of height.

✅ Correct Answer Explanation

For a cylinder, \(V=\pi r^{2}h.\) Differentiating with respect to time, \(\frac{dV}{dt}=\pi r^{2}\frac{dh}{dt}.\) Given, \(r=0.5 m,\frac{dV}{dt}=0.25\pi .\) Hence, \(0.25\pi =\pi (0.5)^{2}\frac{dh}{dt}=\pi (0.25)\frac{dh}{dt}.\) Therefore, \(\frac{dh}{dt}=1 m/min.\) Thus, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect because substituting \(dh/dt=0.5\) does not satisfy the related-rates equation.
Option C → Incorrect because it underestimates the height increase by failing to divide correctly by the cylinder's base area.
Option D → Incorrect because it doubles the required value and does not satisfy the differentiated volume equation.

🎯 Strategy

Used: Substitution

Application: Differentiate the geometric formula, substitute the known radius and volume rate, and isolate the unknown rate of change.

Final Logic: Using

\(\frac{dV}{dt}=\pi r^{2}\frac{dh}{dt},\)
the calculation gives
\(\frac{dh}{dt}=1 m/min,\)
making Option B correct.

🧠 Memory Trick

"Volume Rate ÷ Base Area = Height Rate."

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

16 A manufacturer charges ₹6000 per pen drive for an order of 50 or less, reduced by ₹75 for each drive in excess of 50. The Revenue
\(R(x)=x\left(6000-75(x-50)\right).\)
Maximizing this yields the optimum order size 'x' as:

A 50

B 60

C 65

D 75

💡 Short Explanation

Expand the revenue function. Differentiate with respect to \(x\). Set the derivative equal to zero to obtain the order size giving maximum revenue.

✅ Correct Answer Explanation

Expand the function: \(R(x)=x(9750-75x)=9750x-75x^{2}.\) Differentiate: \(R^{'}(x)=9750-150x.\) Setting \(R^{'}(x)=0,\) gives \(150x=9750,x=65.\) Also, \(R^{''}(x)=-150<0,\) confirming maximum revenue. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect because \(R^{'}(50)=2250\), indicating revenue is still increasing at 50 units.
Option B → Incorrect because \(R^{'}(60)=750\), so the maximum has not yet been reached.
Option D → Incorrect because \(R^{'}(75)=-1500\), indicating revenue is already decreasing beyond the optimum.

🎯 Strategy

Used: Substitution

Application: Expand the revenue equation, differentiate it, and solve the stationary condition. Use the second derivative to verify the maximum.

Final Logic: Since \(R^{'}(65)=0\) and \(R^{''}<0\), maximum revenue occurs at 65 pen drives, making Option C correct

🧠 Memory Trick

"Revenue Peak ⇒ \(R^{'}(x)=0, R^{''}<0\)."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Maximum Revenue Problems

17 Using tangent-based interpretation, what is the exact algebraic equation of the tangent to the curve
\(x^{2}+3y-3=0\)
that is strictly parallel to the line
\(y=4x-5?\)

A \(4x-y+13=0\)

B \(4x+y-13=0\)

C \(x-4y+13=0\)

D \(x+4y-13=0\)

💡 Short Explanation

Differentiate the curve implicitly. Equate the tangent slope to 4. Determine the point of tangency and form the tangent equation.

✅ Correct Answer Explanation

Given \(x^{2}+3y-3=0,\) implicit differentiation gives \(2x+3\frac{dy}{dx}=0,\) so \(\frac{dy}{dx}=-\frac{2x}{3}.\) Since the tangent is parallel to \(y=4x-5,\) its slope is 4. Thus, \(-\frac{2x}{3}=4\) gives \(x=-6.\) Substituting into the curve, \(36+3y-3=03y=-33,y=-11.\) The tangent through \(\left(-6,-11\right)\)with slope 4 is \(y+11=4(x+6),\) or \(4x-y+13=0.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because its slope is −4, not 4, so it is not parallel to the given line.
Option C → Incorrect because rewriting gives slope 1/4, which does not match the required tangent slope.
Option D → Incorrect because rewriting gives slope −1/4, so it cannot represent the required tangent.

🎯 Strategy

Used: Substitution

Application: Find the derivative using implicit differentiation, equate it to the given slope, determine the point, and apply the point-slope form.

Final Logic: The tangent equation simplifies to

\(4x-y+13=0,\)

🧠 Memory Trick

"Parallel Lines ⇒ Equal Slopes."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Tangents and Normals

18 In gradient applications, evaluating the normal gradient to the curve
\(y=x^{3}-x\)
at the specific point (2, 6) gives a slope of:

A 11

B -11

C -1/11

D 1/11

💡 Short Explanation

Differentiate the curve to obtain the tangent slope. Evaluate it at the given point. Take the negative reciprocal to obtain the normal slope. 2Correct Answer Explanation Differentiate the curve: \(\frac{dy}{dx}=3x^{2}-1.\) At \(x=2\), \(\frac{dy}{dx}=3(2)^{2}-1=12-1=11.\) The slope of the normal is the negative reciprocal of the tangent slope: \(m_{n}=-\frac{1}{11}.\) Hence, the normal gradient is \(-\frac{1}{11}\). Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect because 11 is the slope of the tangent, not the normal.
Option B → Incorrect because −11 is neither the tangent slope nor its negative reciprocal.
Option D → Incorrect because the normal slope must be the negative reciprocal of the tangent slope, not the positive reciprocal.

🎯 Strategy

Used: Substitution

Application: Differentiate the curve, substitute the given \(x\)-coordinate, and compute the negative reciprocal to determine the normal slope.

Final Logic: Since the tangent slope is 11, the normal slope is

\(-\frac{1}{11},\)
making Option C correct.

🧠 Memory Trick

"Normal = Negative Reciprocal of Tangent."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Tangents and Normals

19 In derivative-based modelling, a toy firm assesses variable cost as \(x(x+30)\)and fixed cost as 1500. The Marginal Cost (MC) equation evaluated directly from
\(C(x)=x^{2}+30x+1500\)
is:

A \(2x\)

B \(2x+30\)

C \(x+30\)

D \(30x\)

💡 Short Explanation

Marginal cost is the derivative of the total cost function. Differentiate each term separately. The constant fixed cost contributes zero to the derivative.

✅ Correct Answer Explanation

The total cost function is \(C(x)=x^{2}+30x+1500.\) Differentiating, \(MC=C^{'}(x)=2x+30+0.\) The derivative of the constant 1500 is zero because fixed cost does not change with production. Hence, the marginal cost equation is \(2x+30.\) Therefore, Option B is correct. 3Why Other Options Are Incorrect Option A → Incorrect because it omits the derivative of the linear term \(30x\), which contributes 30. Option C → Incorrect because the derivative of \(x^{2}\)is 2x, not x. Option D → Incorrect because \(30x\) is not the derivative of the given cost function and ignores the quadratic term.

🎯 Strategy

Used: Substitution

Application: Differentiate each term of the total cost function separately using standard derivative rules.

Final Logic: The derivative of \(x^{2}+30x+1500\) is

\(2x+30,\)
so Option B is the correct answer.

🧠 Memory Trick

"Marginal Cost = Derivative of Total Cost."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Marginal Cost and Marginal Revenue

20 In applied mathematical problems, a boy 1 m tall walks towards a 5 m lamp post at −0.5 m/s. Using similar triangles to link shadow length 'y' and distance 'x',
\(\frac{y}{x+y}=\frac{1}{5},\)
the rate at which his shadow decreases \(\left(\frac{dy}{dt}\right)\)is exactly:

A \(-\frac{1}{8}\)m/s

B \(-\frac{1}{4}\)m/s

C \(\frac{1}{8}\)m/s

D \(-\frac{1}{2}\)m/s

💡 Short Explanation

Apply similar triangles to relate shadow length and distance. Differentiate the relation with respect to time. Substitute the walking speed to determine the shadow's rate of change.

✅ Correct Answer Explanation

From similar triangles, \(\frac{y}{x+y}=\frac{1}{5}.\) Cross-multiplying, \(5y=x+y4y=xy=\frac{x}{4}.\) Differentiating with respect to time, \(\frac{dy}{dt}=\frac{1}{4}\frac{dx}{dt}.\) Since the boy walks towards the lamp, \(\frac{dx}{dt}=-0.5 m/s.\) Therefore, \(\frac{dy}{dt}=\frac{1}{4}(-0.5)=-\frac{1}{8} m/s.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because it assumes the shadow changes at half the walking speed rather than one-fourth, contradicting the similar triangles relation.
Option C → Incorrect because the shadow is decreasing as the boy approaches the lamp, so the rate must be negative.
Option D → Incorrect because it incorrectly assumes the shadow shrinks at the same rate as the boy's movement, ignoring the proportional relationship.

🎯 Strategy

Used: Substitution

Application: Use similar triangles to derive an algebraic relation between \(x\) and \(y\), differentiate implicitly, and substitute the given walking speed.

Final Logic: Since

\(y=\frac{x}{4},\)
the shadow changes at one-fourth the boy's speed. With
\(\frac{dx}{dt}=-0.5,\)
we obtain
\(\frac{dy}{dt}=-\frac{1}{8} m/s,\)
making Option A correct.

🧠 Memory Trick

"Similar Triangles → Relate → Differentiate → Substitute."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Related Rates of Change (Similar Triangles)

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