CUET UG Mathematics Booster Test 1 - Maxima and Minima Concepts -

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CUET UG Mathematics Booster Test 1 - Maxima and Minima Concepts

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 1 - Maxima and Minima Concepts

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QUESTION 1 OF 20

If the rate of change of a function is given by \(\frac{dy}{dx}=x-4\), at what value of \(x\) does the function reach its local maximum?

Answer Review

1 If the rate of change of a function is given by \(\frac{dy}{dx}=x-4\), at what value of \(x\) does the function reach its local maximum?

A 2

B 4

C 6

D 0

💡 Short Explanation

Local extrema occur at critical points. Set derivative equal to zero. Solve \(x-4=0\).

✅ Correct Answer Explanation

A local maximum or minimum can occur where \(\frac{dy}{dx}=0\) Given \(\frac{dy}{dx}=x-4\) Setting it equal to zero gives \(x=4\). Although further testing shows this is actually a local minimum (since derivative changes from negative to positive), the critical point occurs at \(x=4\). Hence Option B matches the intended answer.

❓ Why Other Options Are Incorrect

Option A → Derivative is not zero.
Option C → Derivative is positive.
Option D → Derivative equals −4.

🎯 Strategy

Used: Substitution

Application: Set derivative equal to zero.

Final Logic: \(x-4=0\Rightarrow x=4\).

🧠 Memory Trick

Critical Point ⇒ Derivative Zero

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Critical Points

2 The position of a moving particle is given by \(\vec{r}(t)=(t^{2}-4t, t)\). The minimum x-coordinate occurs at:

A 2

B 4

C 0

D -2

💡 Short Explanation

x-coordinate is \(t^{2}-4t\). Differentiate and set equal to zero. Minimum occurs at vertex.

✅ Correct Answer Explanation

The x-coordinate is \(x(t)=t^{2}-4t\) Differentiating: \(\frac{dx}{dt}=2t-4\) Setting equal to zero: \(2t-4=0\Rightarrow t=2\) Since the coefficient of \(t^{2}\)is positive, the parabola opens upward, giving a minimum at \(t=2\).

❓ Why Other Options Are Incorrect

Option B → Not the vertex.
Option C → Endpoint, not minimum.
Option D → Not a valid minimizing value.

🎯 Strategy

Used: Substitution

Application: Differentiate coordinate function.

Final Logic: Vertex occurs at \(t=2\).

🧠 Memory Trick

Vertex = \(-b/2a\)

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

3 A critical point in the domain of \(f(x)\)is classified as an extreme point if the graph shows:

A A vertical asymptote at the point

B A constant slope passing through the point

C A point where the function crosses the x-axis

D A local hill or valley at the point

💡 Short Explanation

Extreme points correspond to maxima or minima. Graph appears as hill or valley. Sign change occurs around the point.

✅ Correct Answer Explanation

An extreme point is a point where the function attains a local maximum or minimum. Graphically, these appear as hilltops (local maxima) or valleys (local minima). Hence Option D correctly describes an extreme point.

❓ Why Other Options Are Incorrect

Option A → Asymptotes are not extrema.
Option B → Constant slope does not imply extremum.
Option C → Crossing x-axis is unrelated.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret graph behavior.

Final Logic: Hill or valley ⇒ extremum.

🧠 Memory Trick

Hill = Max, Valley = Min

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Local Extrema

4 Arrange the steps for finding absolute extreme values in a closed interval \(\left[a\ ,\ b\right]\):
1. Identify maximum and minimum among evaluated values
2. Find all critical points in the interval
3. Evaluate function at critical points and endpoints
4. Consider endpoints a and b

A 1, 2, 3, 4

B 2, 3, 4, 1

C 2, 4, 3, 1

D 4, 2, 1, 3

💡 Short Explanation

Find critical points. Include endpoints. Compare all function values.

✅ Correct Answer Explanation

To find absolute extrema: 1. Find critical points. 2. Consider interval endpoints. 3. Evaluate the function at all these points. 4. Select the largest and smallest values. Therefore the correct sequence is 2, 4, 3, 1.

❓ Why Other Options Are Incorrect

Option A → Begins with conclusion.
Option B → Evaluates before including endpoints.
Option D → Incorrect order.

🎯 Strategy

Used: Elimination

Application: Follow standard extrema procedure.

Final Logic: Critical points → Endpoints → Evaluate → Compare.

🧠 Memory Trick

Critical → Ends → Evaluate → Decide

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Absolute Extrema

5 Which conditions are true for identifying turning points?
(I) \(f^{'}(x)\)changes sign across the point
(II) \(f^{'}(x)\)must be non-zero
(III) Function changes from increasing to decreasing or vice-versa

A I and II

B I and III

C II and III

D I, II, and III

💡 Short Explanation

Turning points require sign change. Increasing/decreasing behavior reverses. Derivative need not be non-zero.

✅ Correct Answer Explanation

A turning point occurs when the function changes from increasing to decreasing or vice versa. This corresponds to a sign change in \(f^{'}(x)\). Statement II is false because derivatives at turning points are often zero or may not exist. Thus I and III are correct.

❓ Why Other Options Are Incorrect

Option A → Includes false Statement II.
Option C → Omits sign-change criterion.
Option D → Statement II is incorrect.

🎯 Strategy

Used: Option Grouping

Application: Evaluate each statement individually.

Final Logic: Turning points require sign reversal.

🧠 Memory Trick

Turn = Sign Change

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: First Derivative Test

6 Match geometric features with conditions:

List I	List II
1. Hill top	a. \(f^{'}(x)=0, f^{''}(x)<0\)
2. Valley bottom	b. \(f^{'}(x)=0, f^{''}(x)>0\)
3. Point of inflection	c. \(f^{''}(x)=0\)(test inconclusive initially)

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

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

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

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

💡 Short Explanation

Hill top indicates maximum. Valley indicates minimum. Inflection relates to second derivative.

✅ Correct Answer Explanation

For a local maximum: \(f^{'}(x)=0,f^{''}(x)<0\) For a local minimum: \(f^{'}(x)=0,f^{''}(x)>0\) A point of inflection often satisfies \(f^{''}(x)=0\). Thus the correct matching is Option A.

❓ Why Other Options Are Incorrect

Option B → Reverses maximum and minimum.
Option C → Incorrectly matches inflection.
Option D → Incorrect assignments.

🎯 Strategy

Used: Option Grouping

Application: Apply second derivative test.

Final Logic: \(f^{''}<0\) max, \(f^{''}>0\) min.

🧠 Memory Trick

Smile Up = Min, Smile Down = Max

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

7 Identify the INCORRECT statement regarding open intervals:

A A function may have no maximum in an open interval

B A continuous function on an open interval may lack both max and min

C Critical points can exist within an open interval

D Every monotonic function attains maximum in an open interval

💡 Short Explanation

Open intervals exclude endpoints. Extrema may not exist. Monotonic functions need not attain maxima.

✅ Correct Answer Explanation

A monotonic function on an open interval may fail to attain its largest value because the endpoint is not included. For example, \(f(x)=x\) on \(\left(0\ ,\ 1\right)\)has no maximum. Hence Option D is incorrect.

❓ Why Other Options Are Incorrect

Option A → True statement.
Option B → True for many open intervals.
Option C → Critical points may exist internally.

🎯 Strategy

Used: Extreme Word Filter

Application: Check universal claim.

Final Logic: "Every" makes the statement false.

🧠 Memory Trick

Open Ends Hide Extremes

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Open Intervals and Extrema

8 Let \(f(x)=x^{2}-4x+3\) on \(\left[0\ ,\ 3\right]\). What is the absolute minimum value?

A 0

B 1

C -1

D 4

💡 Short Explanation

Find critical point. Evaluate endpoints and critical point. Select smallest value.

✅ Correct Answer Explanation

\(f^{'}(x)=2x-4\) Critical point: \(x=2\) Evaluate: \(f(0)=3,f(2)=-1,f(3)=0\) The smallest value is \(-1\). Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Endpoint value only.
Option B → Not attained.
Option D → Larger than minimum.

🎯 Strategy

Used: Substitution

Application: Compare endpoint and critical values.

Final Logic: Minimum value = \(-1\).

🧠 Memory Trick

Check Critical + Endpoints

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Absolute Minimum Value

📄 Passage

"The total profit from a grove of orange trees is mathematically modeled by , where are constants and is the number of trees per acre. Maximizing this profit involves finding the critical points where . Similarly, if a ball is thrown upwards from a building, its path forms a parabola, and its maximum height is found exactly where the rate of change of height equals zero."

For the profit function \(P(x)=ax^{2}+bx+c\), the number of trees that maximizes profit is obtained by solving:

A \(P(x)=0\)

B \(P^{'}(x)=0\)

C \(P^{''}(x)=0\)

D \(P(x)=\infty\)

💡 Short Explanation

Maximum candidates occur at critical points. Critical points satisfy derivative zero. Standard optimization method.

✅ Correct Answer Explanation

To maximize profit, first locate critical points by solving \(P^{'}(x)=0\) These points are then tested to determine whether they give a maximum. Therefore Option B is correct according to optimization principles.

❓ Why Other Options Are Incorrect

Option A → Finds roots, not maxima.
Option C → Does not locate critical points.
Option D → Not a valid condition.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Apply optimization rule.

Final Logic: Maximum candidates satisfy \(P^{'}(x)=0\).

🧠 Memory Trick

Max ⇒ First Derivative Zero

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

📄 Passage

If the height of a ball is \(h(x)=-x^{2}+60x\), the maximum height attained is:

A 60

B 900

C 960

D 120

💡 Short Explanation

Maximum occurs at vertex. Set derivative equal to zero. Substitute critical value.

✅ Correct Answer Explanation

\(h(x)=-x^{2}+60xh^{'}(x)=-2x+60\) Setting \(h^{'}(x)=0\): \(x=30\) Then \(h(30)=-(30)^{2}+60(30)=900\) Hence the maximum height is 900, making Option B correct.

❓ Why Other Options Are Incorrect

Option A → Equals coefficient, not height.
Option C → Incorrect evaluation.
Option D → Not the vertex height.

🎯 Strategy

Used: Substitution

Application: Evaluate function at critical point.

Final Logic: Vertex height = 900.

🧠 Memory Trick

Find Vertex, Then Height

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Maximum Value Problems

11 If the shortest distance from a point to a curve is \(d\), what is the area of a square constructed with this distance as its side?

A \(d^{2}\)

B \(2d^{2}\)

C \(4d\)

D \(d\)

💡 Short Explanation

Side length of square is \(d\). Area of square = side². Apply standard geometry formula.

✅ Correct Answer Explanation

The area of a square is given by \(Area=(side)^{2}\) Since the side length equals the shortest distance \(d\), \(Area=d^{2}\) Hence Option A is correct. The other expressions do not represent the area formula for a square.

❓ Why Other Options Are Incorrect

Option B → Not the area formula of a square.
Option C → Dimensionally resembles perimeter.
Option D → Represents length, not area.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Area must have square units.

Final Logic: Square area = side².

🧠 Memory Trick

Square ⇒ Side²

📚 Class IX – NCERT Mathematics – Chapter: Areas of Plane Figures – Topic: Area of Square

12 A squared distance function is \(D(x)=x^{2}-20x+1040\). The minimum occurs at \(x=10\). What is the minimum squared distance?

A 1140

B 1000

C 400

D 940

💡 Short Explanation

Minimum occurs at \(x=10\). Substitute into \(D(x)\). Evaluate carefully.

✅ Correct Answer Explanation

Given \(D(x)=x^{2}-20x+1040\) and minimum at \(x=10\), \(D(10)=100-200+1040=940\) Therefore the minimum squared distance equals 940. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect substitution.
Option B → Arithmetic error.
Option C → Not obtained from evaluation.

🎯 Strategy

Used: Substitution

Application: Evaluate the function at the given minimum point.

Final Logic: \(100-200+1040=940\).

🧠 Memory Trick

Plug Minimum Back In

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Minimum Value Problems

13 If \(f(x)=-x^{2}+6x+3\) on \(\left[0\ ,\ 6\right]\), what is the absolute maximum value?

A 7

B -20

C 12

D 0

💡 Short Explanation

Function is a downward parabola. Maximum occurs at vertex. Evaluate function there.

✅ Correct Answer Explanation

For \(f(x)=-x^{2}+6x+3f^{'}(x)=-2x+6\) Setting \(f^{'}(x)=0\), \(x=3\) Then \(f(3)=-9+18+3=12\) Thus the absolute maximum value is 12. Option C is mathematically correct.

❓ Why Other Options Are Incorrect

Option A → Not the maximum value.
Option B → Below endpoint values.
Option D → Not attained as maximum.

🎯 Strategy

Used: Substitution

Application: Find vertex and evaluate function.

Final Logic: Maximum occurs at \(x=3\).

🧠 Memory Trick

Downward Parabola ⇒ Vertex Maximum

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Absolute Maximum Value

14 For the function \(f(x)=x^{2}-6x-20\), what is the local minimum value?

A 12

B 7

C -29

D -24

💡 Short Explanation

Upward parabola has minimum at vertex. Compute vertex coordinate. Evaluate function there.

✅ Correct Answer Explanation

For \(f(x)=x^{2}-6x-20\) the vertex occurs at \(x=\frac{-(-6)}{2}=3\) Then \(f(3)=9-18-20=-29\) Therefore the local minimum value is \(-29\). Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Not the vertex value.
Option B → Incorrect evaluation.
Option D → Arithmetic mistake.

🎯 Strategy

Used: Substitution

Application: Evaluate the function at the vertex.

Final Logic: Minimum = \(f(3)=-29\).

🧠 Memory Trick

Vertex Gives Minimum

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Local Minimum

15 Assertion (A): A strictly increasing function on an open interval attains its maximum at the right endpoint.
Reason (R): Every monotonic function on a closed interval attains extrema at endpoints.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Open intervals exclude endpoints. Maximum may not exist. Closed interval theorem is true.

✅ Correct Answer Explanation

A strictly increasing function on an open interval need not attain a maximum because the right endpoint is not included. Thus the Assertion is false. The Reason is true because monotonic functions on closed intervals attain extrema at endpoints. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Reason is true.
Option B → Assertion is false.
Option C → Assertion is false.

🎯 Strategy

Used: Extreme Word Filter

Application: Check endpoint inclusion carefully.

Final Logic: Open interval prevents endpoint attainment.

🧠 Memory Trick

Open End = Missing Maximum

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

16 Theorem states that a continuous function on a closed interval:

A Attains absolute maximum and minimum

B Is strictly monotonic

C Is always differentiable

D Never attains extrema at endpoints

💡 Short Explanation

Extreme Value Theorem applies. Continuity on closed interval is sufficient. Both extrema exist.

✅ Correct Answer Explanation

The Extreme Value Theorem states that every continuous function on a closed interval \(\left[a\ ,\ b\right]\)attains both an absolute maximum and an absolute minimum. Hence Option A is correct. Monotonicity and differentiability are not required.

❓ Why Other Options Are Incorrect

Option B → Continuity does not imply monotonicity.
Option C → Continuous functions need not be differentiable.
Option D → Extrema can occur at endpoints.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Recall the theorem statement.

Final Logic: Continuous + closed interval ⇒ extrema exist.

🧠 Memory Trick

Closed + Continuous = Extrema

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Extreme Value Theorem

17 A function has a local minimum at a point where \(f^{''}(x)=0\). Which test fails here?

A First Derivative Test

B Second Derivative Test

C Continuity Test

D Absolute Extrema Test

💡 Short Explanation

Second derivative test requires nonzero value. \(f^{''}(x)=0\) is inconclusive. Other tests may still work.

✅ Correct Answer Explanation

The Second Derivative Test states: \(f^{''}(c)>0\Rightarrow minimumf^{''}(c)<0\Rightarrow maximum\) If \(f^{''}(c)=0\) the test becomes inconclusive. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → May still identify extrema.
Option C → Not related to \(f^{''}(x)\).
Option D → Still applicable in suitable cases.

🎯 Strategy

Used: Elimination

Application: Identify which test specifically depends on \(f^{''}(x)\).

Final Logic: Zero second derivative makes second derivative test fail.

🧠 Memory Trick

\(f^{''}=0\)⇒ Test Inconclusive

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

18 If \(f^{'}(x)\)does not change sign at a critical point, the point is called:

A Point of inflection

B Local maximum

C Absolute minimum

D Non-differentiable point

💡 Short Explanation

No sign change means no extremum. Curve may continue increasing or decreasing. Often indicates inflection behavior.

✅ Correct Answer Explanation

When \(f^{'}(x)\)does not change sign across a critical point, the function does not switch between increasing and decreasing. Hence the point is not a maximum or minimum. Such behavior is commonly associated with a stationary point of inflection. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Requires sign change from positive to negative.
Option C → Requires extremum behavior.
Option D → Critical point may still be differentiable.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret derivative-sign behavior.

Final Logic: No sign change ⇒ no turning point.

🧠 Memory Trick

No Sign Change = No Extremum

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Stationary Points

19 The function \(f(x)=x^{2}\)on \(\left(0\ ,\ 1\right)\)exhibits:

A Both maximum and minimum

B A maximum but no minimum

C Neither maximum nor minimum

D A minimum but no maximum

💡 Short Explanation

Domain is open. Endpoint values are excluded. No extrema are attained.

✅ Correct Answer Explanation

On the open interval \(\left(0\ ,\ 1\right)\), \(0<f(x)<1\) but neither 0 nor 1 is attained because endpoints are excluded. Therefore the function has neither a maximum nor a minimum value. Option C is correct.

❓ Why Other Options Are Incorrect

Option A → No extrema are attained.
Option B → Maximum value does not exist.
Option D → Minimum value does not exist.

🎯 Strategy

Used: Extreme Word Filter

Application: Check endpoint inclusion.

Final Logic: Open interval excludes extremal values.

🧠 Memory Trick

Open Ends, No Extremes

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Absolute Extrema

20 For which interval does \(f(x)=x^{2}\)have a defined minimum?

A \(\left(0\ ,\ 1\right)\)

B \(\left(-\infty ,\infty \right)\)

C \(\left[0\ ,\ 1\right]\)

D \(\left(1\ ,\ \infty \right)\)

💡 Short Explanation

Minimum value of \(x^{2}\)is 0. Achieved at \(x=0\). Any interval containing 0 has a defined minimum.

✅ Correct Answer Explanation

The function \(f(x)=x^{2}\) has minimum value 0 at \(x=0\). On \(\left(-\infty ,\infty \right)\), the minimum exists and equals 0. On \(\left[0\ ,\ 1\right]\), the minimum also exists and equals 0. Therefore both Options B and C are mathematically correct. The question is flawed as a single-correct MCQ.

❓ Why Other Options Are Incorrect

Option A → 0 is excluded, so no minimum exists.
Option D → Smallest value is not attained.
Option C → Correct, but not uniquely correct.

🎯 Strategy

Used: Elimination

Application: Check whether \(x=0\) belongs to the interval.

Final Logic: Minimum exists whenever 0 is included.

🧠 Memory Trick

\(x^{2}\)Minimum at 0

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Absolute Minimum Value

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