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

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CUET UG Mathematics Booster Test 1 - Local Maxima and Minima

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 1 - Local Maxima and Minima

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QUESTION 1 OF 20

Let the area of a rectangle with perimeter 20 be \(A(x)=x(10-x)\). Using the local maxima definition, the maximum area occurs when \(x\) is:

Answer Review

1 Let the area of a rectangle with perimeter 20 be \(A(x)=x(10-x)\). Using the local maxima definition, the maximum area occurs when \(x\) is:

A 2

B 10

C 5

D 4

💡 Short Explanation

Area function is quadratic. Maximum occurs at vertex. Vertex lies at \(x=5\).

✅ Correct Answer Explanation

\(A(x)=x(10-x)=10x-x^{2}\) This is a downward-opening parabola. The maximum value occurs at its vertex. \(A^{'}(x)=10-2x\) Setting \(A^{'}(x)=0\), \(10-2x=0\Rightarrow x=5\) Hence the rectangle has maximum area when \(x=5\). Options A, B, and D do not maximize the area.

❓ Why Other Options Are Incorrect

Option A → Gives area \(16\), not maximum.
Option B → Gives area \(0\).
Option D → Gives area \(24\), less than maximum \(25\).

🎯 Strategy

Used: Substitution

Application: Differentiate the area function and locate the critical point.

Final Logic: Maximum area occurs where \(A^{'}(x)=0\), giving \(x=5\).

🧠 Memory Trick

Rectangle + Fixed Perimeter ⇒ Equal Sides

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Maxima and Minima Problems

2 The probability of an event is modeled by \(P(x)=x(1-x)\). At what value of \(x\) is the probability locally maximized?

A 0.5

B 1.0

C 0.25

D 0

💡 Short Explanation

Probability function is quadratic. Vertex gives maximum probability. Critical point is \(x=0.5\).

✅ Correct Answer Explanation

\(P(x)=x-x^{2}P^{'}(x)=1-2x\) Setting \(P^{'}(x)=0\), \(1-2x=0\Rightarrow x=\frac{1}{2}\) Since the parabola opens downward, this critical point corresponds to a local maximum. Therefore the probability is maximized at \(x=0.5\).

❓ Why Other Options Are Incorrect

Option B → Probability becomes zero.
Option C → Not a critical point.
Option D → Probability equals zero.

🎯 Strategy

Used: Substitution

Application: Differentiate and solve for the stationary point.

Final Logic: \(P^{'}(x)=0\) gives \(x=0.5\).

🧠 Memory Trick

\(x(1-x)\)peaks at Half

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

3 A drone's path is given by the vector \(\vec{r}(t)=(t, 5-t)\). Its magnitude is evaluated near \(t=2.5\). What is the local minimum magnitude?

A 2

B 0

C 25

D 5

💡 Short Explanation

Magnitude squared simplifies calculation. Minimum occurs at \(t=2.5\). Magnitude equals \(\sqrt{12.5}\), not 5.

✅ Correct Answer Explanation

\(∣\vec{r}(t)∣=\sqrt{t^{2}+(5-t)^{2}}=t^{2}+(25-10t+t^{2})=2t^{2}-10t+25\) Minimum occurs at \(t=\frac{10}{4}=2.5\) Then \(∣\vec{r}∣=\sqrt{12.5}\approx 3.54\) This value is not listed. Therefore the provided answer D) 5 is incorrect. The correct magnitude is \(\sqrt{12.5}\).

❓ Why Other Options Are Incorrect

Option A → Not equal to calculated minimum.
Option B → Magnitude cannot be zero.
Option C → Represents neither magnitude nor minimum value.

🎯 Strategy

Used: Substitution

Application: Minimize the squared magnitude.

Final Logic: Actual minimum magnitude is \(\sqrt{12.5}\).

🧠 Memory Trick

Minimize Distance → Minimize Distance²

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

4 Assertion (A): Endpoints of a closed interval are not considered points of local extrema under the standard interior point definition.
Reason (R): The derivative must always equal zero at endpoints.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Local extrema require neighborhoods. Endpoints are not interior points. Derivative need not be zero at endpoints.

✅ Correct Answer Explanation

The standard definition of local extrema requires an interior point with a neighborhood on both sides. Endpoints fail this requirement. However, derivatives at endpoints need not be zero. Hence Assertion is true while Reason is false. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Assertion is true.
Option C → Reason is false.
Option D → Assertion is not false.

🎯 Strategy

Used: Elimination

Application: Check assertion and reason independently.

Final Logic: Interior-point requirement makes A true and R false.

🧠 Memory Trick

Local ⇒ Interior Point

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

5 Which statement is incorrect? (Incorrect Statement)

A If \(f\) has a local maximum at \(c\) and is differentiable, then \(f^{'}(c)=0\)

B \(f^{'}(c)=0\) does not guarantee a local extremum

C If \(f^{''}(c)>0\), then \(c\) is always a point of local minima

D A critical point can occur where \(f^{'}(c)=0\) or undefined

💡 Short Explanation

Second derivative test needs \(f^{'}(c)=0\). Positive second derivative alone is insufficient. Hence statement C is incorrect.

✅ Correct Answer Explanation

The Second Derivative Test states that if \(f^{'}(c)=0\) and \(f^{''}(c)>0\), then \(c\) is a local minimum. Option C omits the essential condition \(f^{'}(c)=0\). Therefore it is incorrect. Options A, B, and D are standard results from derivative theory.

❓ Why Other Options Are Incorrect

Option A → Fermat's theorem supports it.
Option B → Example: \(f(x)=x^{3}\).
Option D → Correct definition of critical point.

🎯 Strategy

Used: Extreme Word Filter

Application: Focus on the word "always".

Final Logic: \(f^{''}(c)>0\) alone does not guarantee a minimum.

🧠 Memory Trick

Need \(f^{'}=0\) before using \(f^{''}\)

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

6 The function \(f(x)=∣x∣\)at \(x=0\) is an example of a:

A Differentiable function, local maximum

B Differentiable function, local minimum

C Non-differentiable function, local minimum

D Non-differentiable function, local maximum

💡 Short Explanation

Graph has a sharp corner. Derivative does not exist at zero. Function attains minimum there.

✅ Correct Answer Explanation

The graph of \(y=∣x∣\)has a cusp at \(x=0\). Since left and right derivatives differ, \(f^{'}(0)\)does not exist. Also, \(∣x∣\geq 0=∣0∣\) for nearby points. Hence \(x=0\) is a local minimum. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Not differentiable and not maximum.
Option B → Not differentiable.
Option D → Value is minimum, not maximum.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Recall graph of \(∣x∣\).

Final Logic: Corner point with minimum value.

🧠 Memory Trick

\(∣x∣\): Corner + Minimum

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

7 Match the conditions with graph behavior:

List I	List II
1. \(f^{'}(c)=0\) and sign changes from \(+\)to \(-\)	a. Smooth peak (maximum)
2. \(f^{'}(c)\)undefined at \(c\)	b. Sharp corner/cusp
3. \(f^{'}(x)\)does not change sign	c. Point of inflection

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

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

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

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

💡 Short Explanation

Positive to negative gives maximum. Undefined derivative suggests cusp. No sign change indicates inflection.

✅ Correct Answer Explanation

A derivative sign change from positive to negative indicates a local maximum. Undefined derivative commonly appears at cusps or corners. If the derivative does not change sign across a critical point, the point is generally an inflection point rather than an extremum. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Reverses maximum and cusp.
Option C → Misclassifies inflection.
Option D → Incorrect matching throughout.

🎯 Strategy

Used: Option Grouping

Application: Match standard derivative-test interpretations.

Final Logic: Sign change determines graph behavior.

🧠 Memory Trick

+\(\rightarrow\)- = Peak, No Change = Inflection

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

8 Which are valid types of points at a critical point? (Multiple Correct)
1. Local maxima
2. Local minima
3. Points of inflection

A 1 and 2

B 2 and 3

C 1 and 3

D 1, 2, and 3

💡 Short Explanation

Critical points include maxima. Critical points include minima. Critical points may be inflection points.

✅ Correct Answer Explanation

A critical point occurs where the derivative is zero or undefined. Such points may correspond to local maxima, local minima, or points of inflection. Therefore all three listed possibilities are valid. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Omits inflection points.
Option B → Omits maxima.
Option C → Omits minima.

🎯 Strategy

Used: Option Grouping

Application: Recall all possible outcomes of a critical point.

Final Logic: Critical points need not always be extrema.

🧠 Memory Trick

Critical ≠ Only Max/Min

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

9 If data shows \(f^{'}(x)\)changes from positive to negative near \(x=c\), then geometrically it is a: (Moving Average/Data)

A Inflection point

B Local minimum

C Absolute minimum

D Local maximum

💡 Short Explanation

Function rises before \(c\). Function falls after \(c\). Peak forms at \(c\).

✅ Correct Answer Explanation

When the derivative changes from positive to negative, the function changes from increasing to decreasing. According to the First Derivative Test, this behavior identifies a local maximum. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Inflection requires no sign change in \(f^{'}\).
Option B → Needs negative-to-positive change.
Option C → Local information alone cannot guarantee absolute minimum.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret derivative sign changes.

Final Logic: Rise then fall implies maximum.

🧠 Memory Trick

+\(\rightarrow\)- = Hilltop

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

📄 Passage

"Let \(f\) be a function defined on an interval and let \(c\) be a point in it. Let \(f\) be twice differentiable at \(c\). If \(f^{'}(c)=0\) and \(f^{''}(c)<0\), then \(c\) is a point of local maxima. If \(f^{'}(c)=0\) and \(f^{''}(c)>0\), then \(c\) is a point of local minima. The test fails if \(f^{'}(c)=0\) and \(f^{''}(c)=0\)."

According to the Second Derivative Test passage, what corresponds to a sign change of the derivative from negative to positive (a valley)?

A \(f^{''}(c)<0\)

B \(f^{''}(c)>0\)

C \(f^{'}(c)\neq 0\)

D Undefined

💡 Short Explanation

Valley indicates local minimum. Second derivative is positive. Graph is concave upward.

✅ Correct Answer Explanation

A change from decreasing to increasing indicates a local minimum. By the Second Derivative Test, if \(f^{'}(c)=0,f^{''}(c)>0,\) then the graph is concave upward and \(c\) is a local minimum. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Indicates local maximum.
Option C → Critical point requires \(f^{'}(c)=0\).
Option D → Not implied by the passage.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Connect valley shape with concavity.

Final Logic: Valley ⇒ concave upward ⇒ \(f^{''}(c)>0\).

🧠 Memory Trick

Positive Second Derivative = Smile = Minimum

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

📄 Passage

Based on the passage, if the test fails because \(f^{''}(c)=0\), what should be done to determine the nature of the point?

A Use the first derivative test

B Conclude it is a local maximum

C Conclude it is a local minimum

D Conclude no extrema exists

💡 Short Explanation

Second derivative test becomes inconclusive. Nature of point remains unknown. First derivative test is applied next.

✅ Correct Answer Explanation

When \(f^{'}(c)=0\) and \(f^{''}(c)=0\), the Second Derivative Test fails. No conclusion about maxima, minima, or inflection can be drawn directly. The standard approach is to apply the First Derivative Test and inspect sign changes of \(f^{'}(x)\)around \(c\). Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Cannot conclude maximum when the test fails.
Option C → Cannot conclude minimum without further analysis.
Option D → Extrema may still exist.

🎯 Strategy

Used: Elimination

Application: Remove options making unjustified conclusions.

Final Logic: Failure of one test requires another test.

🧠 Memory Trick

\(f^{''}=0\)⇒ Check \(f^{'}\)

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

12 Arrange the sequential behaviors of \(f^{''}(x)\)as \(x\) moves across an inflection point \(c\), given \(f^{''}(x)>0\) on the left:
1. \(f^{''}(x)\)remains positive on the right of \(c\)
2. \(f^{''}(c)=0\)
3. \(f^{''}(x)\)approaches from the left \(\left(x\ <\ c\right)\)

A 1, 2, 3

B 3, 2, 1

C 2, 3, 1

D 1, 3, 2

💡 Short Explanation

Observe left behavior first. Reach the critical point. Then examine right behavior.

✅ Correct Answer Explanation

The logical sequence while moving across \(c\) is: 1. Observe \(f^{''}(x)\)for \(x<c\). 2. Reach \(c\) where \(f^{''}(c)=0\). 3. Examine \(f^{''}(x)\)for \(x>c\). Thus the correct order is \(3,2,1\). Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Starts from the right side.
Option C → Places center before left behavior.
Option D → Incorrect chronological order.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Follow the movement from left to right.

Final Logic: Left → Center → Right.

🧠 Memory Trick

L-C-R (Left, Center, Right)

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Inflection and Concavity

13 For the function \(f(x)=x^{3}-3x^{2}\), the left-hand behavior near the critical point \(x=2\) yields a derivative sign that is:

A Positive

B Zero

C Negative

D Undefined

💡 Short Explanation

Differentiate function. Test value slightly less than 2. Derivative becomes negative.

✅ Correct Answer Explanation

\(f^{'}(x)=3x^{2}-6x=3x(x-2)\) Choose \(x=1.5\)(left of 2): \(f^{'}(1.5)=3(1.5)(-0.5)<0\) Hence the derivative is negative on the left side of \(x=2\). Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Sign is not positive.
Option B → Only true exactly at \(x=2\).
Option D → Derivative exists.

🎯 Strategy

Used: Substitution

Application: Test a value immediately left of 2.

Final Logic: Left of 2 gives \(f^{'}(x)<0\).

🧠 Memory Trick

\(x(x-2)\): Left of 2 ⇒ Negative

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

14 On the graph of \(f(x)=x^{3}-3x^{2}\), the right-hand behavior of the derivative near the critical point \(x=2\) is:

A Negative (decreasing slope)

B Positive (increasing slope)

C Zero

D Constant

💡 Short Explanation

Test point slightly greater than 2. Derivative becomes positive. Function rises after \(x=2\).

✅ Correct Answer Explanation

\(f^{'}(x)=3x(x-2)\) Take \(x=2.5\): \(f^{'}(2.5)=3(2.5)(0.5)>0\) Thus the derivative is positive immediately to the right of \(x=2\). Therefore Option B is correct. The supplied answer A is incorrect.

❓ Why Other Options Are Incorrect

Option A → Opposite sign.
Option C → Only at \(x=2\).
Option D → Derivative varies with \(x\).

🎯 Strategy

Used: Substitution

Application: Evaluate derivative at a nearby right-hand value.

Final Logic: Right of 2 ⇒ \(f^{'}(x)>0\).

🧠 Memory Trick

After 2 ⇒ Positive

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

15 Find the local maximum value of the polynomial \(f(x)=x^{3}-3x^{2}+3x+1\).

A -2

B 0

C 1

D 2

💡 Short Explanation

Rewrite polynomial. Derivative vanishes at one point. No local maximum exists.

✅ Correct Answer Explanation

\(f(x)=x^{3}-3x^{2}+3x+1=(x-1)^{3}+2f^{'}(x)=3(x-1)^{2}\) Since \(f^{'}(x)\geq 0\) and never changes sign, the function is increasing everywhere. The critical point \(x=1\) is a stationary inflection point, not a local maximum. Hence none of the given options is correct.

❓ Why Other Options Are Incorrect

Option A → Not attained as local maximum.
Option B → Not a local maximum value.
Option D → Function has no local maximum.

🎯 Strategy

Used: Substitution

Application: Differentiate and inspect sign changes.

Final Logic: No sign change ⇒ No local maximum.

🧠 Memory Trick

No Sign Change = No Extremum

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

16 The function \(f(x)=x^{3}-6x^{2}+9x\) has critical points at \(x=1\), \(x=3\), and \(x=0\). Which of these is a point of local maximum?

A \(x=1\)

B \(x=3\)

C None

D \(x=0\)

💡 Short Explanation

Differentiate function. Analyze derivative sign. Sign changes positive to negative at 0.

✅ Correct Answer Explanation

\(f^{'}(x)=3x^{2}-12x+9=3(x-1)(x-3)\) Critical points are actually \(x=1\) and \(x=3\), not \(x=0\). At \(x=1\), sign changes from positive to negative, giving a local maximum. Thus the correct answer should be A) \(x=1\). The supplied answer is incorrect.

❓ Why Other Options Are Incorrect

Option B → Local minimum.
Option C → A local maximum exists.
Option D → Not a critical point.

🎯 Strategy

Used: Substitution

Application: Factor derivative and examine sign chart.

Final Logic: Positive → Negative at \(x=1\).

🧠 Memory Trick

Max at First Critical Point

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

17 Assertion (A): \(f(x)=x^{3}\)has no local extrema in the open interval \(\left(-\infty ,\infty \right)\).
Reason (R): \(f^{'}(x)=3x^{2}\), which does not change sign at \(x=0\).

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Derivative is non-negative. Sign does not change across zero. Hence no local extrema exist.

✅ Correct Answer Explanation

\(f^{'}(x)=3x^{2}\) is positive on both sides of \(0\) and equals zero only at \(0\). Since the derivative does not change sign, \(x=0\) is not a maximum or minimum. Thus the assertion is true, the reason is true, and the reason explains the assertion. Option C is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used: Elimination

Application: Check truth of assertion and reason separately.

Final Logic: No sign change ⇒ No extremum.

🧠 Memory Trick

\(x^{3}\): Flat but Not Extreme

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

18 If a curve is perfectly horizontal (flat) over an interval \(I\), its first derivative is:

A Positive

B Negative

C Zero

D Undefined

💡 Short Explanation

Flat graph has no slope. Rate of change is zero. Derivative vanishes throughout.

✅ Correct Answer Explanation

A horizontal graph represents a constant function. The slope of every tangent line is zero. \(f^{'}(x)=0\) for all \(x\) in the interval. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Indicates increasing function.
Option B → Indicates decreasing function.
Option D → Derivative exists and equals zero.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret geometric meaning of flatness.

Final Logic: Flat graph ⇒ zero slope.

🧠 Memory Trick

Horizontal = Zero Derivative

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

19 A continuous shift in a function's nature from increasing to decreasing guarantees the existence of a:

A Local maximum

B Local minimum

C Inflection point

D Vertical asymptote

💡 Short Explanation

Function rises before point. Function falls after point. Peak is formed.

✅ Correct Answer Explanation

When a function changes from increasing to decreasing, the derivative changes from positive to negative. By the First Derivative Test, this behavior guarantees a local maximum. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Requires decreasing-to-increasing change.
Option C → Does not require increase-to-decrease behavior.
Option D → Unrelated to monotonicity.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret sign change of derivative.

Final Logic: Increase → Decrease ⇒ Maximum.

🧠 Memory Trick

Rise Then Fall = Maximum

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

20 In an economic cost minimization model, marginal cost changes from negative to positive at \(x=17\). Selling 17 units represents:

A Maximum profit

B Minimum cost

C Maximum cost

D Break-even

💡 Short Explanation

Marginal cost acts like derivative. Negative to positive sign change. Indicates local minimum.

✅ Correct Answer Explanation

Marginal cost is the derivative of the cost function. A sign change from negative to positive means the cost function changes from decreasing to increasing. By the First Derivative Test, this identifies a local minimum. Hence selling 17 units corresponds to minimum cost.

❓ Why Other Options Are Incorrect

Option A → Profit is not being analyzed.
Option C → Maximum requires positive-to-negative change.
Option D → Break-even concerns profit and revenue equality.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Treat marginal cost as derivative of cost.

Final Logic: Negative → Positive derivative ⇒ Minimum.

🧠 Memory Trick

− to + = Minimum

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

CUET UG Mathematics Booster Test 1 - Local Maxima and Minima

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