CUET UG Mathematics Booster Test 2 - Local Maxima and Minima -

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CUET UG Mathematics Booster Test 2 - 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 2 - Local Maxima and Minima

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QUESTION 1 OF 20

Let \(f(x)=x^{3}-3x^{2}+2\). At what point does \(f(x)\)have a local maximum? (Integral)

Answer Review

1 Let \(f(x)=x^{3}-3x^{2}+2\). At what point does \(f(x)\)have a local maximum? (Integral)

A \(x=0\)

B \(x=2\)

C \(x=1\)

D No local maximum exists

💡 Short Explanation

Find critical points using \(f^{'}(x)=0\). Check sign change of derivative. Positive to negative indicates local maximum.

✅ Correct Answer Explanation

\(f(x)=x^{3}-3x^{2}+2f^{'}(x)=3x^{2}-6x=3x(x-2)\) Critical points are \(x=0\) and \(x=2\). For \(x<0\), \(f^{'}(x)>0\); for \(0<x<2\), \(f^{'}(x)<0\). Thus the derivative changes from positive to negative at \(x=0\), giving a local maximum. At \(x=2\), the change is negative to positive, giving a local minimum.

❓ Why Other Options Are Incorrect

Option B → \(x=2\) is a local minimum, not a maximum.
Option C → \(x=1\) is not a critical point.
Option D → A local maximum exists at \(x=0\).

🎯 Strategy

Used: Substitution

Application: Differentiate and analyze derivative signs around critical points.

Final Logic: \(+\rightarrow -\)sign change at \(x=0\) confirms local maximum.

🧠 Memory Trick

+\(\rightarrow\)- = Max

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

2 Consider the region bounded by the curve \(f(x)=x^{2}-4x+5\) and the x-axis. The function achieves its local minimum value at: (Graph/Region-based)

A \(x=0\)

B \(x=4\)

C \(x=1\)

D \(x=2\)

💡 Short Explanation

Differentiate the quadratic. Set derivative equal to zero. Vertex gives minimum point.

✅ Correct Answer Explanation

\(f(x)=x^{2}-4x+5f^{'}(x)=2x-4\) Setting \(f^{'}(x)=0\), \(2x-4=0\Rightarrow x=2\) Since the coefficient of \(x^{2}\)is positive, the parabola opens upward. Therefore the vertex at \(x=2\) is the local minimum point.

❓ Why Other Options Are Incorrect

Option A → Not a critical point.
Option B → Not where derivative is zero.
Option C → Does not minimize the function.

🎯 Strategy

Used: Substitution

Application: Find derivative and solve \(f^{'}(x)=0\).

Final Logic: Vertex of upward parabola occurs at \(x=2\).

🧠 Memory Trick

Vertex = Minimum for Upward Parabola

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

3 For \(f(x)=x^{2}-4x+5\), the local minimum is at \(x=2\). Which interval around \(x=2\) contains no other critical points? (MCQ)

A \(\left(1\ ,\ 3\right)\)

B \(\left(0\ ,\ 4\right)\)

C \(\left(2\ ,\ 5\right)\)

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

💡 Short Explanation

Function has one critical point. Neighborhood should contain only that point. Small interval around 2 works.

✅ Correct Answer Explanation

\(f^{'}(x)=2x-4\) The only critical point is \(x=2\). A neighborhood such as \(\left(1\ ,\ 3\right)\)contains \(x=2\) and no additional critical points. It is the standard open interval used when studying local extrema.

❓ Why Other Options Are Incorrect

Option B → Larger interval than necessary.
Option C → Does not contain points on both sides of \(2\).
Option D → Entire domain, not a local neighborhood.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Identify the smallest neighborhood around the critical point.

Final Logic: Local analysis requires an open interval around \(x=2\).

🧠 Memory Trick

Neighborhood = Small Open Interval

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

4 Assertion (A): For absolute extrema on a closed interval \(\left[a\ ,\ b\right]\), endpoints must be checked along with interior critical points.
Reason (R): Absolute extrema can occur at boundaries where derivative need not be zero.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Absolute extrema may occur at endpoints. Endpoints need not satisfy \(f^{'}(x)=0\). Both statements are correct.

✅ Correct Answer Explanation

The Closed Interval Method requires evaluating all interior critical points and both endpoints. Absolute maximum or minimum may occur at the boundaries even if the derivative is not zero there. Thus both Assertion and Reason are true, and the Reason correctly explains the Assertion.

❓ Why Other Options Are Incorrect

Option A → Both statements are true.
Option B → Reason is also true.
Option D → Assertion is true.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Apply the closed interval theorem.

Final Logic: Endpoints are always checked for absolute extrema.

🧠 Memory Trick

Closed Interval = Critical Points + Endpoints

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

5 If \(f^{'}(c)=0\), which statements are definitely true? (Multiple Correct)
1. \(c\) is guaranteed extremum
2. Tangent at \(c\) is horizontal
3. Function changes direction at \(c\)

A 1 and 2

B 2 only

C 1 and 3

D 1, 2, and 3

💡 Short Explanation

Zero derivative means horizontal tangent. Extremum is not guaranteed. Direction change may not occur.

✅ Correct Answer Explanation

If \(f^{'}(c)=0\), the tangent line is horizontal at \(c\). However, the point may be a maximum, minimum, or inflection point. Therefore neither a guaranteed extremum nor a direction change follows. Only Statement 2 is always true.

❓ Why Other Options Are Incorrect

Option A → Statement 1 is false.
Option C → Statements 1 and 3 are not guaranteed.
Option D → All three are not always true.

🎯 Strategy

Used: Elimination

Application: Identify what necessarily follows from \(f^{'}(c)=0\).

Final Logic: Horizontal tangent is the only guaranteed result.

🧠 Memory Trick

\(f^{'}=0\)⇒ Flat, Not Always Extremum

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

6 Out of 5 functions, 2 have local minima at \(x=c\) but are not differentiable there. What is the probability of selecting such a function? (Probability)

A \(1/5\)

B \(2/5\)

C \(3/5\)

D 0

💡 Short Explanation

Favorable outcomes = 2. Total outcomes = 5. Probability = favorable/total.

✅ Correct Answer Explanation

Probability is calculated as: \(P=\frac{Favourable Outcomes}{Total Outcomes}\) There are 2 functions satisfying the condition out of 5 total functions. \(P=\frac{2}{5}\) Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Uses one favorable case.
Option C → Overcounts favorable outcomes.
Option D → Probability is not zero.

🎯 Strategy

Used: Substitution

Application: Apply probability formula directly.

Final Logic: \(2/5\) satisfies the given data.

🧠 Memory Trick

Probability = Favorable ÷ Total

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Classical Probability

7 Match conditions with graph behavior: Match the Following

List I	List II
1. \(f^{'}(c)=0\) and sign changes from \(+\)to \(-\)	a. Maxima indicator
2. \(f^{'}(c)\)undefined at a continuous point	b. Sharp corner/cusp
3. \(f^{'}(c)=0\) but no sign change	c. Point of inflection

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

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

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

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

💡 Short Explanation

Sign change \(+\rightarrow -\)indicates maximum. Undefined derivative may create cusp. No sign change suggests inflection.

✅ Correct Answer Explanation

A positive-to-negative derivative sign change indicates a local maximum. A continuous point with undefined derivative often forms a cusp or corner. If the derivative is zero but does not change sign, the point is generally an inflection point. Hence Option A matches correctly.

❓ Why Other Options Are Incorrect

Option B → Incorrectly swaps maximum and cusp.
Option C → Wrongly associates inflection with cusp.
Option D → Misidentifies maximum condition.

🎯 Strategy

Used: Option Grouping

Application: Match standard derivative-test outcomes.

Final Logic: Sign-change patterns uniquely determine behavior.

🧠 Memory Trick

+\(\rightarrow\)- = Max, Undefined = Cusp

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

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. Inflection points may also be critical.

✅ Correct Answer Explanation

A critical point occurs where \(f^{'}(x)=0\) or does not exist. Such points may correspond to local maxima, local minima, or points of inflection. Therefore all three listed possibilities are valid critical-point classifications.

❓ 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 critical-point outcomes.

Final Logic: Critical points can represent maxima, minima, or inflection.

🧠 Memory Trick

Critical ≠ Always Extremum

📚 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:

A Inflection point

B Local minimum

C Absolute minimum

D Local maximum

💡 Short Explanation

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

✅ Correct Answer Explanation

A derivative sign change from positive to negative indicates the function changes from increasing to decreasing. By the First Derivative Test, such a point is a local maximum. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Requires curvature change.
Option B → Requires \(-\rightarrow +\)sign change.
Option C → Not necessarily absolute.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret derivative sign transition.

Final Logic: \(+\rightarrow -\)indicates local maximum.

🧠 Memory Trick

Rise Then Fall = Peak

📚 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 corresponds to local minimum. Second derivative is positive. Curve is concave upward.

✅ Correct Answer Explanation

A sign change from negative to positive means the function decreases then increases, producing a local minimum. According to the Second Derivative Test, if \(f^{'}(c)=0\) and \(f^{''}(c)>0\), then the point is a local minimum. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Indicates local maximum.
Option C → Contradicts critical-point condition.
Option D → Not implied by the passage.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Connect valley shape with second derivative sign.

Final Logic: Local minimum ⇒ \(f^{''}(c)>0\).

🧠 Memory Trick

Positive Second Derivative = Valley

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

11 Which of the following statements is incorrect regarding points of inflection?

A \(f^{''}(x)\)does not change sign through the point

B The tangent crosses the curve at this point

C It is neither a local maximum nor a local minimum

D The first derivative is always strictly greater than zero everywhere

💡 Short Explanation

Inflection points involve change in concavity. They need not have positive derivative everywhere. Derivative behavior varies with function.

✅ Correct Answer Explanation

A point of inflection is where concavity changes, usually indicated by a sign change in \(f^{''}(x)\). The tangent often crosses the curve and the point is generally not a local extremum. However, there is no requirement that \(f^{'}(x)\)be strictly positive everywhere. Hence Option D is incorrect.

❓ Why Other Options Are Incorrect

Option A → Intended inflection-point discussion concerns second derivative behavior.
Option B → Common geometric feature of many inflection points.
Option C → Inflection points are generally not maxima or minima.

🎯 Strategy

Used: Elimination

Application: Remove statements matching standard inflection-point properties.

Final Logic: Only Option D states an unnecessary condition.

🧠 Memory Trick

Inflection = Curvature Change, not Slope Rule

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

12 Moving average data tracks values at \(x=c\) as increasing before and after with a change in curvature. What occurs geometrically at \(c\)?

A Local Max

B Local Min

C Inflection point

D Sharp corner

💡 Short Explanation

Function increases on both sides. Curvature changes sign. No turning point occurs.

✅ Correct Answer Explanation

If the function is increasing before and after \(c\), there is no local maximum or minimum. A change in curvature while maintaining increasing behavior indicates a point of inflection. Therefore Option C correctly describes the geometry at \(c\).

❓ Why Other Options Are Incorrect

Option A → Requires increasing-to-decreasing transition.
Option B → Requires decreasing-to-increasing transition.
Option D → Curvature change does not imply a corner.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match curvature change with geometric interpretation.

Final Logic: Same trend + curvature change ⇒ inflection point.

🧠 Memory Trick

Same Trend, New Curvature = Inflection

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

13 To verify \(x=c\) is a local maximum, arrange derivative signs from left to right:
1. Zero or undefined at \(c\)
2. Negative to the right of \(c\)
3. Positive to the left of \(c\)

A 1, 2, 3

B 3, 1, 2

C 2, 1, 3

D 1, 3, 2

💡 Short Explanation

Positive derivative before point. Zero or undefined at critical point. Negative derivative after point.

✅ Correct Answer Explanation

For a local maximum, the function increases before \(c\) and decreases after \(c\). Thus: Positive derivative on the left. Zero/undefined at the critical point. Negative derivative on the right. Hence the correct order is \(3,1,2\), making Option B correct.

❓ Why Other Options Are Incorrect

Option A → Starts at critical point.
Option C → Reverses sign sequence.
Option D → Places critical point before left behavior.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Follow left-to-right sign change.

Final Logic: \(+\rightarrow 0\rightarrow -\)gives maximum.

🧠 Memory Trick

+ 0 − = Max

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

14 For \(f^{'}(x)=x-2\), the right-hand behavior (for \(x>2\)) yields a derivative that is:

A Negative

B Zero

C Undefined

D Positive

💡 Short Explanation

Choose any value greater than 2. Substitute into derivative. Result is positive.

✅ Correct Answer Explanation

For \(x>2\), \(f^{'}(x)=x-2>0\) For example, at \(x=3\), \(f^{'}(3)=1>0\) Therefore the derivative is positive on the right side of \(x=2\). Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → True only for \(x<2\).
Option B → Occurs only at \(x=2\).
Option C → Derivative exists everywhere.

🎯 Strategy

Used: Substitution

Application: Test a value greater than 2.

Final Logic: \(x>2\Rightarrow x-2>0\).

🧠 Memory Trick

Right of 2 = Positive

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

15 A cubic function has local extrema only if the discriminant of its derivative (a quadratic) satisfies:

A \(D>0\)

B \(D=0\)

C \(D<0\)

D \(D\leq 0\)

💡 Short Explanation

Derivative is quadratic. Two distinct critical points are needed. Positive discriminant guarantees them.

✅ Correct Answer Explanation

A cubic function can have both local maximum and local minimum only when its derivative quadratic has two distinct real roots. For a quadratic, this requires: \(D=b^{2}-4ac>0\) Therefore Option A is the necessary condition for local extrema.

❓ Why Other Options Are Incorrect

Option B → Gives repeated root only.
Option C → No real critical points.
Option D → Does not ensure two distinct roots.

🎯 Strategy

Used: Elimination

Application: Use quadratic discriminant rule.

Final Logic: Two real critical points require \(D>0\).

🧠 Memory Trick

Two Turns ⇒ \(D>0\)

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

16 A square piece of tin of side \(18\) cm is made into an open box by cutting equal squares of side \(x\) from each corner. The volume is \(V(x)=x(18-2x)^{2}\). Setting \(dV/dx=0\) gives critical points \(x=3\) and \(x=9\). The area of the cut square that maximizes the volume is:

A 9

B 81

C 18

D 3

💡 Short Explanation

Maximum volume occurs at \(x=3\). Area of cut square is \(x^{2}\). \(3^{2}=9\).

✅ Correct Answer Explanation

The critical points are \(x=3\) and \(x=9\). Since \(x=9\) makes the box collapse, the feasible maximum occurs at \(x=3\). The area of each cut square is: \(x^{2}=3^{2}=9\) Thus Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Corresponds to \(9^{2}\), not feasible.
Option C → Not an area value.
Option D → Side length, not area.

🎯 Strategy

Used: Substitution

Application: Use maximizing value \(x=3\).

Final Logic: Area \(=x^{2}=9\).

🧠 Memory Trick

Max at 3 ⇒ Area 9

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

17 Consider \(f(x)=sin⁡x\) on the interval \(\left[0\ ,\ 2\pi \right]\). How many local minima does the function possess in this interval?

A 0

B 1

C 2

D 3

💡 Short Explanation

Minimum occurs at \(x=\frac{3\pi }{2}\). Only one interior valley exists. Hence one local minimum.

✅ Correct Answer Explanation

For \(f(x)=sin⁡xf^{'}(x)=cos⁡x\) Critical points are \(x=\frac{\pi }{2}\)and \(x=\frac{3\pi }{2}\). The point \(x=\frac{3\pi }{2}\)is a local minimum. Therefore there is exactly one local minimum in \(\left[0\ ,\ 2\pi \right]\). Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Minimum exists.
Option C → Counts extra points incorrectly.
Option D → Too many minima.

🎯 Strategy

Used: Substitution

Application: Analyze sine graph over one cycle.

Final Logic: One valley ⇒ one local minimum.

🧠 Memory Trick

Sine: Peak at \(\pi /2\), Valley at \(3\pi /2\)

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

18 If \(f^{'}(x)=0\) for an entire interval \(I\), then throughout this interval the function:

A Is strictly increasing

B Is constant

C Has infinite extrema

D Is strictly decreasing

💡 Short Explanation

Zero derivative means no change. Function remains flat. Value stays constant.

✅ Correct Answer Explanation

A standard theorem states that if \(f^{'}(x)=0\) throughout an interval, then \(f(x)\)is constant on that interval. There is no increase or decrease. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Requires positive derivative.
Option C → Not implied by theorem.
Option D → Requires negative derivative.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Connect derivative with rate of change.

Final Logic: Zero rate ⇒ constant function.

🧠 Memory Trick

\(f^{'}=0\Rightarrow f=\)Constant

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

19 A water tank's volume is analyzed over time. The derivative of height with respect to time changes from positive to negative. The water level at that instant is:

A Minimum

B Spilling

C Emptying

D Maximum

💡 Short Explanation

Height rises before instant. Height falls afterward. Peak level is reached.

✅ Correct Answer Explanation

A derivative changing from positive to negative means the function changes from increasing to decreasing. By the First Derivative Test, this indicates a local maximum. Hence the water level is maximum at that instant.

❓ Why Other Options Are Incorrect

Option A → Requires negative-to-positive change.
Option B → Not mathematically implied.
Option C → Describes later behavior only.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret derivative sign change.

Final Logic: \(+\rightarrow -\)indicates maximum.

🧠 Memory Trick

Rise Then Fall = Maximum

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

20 In a manufacturing model, if the marginal cost curve transitions from negative to positive at \(x=17\), selling 17 units represents the:

A Max Profit

B Max Cost

C Minimum Cost

D Break-even

💡 Short Explanation

Marginal cost acts as derivative. Negative-to-positive sign change occurs. Cost reaches minimum.

✅ Correct Answer Explanation

Marginal cost is the derivative of the total cost function. A change from negative to positive means the cost function decreases before \(x=17\) and increases after \(x=17\). By the First Derivative Test, \(x=17\) corresponds to a minimum cost point.

❓ Why Other Options Are Incorrect

Option A → Profit is not under consideration.
Option B → Maximum requires positive-to-negative change.
Option D → Break-even relates to profit and revenue.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret marginal cost as derivative.

Final Logic: \(-\rightarrow +\)indicates minimum.

🧠 Memory Trick

Negative to Positive = Minimum

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

CUET UG Mathematics Booster Test 2 - Local Maxima and Minima

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