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

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

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QUESTION 1 OF 20

If a graph shows a sharp, non-differentiable peak at \(x=c\) where the function is continuous, what can be said about \(f(c)\)?

Answer Review

1 If a graph shows a sharp, non-differentiable peak at \(x=c\) where the function is continuous, what can be said about \(f(c)\)?

A It cannot be a maximum value

B \(f^{'}(c)=0\) must hold

C It can be a local maximum if \(f(x)\leq f(c)\)in a neighborhood of \(c\)

D The second derivative test will identify it

💡 Short Explanation

Local maxima need not be differentiable. Compare nearby function values. Sharp peaks can still be maxima.

✅ Correct Answer Explanation

A local maximum means \(f(x)\leq f(c)\) for all nearby \(x\). Differentiability is not required. A sharp peak such as \(f(x)=∣x∣\)reflected downward can have a local maximum despite the derivative not existing. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Non-differentiability does not prevent maxima.
Option B → Derivative may not exist.
Option D → Second derivative test requires differentiability.

🎯 Strategy

Used: Elimination

Application: Use the definition of local maximum.

Final Logic: Maximum depends on nearby values, not differentiability.

🧠 Memory Trick

Peak ≠ Smooth Required

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

2 If marginal cost is \(MC=\frac{dC}{dx}=x^{2}-6x+9\), at what \(x\) does the rate change indicate a local minimum cost?

A 2

B 0

C 6

D 3

💡 Short Explanation

Critical points occur when derivative is zero. \(MC=(x-3)^{2}\). Minimum occurs at \(x=3\).

✅ Correct Answer Explanation

Given \(MC=x^{2}-6x+9=(x-3)^{2}\) This parabola opens upward. Its minimum value occurs at the vertex: \(x=3\) Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Not the vertex.
Option B → Gives larger value.
Option C → Gives larger value.

🎯 Strategy

Used: Substitution

Application: Rewrite as a perfect square.

Final Logic: \({\left(x-3\right)}^{2}\)is minimum at \(x=3\).

🧠 Memory Trick

Perfect Square ⇒ Vertex at Center

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

3 Steps for Second Derivative Test:
1. Check \(f^{''}(c)>0\) or \(f^{''}(c)<0\)
2. Find \(f^{'}(x)\)and set \(f^{'}(x)=0\) to get critical points
3. Conclude maxima/minima
4. Compute \(f^{''}(x)\)

A 2, 4, 1, 3

B 4, 2, 1, 3

C 2, 1, 4, 3

D 1, 2, 3, 4

💡 Short Explanation

Find critical points first. Compute second derivative. Use sign to classify.

✅ Correct Answer Explanation

The second derivative test follows: 1. Find critical points using \(f^{'}(x)=0\). 2. Compute \(f^{''}(x)\). 3. Check sign of \(f^{''}(c)\). 4. Conclude maximum or minimum. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Second derivative computed too early.
Option C → Classification before calculation.
Option D → Incorrect logical sequence.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Follow standard theorem procedure.

Final Logic: Critical point → Second derivative → Decision.

🧠 Memory Trick

First Find, Then Test

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

4 Match the Following

List I	List II
1. \(f(x)=x^{3}\)	a. Local min at \(x=0\)
2. \(f(x)=x^{2}\)	b. Neither max nor min, point of inflection at \(x=0\)
3. \(f(x)=-x^{2}\)	c. Local max at \(x=0\)

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

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

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

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

💡 Short Explanation

\(x^{3}\)has inflection at origin. \(x^{2}\)has minimum. \(-x^{2}\)has maximum.

✅ Correct Answer Explanation

For \(x^{3}\), the origin is a stationary point of inflection. For \(x^{2}\), the vertex is a local minimum. For \(-x^{2}\), the vertex is a local maximum. Therefore the matching in Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Misclassifies \(x^{3}\).
Option C → Reverses maximum and inflection.
Option D → Incorrect assignments.

🎯 Strategy

Used: Option Grouping

Application: Recall standard graph shapes.

Final Logic: Cubic-inflection, upward parabola-minimum, downward parabola-maximum.

🧠 Memory Trick

Cubic–Inflection, Up–Min, Down–Max

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

5 A curve shows a clear turning peak. The absolute maximum occurs where:

A \(f^{'}(x)>0\)

B \(f^{'}(x)<0\)

C \(f^{'}(x)=0\) and sign changes from positive to negative

D \(f^{''}(x)=0\)

💡 Short Explanation

Maximum occurs at turning peak. Derivative becomes zero. Sign changes from + to −.

✅ Correct Answer Explanation

A local or absolute maximum at a smooth turning point occurs when \(f^{'}(x)=0\) and the derivative changes from positive to negative. This shows the function rises before the point and falls after it. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Increasing region only.
Option B → Decreasing region only.
Option D → Not sufficient for maximum.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Use First Derivative Test.

Final Logic: \(+\rightarrow -\)sign change implies maximum.

🧠 Memory Trick

Rise Then Fall = Max

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

6 Assertion (A): At the top of a smooth hill, \(f^{'}(c)=0\).
Reason (R): A local maximum implies \(f^{'}(c)=0\) and concave downward \(\left(f^{''}(c)<0\right)\).

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Smooth hill indicates maximum. Derivative vanishes there. Second derivative is negative.

✅ Correct Answer Explanation

At a smooth local maximum, \(f^{'}(c)=0\) and typically \(f^{''}(c)<0.\) Thus both the Assertion and Reason are true, and the Reason correctly explains why the derivative vanishes at the hilltop.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Apply derivative tests for maxima.

Final Logic: Smooth maximum ⇒ \(f^{'}(c)=0\).

🧠 Memory Trick

Hilltop = Flat Tangent

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

7 For \(f(x)=x\) on \(\left(0\ ,\ 1\right)\):
(I) No absolute maximum
(II) No absolute minimum
(III) Has a turning point

A I and II

B I and III

C II and III

D I, II, and III

💡 Short Explanation

Open interval excludes endpoints. Function is strictly increasing. No turning point exists.

✅ Correct Answer Explanation

The function \(f(x)=x\) on \(\left(0\ ,\ 1\right)\)never attains 0 or 1 because endpoints are excluded. Hence it has neither an absolute maximum nor an absolute minimum. Being linear, it has no turning point. Therefore I and II only are correct.

❓ Why Other Options Are Incorrect

Option B → Turning point absent.
Option C → Statement I is true.
Option D → Statement III is false.

🎯 Strategy

Used: Option Grouping

Application: Check each statement individually.

Final Logic: Open interval removes extrema.

🧠 Memory Trick

Open Ends, No Extremes

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

8 For a continuous function on \(\left[a\ ,\ b\right]\):

A Absolute maximum may occur at endpoints

B Absolute minimum must occur where \(f^{'}(x)=0\)

C Function attains maximum at least once

D Extrema can occur at non-differentiable endpoints

💡 Short Explanation

Minima may occur at endpoints. Derivative need not be zero. Extreme Value Theorem applies.

✅ Correct Answer Explanation

An absolute minimum can occur at an endpoint where the derivative may not be zero or may not exist. Therefore Option B is incorrect. Continuous functions on closed intervals always attain maximum and minimum values.

❓ Why Other Options Are Incorrect

Option A → True statement.
Option C → Guaranteed by theorem.
Option D → Endpoint extrema are possible.

🎯 Strategy

Used: Extreme Word Filter

Application: Check the word "must".

Final Logic: Absolute minimum need not satisfy \(f^{'}(x)=0\).

🧠 Memory Trick

Endpoint Extremes Matter

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

9 If profit \(P(x)=-x^{2}+4x\), when is it maximized?

A 2

B 4

C -2

D 0

💡 Short Explanation

Downward parabola. Maximum at vertex. Compute critical point.

✅ Correct Answer Explanation

\(P^{'}(x)=-2x+4\) Setting \(P^{'}(x)=0\), \(x=2.\) Since the coefficient of \(x^{2}\)is negative, the parabola opens downward, so the vertex gives the maximum profit.

❓ Why Other Options Are Incorrect

Option B → Not the vertex.
Option C → Outside maximizing point.
Option D → Profit not maximum.

🎯 Strategy

Used: Substitution

Application: Find derivative and critical point.

Final Logic: Downward parabola maximized at vertex.

🧠 Memory Trick

Max Profit = Vertex

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

10 For a cylinder of maximum curved surface area inscribed in a cone, the radius ratio \(r/R=1/2\). What is the ratio \(R:r\)?

A 1:1

B 1:2

C 2:1

D 3:1

💡 Short Explanation

Given \(r/R=1/2\). Invert both sides. Obtain required ratio.

✅ Correct Answer Explanation

Given \(\frac{r}{R}=\frac{1}{2}\) Taking reciprocals, \(\frac{R}{r}=2.\) Hence \(R:r=2:1.\) Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Ratios are unequal.
Option B → Reversed ratio.
Option D → Not supported by data.

🎯 Strategy

Used: Substitution

Application: Convert the given ratio.

Final Logic: Reciprocal gives \(2:1\).

🧠 Memory Trick

Half One Way = Double Back

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

11 What is the calculated shortest distance of the point (0,3) from the upward parabola \(y=x^{2}\)?

A 2

B √2

C 1

D √5

💡 Short Explanation

Distance from \(\left(0\ ,\ 3\right)\)to \(\left(x\ ,\ x^{2}\right)\). Minimize squared distance. Minimum occurs at \(x=1\).

✅ Correct Answer Explanation

Distance squared from \(\left(0\ ,\ 3\right)\)to \(\left(x\ ,\ x^{2}\right)\): \(D^{2}=x^{2}+(x^{2}-3)^{2}\) Differentiating gives critical points \(x=0,\pm \sqrt{\frac{5}{2}}\). Evaluating distances shows the minimum occurs at \(x=\pm \sqrt{\frac{5}{2}}\), giving \(D^{2}=\frac{11}{4}\) and \(D=\frac{\sqrt{11}}{2}.\) Thus none of the options is exact. The provided answer \(\sqrt{5}\)is incorrect.

❓ Why Other Options Are Incorrect

Option A → Not obtained from minimum-distance calculation.
Option B → Too small compared with actual minimum distance.
Option C → Does not satisfy distance formula.

🎯 Strategy

Used: Substitution

Application:

Form distance-squared function and minimize.

Final Logic:

Minimum distance comes from minimizing \(D^{2}\).

🧠 Memory Trick

Distance → Minimize \(D^{2}\)

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

12 Two positive numbers whose sum is strictly 15 have the absolute minimum sum of squares when they are equal. What is the probability of randomly selecting one of these optimal numbers from the finite set {1,2,3,…,15}?

A 1/15

B 2/15

C 1/5

D 1/3

💡 Short Explanation

Equal numbers minimize sum of squares. Numbers are \(7.5\) and \(7.5\). No integer optimal number exists.

✅ Correct Answer Explanation

For two positive numbers with sum 15, \(x+y=15\) the minimum of \(x^{2}+y^{2}\) occurs at \(x=y=7.5.\) Since \(7.5\) is not in the set \(\left\{1,…,15\right\}\), the probability question is mathematically inconsistent. The listed answer \(1/15\) is incorrect.

❓ Why Other Options Are Incorrect

Option A → No optimal integer exists.
Option C → Unsupported by optimization.
Option D → No basis from the data.

🎯 Strategy

Used: Elimination

Application:

Check whether optimal value belongs to the set.

Final Logic:

Optimal number is not an integer.

🧠 Memory Trick

Equal Split ⇒ Minimum Square Sum

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

13 Find the total area of the largest mathematically possible rectangle that can be inscribed inside a circle of fixed radius R.

A \(R^{2}\)

B \(2R^{2}\)

C \(\pi R^{2}\)

D \(4R^{2}\)

💡 Short Explanation

Maximum rectangle is a square. Diagonal equals circle diameter. Area becomes \(2R^{2}\).

✅ Correct Answer Explanation

The rectangle of maximum area inscribed in a circle is a square. If the circle radius is \(R\), the square diagonal equals \(2R\). \(s\sqrt{2}=2Rs=\sqrt{2}R\) Therefore \(Area=s^{2}=2R^{2}.\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Underestimates area.
Option C → Circle area, not rectangle area.
Option D → Exceeds possible maximum.

🎯 Strategy

Used: Substitution

Application:

Use square as optimal rectangle.

Final Logic:

Maximum rectangle area = \(2R^{2}\).

🧠 Memory Trick

Largest Rectangle = Square

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

14 The absolute minimum value of the function \(f(x)=x^{2}-4x+5\) computed on the closed interval [0,4] occurs specifically at:

A x = 0

B x = 2

C x = 4

D x = 1

💡 Short Explanation

Find critical point. Compare with endpoints. Smallest value occurs at \(x=2\).

✅ Correct Answer Explanation

\(f^{'}(x)=2x-4\) Setting \(f^{'}(x)=0\), \(x=2.\) Evaluate: \(f(0)=5,f(2)=1,f(4)=5.\) The least value is attained at \(x=2\). Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Gives value 5.
Option C → Gives value 5.
Option D → Gives value 2.

🎯 Strategy

Used: Substitution

Application:

Check critical point and endpoints.

Final Logic:

Minimum occurs at \(x=2\).

🧠 Memory Trick

Vertex Gives Minimum

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

15 A given continuous function is strictly decreasing on the interval [a,b]. What represents its absolute minimum value?

A \(f(b)\)

B \(f(a)\)

C Any interior value

D Cannot be determined

💡 Short Explanation

Function decreases throughout interval. Smallest output occurs at right endpoint. Endpoint theorem applies.

✅ Correct Answer Explanation

For a strictly decreasing function, \(x_{1}<x_{2}\Rightarrow f(x_{1})>f(x_{2}).\) Therefore values continually decrease as \(x\) moves from \(a\) to \(b\). The smallest value must occur at the largest input, namely \(x=b\). Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Represents maximum value.
Option C → Interior values exceed \(f(b)\).
Option D → Determinable from monotonicity.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use decreasing-function property.

Final Logic:

Decreasing ⇒ minimum at right endpoint.

🧠 Memory Trick

Decreasing → Min at Right End

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

16 Assertion (A): By Theorem 6, if a function \(f\) is differentiable on an interval and attains an absolute maximum at an interior point \(c\), then conclusively:
Reason (R): \(f^{'}(c)=0\)

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

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

C \(f^{'}(c)\)does not exist

D \(f^{'}(c)=0\)

💡 Short Explanation

Interior extreme point is given. Function is differentiable. Fermat's theorem applies.

✅ Correct Answer Explanation

Fermat's Theorem states that if a function is differentiable at an interior point where it attains a local or absolute maximum/minimum, then \(f^{'}(c)=0.\) Since the function is differentiable and has an absolute maximum at interior point \(c\), the derivative must vanish. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Positive derivative implies increasing behavior.
Option B → Negative derivative implies decreasing behavior.
Option C → Contradicts differentiability assumption.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Apply Fermat's Theorem directly.

Final Logic: Differentiable interior extremum ⇒ \(f^{'}(c)=0\).

🧠 Memory Trick

Interior Max/Min ⇒ Derivative Zero

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Fermat's Theorem

17 For the function \(f(x)=∣x∣\), the mathematically critical point at \(x=0\) is classified strictly because:

A \(f^{'}(0)=0\)

B \(f^{'}(0)\)is undefined

C The function is not differentiable at \(x=0\)

D \(f^{''}(0)=0\)

💡 Short Explanation

Left and right derivatives differ. Function remains continuous. Differentiability fails at origin.

✅ Correct Answer Explanation

For \(f(x)=∣x∣,\) the left derivative at \(0\) is \(-1\) and the right derivative is \(+1\). Since these are unequal, \(f^{'}(0)\)does not exist. Thus the point is critical because the function is not differentiable there. Option C is the best description.

❓ Why Other Options Are Incorrect

Option A → Derivative is not zero.
Option B → True consequence, but classification is due to non-differentiability.
Option D → Second derivative is irrelevant here.

🎯 Strategy

Used: Elimination

Application: Check derivative definition at \(x=0\).

Final Logic: Unequal one-sided derivatives ⇒ non-differentiable.

🧠 Memory Trick

\(∣x∣\)Has a Corner at 0

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Non-Differentiable Points

18 On a continuous function graph, if a critical point is logically neither a maxima nor a minima, it visually resembles:

A A sharp, jagged corner

B A vertical asymptotic line

C A disconnected jump or hole

D A point of inflection where the curve flattens and continues its trend

💡 Short Explanation

Derivative may become zero. Function keeps same trend. No turning occurs.

✅ Correct Answer Explanation

A critical point that is neither a maximum nor a minimum often occurs at a stationary point of inflection. Here the tangent may be horizontal, but the function continues increasing or decreasing on both sides. Therefore Option D correctly describes the graphical behavior.

❓ Why Other Options Are Incorrect

Option A → Represents non-differentiable corner.
Option B → Not a critical point of the function.
Option C → Contradicts continuity.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Relate graph shape to derivative behavior.

Final Logic: No sign change ⇒ stationary inflection.

🧠 Memory Trick

Flat but No Turn = Inflection

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

19 Which of the following continuous, smooth functions completely lacks a maximum value anywhere in its entire domain?

A \(f(x)=x^{2}\)

B \(f(x)=-x^{2}\)

C \(f(x)=sin⁡x\)

D \(f(x)=1\)

💡 Short Explanation

Function is unbounded above. Values increase indefinitely. No largest value exists.

✅ Correct Answer Explanation

For \(f(x)=x^{2},\) as \(∣x∣\rightarrow \infty\), \(f(x)\rightarrow \infty .\) Therefore there is no largest function value. Option A is correct. The other functions possess maximum values: \(-x^{2}\)has maximum 0, \(\sin\,x\) has maximum 1, and constant function 1 has maximum 1.

❓ Why Other Options Are Incorrect

Option B → Maximum value equals 0.
Option C → Maximum value equals 1.
Option D → Constant value is both maximum and minimum.

🎯 Strategy

Used: Extreme Word Filter

Application: Check existence of a largest value.

Final Logic: Unbounded above ⇒ no maximum.

🧠 Memory Trick

\(x^{2}\)Climbs Forever

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

20 Which of the following mathematically defined functions distinctly does NOT have a minimum value?

A \(f(x)=x^{2}\)

B \(f(x)=-x^{2}\)

C \(f(x)=∣x∣\)

D \(f(x)=0\)

💡 Short Explanation

Function is unbounded below. Values decrease without limit. No smallest value exists.

✅ Correct Answer Explanation

For \(f(x)=-x^{2},\) as \(∣x∣\rightarrow \infty\), \(f(x)\rightarrow -\infty .\) Hence the function has no minimum value. Option B is correct. In contrast, \(x^{2}\), \(∣x∣\), and the constant function 0 all possess minimum values equal to 0.

❓ Why Other Options Are Incorrect

Option A → Minimum value is 0.
Option C → Minimum value is 0.
Option D → Minimum value is 0.

🎯 Strategy

Used: Elimination

Application: Check whether a lowest value exists.

Final Logic: Unbounded below ⇒ no minimum.

🧠 Memory Trick

\(-x^{2}\)Falls Forever

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

CUET UG Mathematics Booster Test 2 - Maxima and Minima Concepts

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