CUET UG Mathematics Booster Test 1 - Second Derivative and Advanced Tests -

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CUET UG Mathematics Booster Test 1 - Second Derivative and Advanced Tests

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 1 - Second Derivative and Advanced Tests

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QUESTION 1 OF 20

At a critical point \(c\), if \(f^{'}(c)=0\) and \(f^{''}(c)<0\), the actual local maximum value is obtained by evaluating:

Answer Review

1 At a critical point \(c\), if \(f^{'}(c)=0\) and \(f^{''}(c)<0\), the actual local maximum value is obtained by evaluating:

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

B \(f(c)\)

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

D \({lim⁡}_{x\rightarrow c}f(x)\)

💡 Short Explanation

Local maximum value means function value. Derivatives only identify nature of the point. Evaluate \(f(c)\)to obtain the maximum value.

✅ Correct Answer Explanation

The Second Derivative Test determines whether a critical point is a maximum or minimum. If \(f^{'}(c)=0\) and \(f^{''}(c)<0\), then \(c\) is a point of local maximum. The actual maximum value is the function value \(f(c)\), not the derivative values. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → \(f^{'}(c)=0\) only identifies a critical point.
Option C → \(f^{''}(c)\)indicates concavity, not the function value.
Option D → The limit equals \(f(c)\)only under continuity, but the maximum value is directly \(f(c)\).

🎯 Strategy

Used: Elimination

Application: Distinguish between function values and derivative values.

Final Logic: Maximum value means evaluating the function itself.

🧠 Memory Trick

Maximum Value = Function Value

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

2 Match the following.

List I	List II
1. (f''(c) > 0)	a. Valley (Local Minimum)
2. (f''(c) < 0)	b. Hill (Local Maximum)
3. Graph is concave upward	c. Positive curvature
4. Graph is concave downward	d. Negative curvature

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

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

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

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

💡 Short Explanation

A positive second derivative indicates upward curvature. A negative second derivative indicates downward curvature. Upward curvature gives a local minimum, while downward curvature gives a local maximum.

✅ Correct Answer Explanation

The Second Derivative Test is used to determine the nature of a stationary point after finding the critical point. If f''(c) > 0 then the graph is concave upward (curves upward like a cup). Therefore, the point represents a local minimum (valley). Hence, 1 → a and 3 → c If f''(c) < 0 then the graph is concave downward (curves downward like a cap). Therefore, the point represents a local maximum (hill). Hence, 2 → b and 4 → d Therefore, the correct matching is: 1 → a 2 → b 3 → c 4 → d Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because it reverses the conditions for maxima and minima.
Option C → Incorrect because concave upward corresponds to positive curvature, not a hill.
Option D → Incorrect because it incorrectly associates positive second derivative with a maximum.

🎯 Strategy

Used

Option Grouping
Application
Identify the sign of the second derivative first.
If f''(c) is positive, the graph opens upward and the point is a local minimum.
If f''(c) is negative, the graph opens downward and the point is a local maximum.
Final Logic
Positive second derivative → Concave upward → Valley → Local minimum.
Negative second derivative → Concave downward → Hill → Local maximum.

🧠 Memory Trick

"Downward Cap → Maximum"

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

3 When \(f^{''}(c)=0\) for a polynomial \(f(x)\), which are true?
I. It might be a point of inflection
II. The second derivative test fails
III. The value of \(x\) must be 0

A I only

B III only

C I & III

D I & II

💡 Short Explanation

Zero second derivative makes the test inconclusive. Such points may be inflection points. No requirement that \(x=0\).

✅ Correct Answer Explanation

If \(f^{''}(c)=0\), the Second Derivative Test fails because no conclusion about maxima or minima can be drawn. The point may be an inflection point. Statement III is false because \(c\) can be any number, not necessarily zero. Therefore I and II are correct.

❓ Why Other Options Are Incorrect

Option A → Omits the failure of the test.
Option B → Statement III is false.
Option C → Includes false statement III.

🎯 Strategy

Used: Elimination

Application: Check each statement individually.

Final Logic: \(f^{''}(c)=0\)⇒ test fails and inflection is possible.

🧠 Memory Trick

Second Derivative Zero = No Conclusion

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

4 Which statement is incorrect when reverting to the first derivative test if \(f^{''}(c)=0\)?

A It strictly requires \(f\) to be continuous

B It checks sign change of \(f^{'}(x)\)around \(c\)

C It indicates maxima if \(f^{'}(x)\)changes from positive to negative

D It indicates minima if \(f^{'}(x)\)changes from negative to positive

💡 Short Explanation

First Derivative Test uses sign changes. Positive-to-negative gives maxima. Negative-to-positive gives minima.

✅ Correct Answer Explanation

The First Derivative Test analyzes the sign of \(f^{'}(x)\)around a critical point. It does not strictly require continuity as stated in Option A. Options B, C, and D correctly describe the First Derivative Test. Hence Option A is the incorrect statement.

❓ Why Other Options Are Incorrect

Option B → Correct description of the test.
Option C → Positive-to-negative indicates maximum.
Option D → Negative-to-positive indicates minimum.

🎯 Strategy

Used: Elimination

Application: Compare each statement with the First Derivative Test.

Final Logic: Only Option A overstates a requirement.

🧠 Memory Trick

Sign Change Reveals Nature

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

5 For a rectangle with area \(A(x)=x(10-x)\), maximizing the area requires verifying that the second derivative is:

A Positive

B Negative

C Zero

D Undefined

💡 Short Explanation

Maximum occurs at a downward-opening parabola. \(A^{''}(x)=-2\). Negative second derivative confirms maximum.

✅ Correct Answer Explanation

\(A(x)=x(10-x)=10x-x^{2}A^{'}(x)=10-2xA^{''}(x)=-2\) Since \(A^{''}(x)\)is negative, the graph is concave downward and the critical point corresponds to a maximum area. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Would indicate a minimum.
Option C → Makes the test inconclusive.
Option D → The second derivative exists.

🎯 Strategy

Used: Substitution

Application: Differentiate and evaluate the second derivative.

Final Logic: Negative second derivative confirms maximum.

🧠 Memory Trick

Max → Downward Curve

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

6 If \(f^{'}(c)=0\) and \(f^{''}(c)>0\), the smooth curve has at \(c\):

A Maxima

B Inflection

C Minima

D Corner point

💡 Short Explanation

Positive second derivative means concave up. Graph forms a valley. Critical point becomes a minimum.

✅ Correct Answer Explanation

The Second Derivative Test states that if \(f^{'}(c)=0\) and \(f^{''}(c)>0\), the curve is concave upward at \(c\). Such points correspond to local minima because nearby values are larger than \(f(c)\). Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Requires \(f^{''}(c)<0\).
Option B → Inflection is not guaranteed.
Option D → Smooth curves have no corner.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match positive curvature with graph shape.

Final Logic: Concave up ⇒ minimum.

🧠 Memory Trick

Cup Up = Minimum

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

7 In a graph, the highest peak of a function where \(f^{'}(c)=0\) represents a:

A Absolute minimum

B Point of inflection

C Sharp corner

D Local maxima

💡 Short Explanation

Highest nearby point is a maximum. Slope becomes zero at the peak. Graph changes from rising to falling.

✅ Correct Answer Explanation

A highest peak corresponds to a local maximum because nearby function values are smaller. At a smooth peak, \(f^{'}(c)=0\) and the derivative changes from positive to negative. Therefore Option D correctly identifies the point.

❓ Why Other Options Are Incorrect

Option A → Refers to the lowest point.
Option B → Does not represent a peak.
Option C → No sharp corner is specified.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret the graph shape.

Final Logic: Peak ⇒ local maximum.

🧠 Memory Trick

Peak = Maximum

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

8 If slope data decreases, reaches zero, then increases, the turning point is:

A Local minimum

B Local maximum

C Saddle point

D Linear state

💡 Short Explanation

Function decreases before the point. Function increases after the point. Valley indicates minimum.

✅ Correct Answer Explanation

A change from decreasing to increasing means the derivative changes from negative to positive. Such behavior creates a valley-shaped turning point, which is a local minimum. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Requires positive-to-negative change.
Option C → No turning behavior.
Option D → Not a turning point.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Relate slope behavior to extrema.

Final Logic: Decrease → Increase ⇒ minimum.

🧠 Memory Trick

Down Then Up = Minimum

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

9 If \(f^{''}(x)\)is very complex, the probability that using the first derivative test is more efficient is:

A 0

B 1 (Highly Certain)

C 0.5

D 0.25

💡 Short Explanation

Complex second derivatives are difficult to analyze. First Derivative Test may be simpler. Problem assumes certainty.

✅ Correct Answer Explanation

The question is conceptual rather than statistical. If evaluating \(f^{''}(x)\)is extremely complicated, using the First Derivative Test is generally the more efficient approach. Hence the intended answer is Option B, representing complete certainty under the given assumption.

❓ Why Other Options Are Incorrect

Option A → Contradicts the stated condition.
Option C → No equal likelihood is implied.
Option D → Arbitrary probability.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Use the scenario described.

Final Logic: Complex second derivative ⇒ prefer First Derivative Test.

🧠 Memory Trick

Complex \(f^{''}\)⇒ Use \(f^{'}\)

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: First and Second Derivative Tests

10 To test a critical point \(c\) efficiently, evaluating \(f^{''}(c)\)gives:

A Nature of extremum directly

B Only slope

C No information

D Always zero

💡 Short Explanation

Second Derivative Test identifies maxima/minima. Positive means minimum. Negative means maximum.

✅ Correct Answer Explanation

When \(f^{'}(c)=0\), evaluating \(f^{''}(c)\)directly reveals the nature of the critical point. A positive value indicates a local minimum, while a negative value indicates a local maximum. Thus Option A correctly describes the usefulness of the Second Derivative Test.

❓ Why Other Options Are Incorrect

Option B → Slope comes from the first derivative.
Option C → The second derivative provides useful information.
Option D → \(f^{''}(c)\)need not be zero.

🎯 Strategy

Used: Elimination

Application: Compare the role of first and second derivatives.

Final Logic: Second derivative determines nature of extremum.

🧠 Memory Trick

\(f^{''}\)Decides Max or Min

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

11 A particle's distance from the origin is \(d(t)=∣t∣\). The minimum distance occurs at \(t=0\). Is the distance function differentiable at \(t=0\)?

A Yes, smoothly differentiable

B It is continuous and differentiable

C Yes, \(d^{'}(0)=0\)

D No, it has a sharp corner

💡 Short Explanation

\(d(t)=∣t∣\)forms a V-shaped graph. Left and right derivatives differ at \(t=0\). Function has a minimum but is not differentiable.

✅ Correct Answer Explanation

The graph of \(d(t)=∣t∣\)has a sharp corner at \(t=0\). The left derivative is \(-1\) and the right derivative is \(+1\), so the derivative does not exist there. Despite this, \(d(0)=0\) is the minimum distance. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → The function is not smoothly differentiable at \(t=0\).
Option B → It is continuous but not differentiable.
Option C → \(d^{'}(0)\)does not exist, so it cannot equal zero.

🎯 Strategy

Used: Elimination

Application: Check differentiability using left and right derivatives.

Final Logic: Unequal one-sided derivatives imply non-differentiability.

🧠 Memory Trick

V-shape ⇒ Vertex ⇒ No Derivative

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Non-Differentiable Extrema

12 Assertion (A): A function can have a minimum without being differentiable at that point.
Reason (R): \(f(x)=∣x∣\)has a local minimum at \(x=0\), but \(f^{'}(0)\)does not exist.

A Both true, R explains A

B Both true, R false

C A false, R true

D Both false

💡 Short Explanation

Extrema do not require differentiability. \(∣x∣\)gives a standard example. Reason directly supports the assertion.

✅ Correct Answer Explanation

The assertion is true because local minima can occur at points where the derivative does not exist. The example \(f(x)=∣x∣\)has a local minimum at \(x=0\), yet \(f^{'}(0)\)is undefined. Hence the reason is true and correctly explains the assertion.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Verify assertion and reason separately.

Final Logic: The example directly proves the statement.

🧠 Memory Trick

Minimum Can Exist Without Slope

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Necessary Conditions for Extrema

13 Arrange the steps used to analyze at \(x=c\):
1. \(f^{'}(c)=0\) marks a critical point
2. \(f^{''}(c)=0\) shows the second derivative test fails
3. The first derivative test determines the nature (e.g., local minima)

A 3, 2, 1

B 1, 2, 3

C 2, 1, 3

D 3, 1, 2

💡 Short Explanation

Identify critical point first. Check second derivative. Use first derivative test if necessary.

✅ Correct Answer Explanation

The standard procedure begins by locating a critical point where \(f^{'}(c)=0\). Next evaluate \(f^{''}(c)\). If \(f^{''}(c)=0\), the Second Derivative Test fails. Finally, apply the First Derivative Test to determine whether the point is a maximum, minimum, or neither.

❓ Why Other Options Are Incorrect

Option A → Begins with the final step.
Option C → Failure cannot be checked before identifying a critical point.
Option D → Incorrect logical sequence.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Follow the standard extrema-testing process.

Final Logic: Critical point → Second derivative → First derivative test.

🧠 Memory Trick

Critical → Check → Confirm

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

📄 Passage

Finding extreme values requires locating critical points where \(f^{'}(x)=0\) or \(f^{'}(x)\)is undefined. In higher-degree functions, repeated roots of \(f^{'}(x)\)can occur. If \(f^{'}(x)\)does not change sign around such a point, it is neither maxima nor minima but a point of inflection where concavity changes.

According to the passage, an inflection point indicates:

A \(f^{'}(x)\)changes from positive to negative

B The graph crosses the x-axis

C The curve's concavity changes

D The function becomes undefined

💡 Short Explanation

Inflection points involve curvature change. They are not necessarily extrema. Concavity switches sign.

✅ Correct Answer Explanation

The passage explicitly states that an inflection point is where the concavity changes. Such points may occur when the derivative does not change sign, meaning no maximum or minimum is formed. Hence Option C directly matches the definition.

❓ Why Other Options Are Incorrect

Option A → Indicates a local maximum.
Option B → Crossing the x-axis is unrelated.
Option D → Inflection points do not require discontinuity.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match the passage statement directly.

Final Logic: Inflection = change in concavity.

🧠 Memory Trick

Inflection = Curvature Flip

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

📄 Passage

Higher degree polynomials often yield such points due to:

A Negative boundaries

B Constant values

C Linear constraints

D Repeated roots

💡 Short Explanation

Repeated roots make derivative touch zero. Sign may remain unchanged. Stationary inflection can result.

✅ Correct Answer Explanation

The passage specifically identifies repeated roots of \(f^{'}(x)\)as a source of such points. When the derivative has an even-multiplicity root, its sign may not change, producing neither a maximum nor a minimum. Thus Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Boundaries are unrelated.
Option B → Constants do not create critical points.
Option C → Linear constraints are irrelevant.

🎯 Strategy

Used: Elimination

Application: Use the exact wording from the passage.

Final Logic: Repeated roots lead to stationary inflection points.

🧠 Memory Trick

Repeated Root = Repeat Sign

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

16 If \(f^{'}(x)=(x-1)^{2}\), then at \(x=1\) there is no sign change in \(f^{'}(x)\). This point is a:

A Point of inflection

B Local maxima

C Local minima

D Discontinuity

💡 Short Explanation

Derivative remains non-negative. No sign change means no turning point. Stationary inflection occurs.

✅ Correct Answer Explanation

Since \(\left(x-1)^{2}\geq 0\right.\)on both sides of \(x=1\), the function remains increasing. Therefore there is no transition between increasing and decreasing behavior. The point is not a maximum or minimum; it is treated as a stationary point of inflection.

❓ Why Other Options Are Incorrect

Option B → Requires positive-to-negative sign change.
Option C → Requires negative-to-positive sign change.
Option D → Function remains continuous and differentiable.

🎯 Strategy

Used: Option Grouping

Application: Analyze sign behavior around the critical point.

Final Logic: No sign change ⇒ no extremum.

🧠 Memory Trick

No Sign Flip = No Turning

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

17 For a graph that is concave up \(\left(f^{''}(x)>0\right)\), the tangent line at any point lies:

A Above the curve

B Below the curve

C Intersecting the curve fully

D Perpendicular to the curve

💡 Short Explanation

Concave-up graphs bend upward. Tangent line stays beneath nearby points. Characteristic property of convex curves.

✅ Correct Answer Explanation

For a concave-up graph, the curve lies above its tangent line near the point of tangency. Equivalently, the tangent line lies below the graph. This follows from positive curvature (\(f^{''}(x)>0\)). Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → True for concave-down graphs.
Option C → Not a defining property.
Option D → Tangency does not imply perpendicularity.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Visualize the shape of a concave-up curve.

Final Logic: Cup-shaped graph sits above its tangent.

🧠 Memory Trick

Cup Up, Tangent Down

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

18 A "little hill" (concave down) corresponds to:

A Absolute minima

B Inflection point

C Local maxima

D Asymptote

💡 Short Explanation

Hilltop shape indicates maximum. Concavity is downward. Nearby values are smaller.

✅ Correct Answer Explanation

A local maximum resembles a small hill on the graph. At such points the function reaches a peak relative to nearby values, and the curve is typically concave downward. Hence Option C correctly identifies the geometric interpretation.

❓ Why Other Options Are Incorrect

Option A → Minima correspond to valleys.
Option B → Inflection points are not peaks.
Option D → Asymptotes are unrelated.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match graph shape with extrema.

Final Logic: Hilltop ⇒ local maximum.

🧠 Memory Trick

Hill = High = Maximum

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

19 Find the absolute maximum value of the function \(f(x)=-x^{2}+10x\) on \(\left[0\ ,\ 10\right]\).

A 0

B 1

C -1

D 25

💡 Short Explanation

Downward-opening parabola. Vertex gives maximum value. Occurs at \(x=5\).

✅ Correct Answer Explanation

\(f(x)=-x^{2}+10x\) The vertex occurs at \(x=\frac{-b}{2a}=\frac{-10}{2(-1)}=5f(5)=-(25)+50=25\) Since the parabola opens downward, the vertex gives the absolute maximum. Thus Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Endpoint value only.
Option B → Not obtained from the function.
Option C → Incorrect evaluation.

🎯 Strategy

Used: Substitution

Application: Evaluate the vertex of the parabola.

Final Logic: Downward parabola ⇒ vertex is maximum.

🧠 Memory Trick

Vertex = Maximum for Downward Parabola

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

20 Identifying intervals where \(f^{''}(x)>0\) helps in curve sketching by showing where the function is:

A Increasing its slope (concave up)

B Decreasing its slope (concave down)

C Constant

D Undefined

💡 Short Explanation

Positive second derivative means slope increases. Graph bends upward. Indicates concave-up intervals.

✅ Correct Answer Explanation

Since \(f^{''}(x)\)measures the rate of change of slope, \(f^{''}(x)>0\) means the slope is increasing. Consequently, the graph bends upward and is concave up. This information is fundamental when sketching curves and identifying shape changes.

❓ Why Other Options Are Incorrect

Option B → Corresponds to \(f^{''}(x)<0\).
Option C → Requires zero derivative behavior.
Option D → No relation to positive second derivative.

🎯 Strategy

Used: Substitution

Application: Interpret the meaning of the second derivative.

Final Logic: Positive second derivative ⇒ increasing slope.

🧠 Memory Trick

\(f^{''}>0\)⇒ Slope Growing

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

CUET UG Mathematics Booster Test 1 - Second Derivative and Advanced Tests

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