CUET UG Mathematics Booster Test 2 - Second Derivative and Advanced Tests
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QUESTION 1 OF 20
In an optimization problem, if the second derivative of a profit function satisfies f''(c) < 0, then producing 240 items yields:
QUESTION 2 OF 20
Match the following.
| List I | List II |
|---|---|
| 1. (f''(c) < 0) | a. Hill shape (Local Maximum) |
| 2. (f''(c) > 0) | b. Valley shape (Local Minimum) |
| 3. Concave downward graph | c. Negative curvature |
| 4. Concave upward graph | d. Positive curvature |
QUESTION 3 OF 20
For f(x)=xΒ³, testing yields f''(0)=0. Which statements hold?
I. f'(x) never changes its sign
II. x=0 is a point of inflection
III. x=0 is a local maxima
QUESTION 4 OF 20
Which is an INCORRECT restriction of the first derivative test at a critical point c?
QUESTION 5 OF 20
For a twice differentiable function, if f'(c)=0 and f''(c)>0, the graph forms a:
QUESTION 6 OF 20
The area of a circle is A=ΟrΒ². Then A''(r)=2Ο>0. Hence the curve is:
QUESTION 7 OF 20
If profit changes show slope shifting from positive to negative, it indicates:
QUESTION 8 OF 20
A function has exactly two turning points (one max, one min). The probability of selecting the extremum where f''(c)>0 is:
QUESTION 9 OF 20
Let f(x)=xΒ². To confirm local minimum, evaluating f''(x) applies:
QUESTION 10 OF 20
Using the first derivative instead of complicated second derivative in distance problems is preferred for:
QUESTION 11 OF 20
Calculate the local minimum value of the non-differentiable function f(x)=|xβ3|.
QUESTION 12 OF 20
Assertion (A): x=0 is a critical point for f(x)=|x|.
Reason (R): f'(0) does not exist.
QUESTION 13 OF 20
Arrange the execution order to evaluate extrema using the 2nd derivative test:
1. Evaluate f''(c)
2. Find f'(x)
3. Determine critical point c
QUESTION 14 OF 20
QUESTION 15 OF 20
QUESTION 16 OF 20
For f(x)=xΒ³, the critical point at x=0 is a:
QUESTION 17 OF 20
If f''(x)>0 for all real x, the graph is:
QUESTION 18 OF 20
A downward parabola has f''(x)<0. The vertex corresponds to:
QUESTION 19 OF 20
An open-top box is made from a square sheet of side m by cutting squares of side x from corners. The volume is maximized when:
QUESTION 20 OF 20
For f'(x)>0 for all x, the function is:
Test Complete!
Answer Review
1 In an optimization problem, if the second derivative of a profit function satisfies f''(c) < 0, then producing 240 items yields:
f''(c) < 0 indicates concave downward behavior. Critical point corresponds to a peak. Peak point gives local maximum profit.
According to the Second Derivative Test, if f'(c)=0 and f''(c)<0, the graph is concave downward near c, producing a local maximum. Hence producing 240 items gives maximum profit. Option A is unrelated, Option B represents minima, and Option D has no connection with derivative tests.
- Option A β Zero profit is not implied by f''(c)<0.
- Option B β Minimum profit occurs when f''(c)>0.
- Option D β Infinite profit cannot be concluded from derivative information.
Used: Elimination
Application: Identify the meaning of negative second derivative and eliminate options unrelated to maxima.
Final Logic: f''(c)<0 β concave down β local maximum.
"Downward cup = Maximum up top."
2 Match the following.
| List I | List II |
|---|---|
| 1. (f''(c) < 0) | a. Hill shape (Local Maximum) |
| 2. (f''(c) > 0) | b. Valley shape (Local Minimum) |
| 3. Concave downward graph | c. Negative curvature |
| 4. Concave upward graph | d. Positive curvature |
A negative second derivative indicates a hill shape. A positive second derivative indicates a valley shape. Concavity is determined by the sign of the second derivative.
The sign of the second derivative determines the concavity of the graph and the nature of the stationary point. If f''(c) < 0 then the graph is concave downward, giving a hill-shaped curve and indicating a local maximum. Therefore, 1 β a 3 β c If f''(c) > 0 then the graph is concave upward, giving a valley-shaped curve and indicating a local minimum. Therefore, 2 β b 4 β d Hence, the correct matching is: 1 β a 2 β b 3 β c 4 β d Therefore, Option A is correct.
- Option B β Incorrect because it interchanges positive and negative curvature.
- Option C β Incorrect because it reverses the hill and valley interpretations.
- Option D β Incorrect because both the concavity and curvature relationships are reversed.
Used
- Option Grouping
- Application
- Determine the sign of the second derivative first, then identify the graph's concavity and corresponding shape.
- Final Logic
- Negative second derivative β Concave downward β Hill shape β Local maximum.
- Positive second derivative β Concave upward β Valley shape β Local minimum.
"Up β Valley β Minimum"
3 For f(x)=xΒ³, testing yields f''(0)=0. Which statements hold?
I. f'(x) never changes its sign
II. x=0 is a point of inflection
III. x=0 is a local maxima
f'(x)=3xΒ²β₯0. Derivative does not change sign. x=0 is an inflection point.
For f(x)=xΒ³, f'(x)=3xΒ², which remains non-negative and does not change sign around x=0. Also f''(x)=6x changes sign through zero, making x=0 a point of inflection. Since there is no sign change in f'(x), x=0 is neither a local maximum nor a local minimum.
- Option B β Statement III is false.
- Option C β Statement III is false though I is true.
- Option D β x=0 is not a local maximum.
Used: Substitution
Application: Compute first and second derivatives directly.
Final Logic: No sign change in f', but sign change in f'' β inflection point.
"xΒ³ bends, not peaks."
4 Which is an INCORRECT restriction of the first derivative test at a critical point c?
Critical points include non-differentiable points. First derivative sign can still be examined nearby. Hence the statement is incorrect.
The first derivative test studies sign changes of f'(x) around a critical point. A critical point may occur where the derivative does not exist, such as |x| at x=0. Therefore the test does not completely fail. Options A, C, and D correctly describe features of the first derivative test.
- Option A β Continuity is generally needed for meaningful local behavior.
- Option C β The test relies on sign changes of f'.
- Option D β It is used to identify local maxima and minima.
Used: Elimination
Application: Recall that non-differentiable critical points can still be tested.
Final Logic: "Completely fails" is an extreme and incorrect statement.
"Corners can still be tested."
5 For a twice differentiable function, if f'(c)=0 and f''(c)>0, the graph forms a:
Positive second derivative means concave up. Critical point lies at lowest nearby value. Shape resembles a valley.
If f'(c)=0 and f''(c)>0, the Second Derivative Test confirms a local minimum. A local minimum appears as a valley on the graph. Option A corresponds to maxima, while Options B and D are unrelated to the second derivative criterion.
- Option A β Peak indicates local maximum.
- Option B β No evidence of a straight line.
- Option D β Differentiability excludes discontinuity.
Used: Contextual/Tonal Matching
Application: Match positive curvature with valley-shaped geometry.
Final Logic: f''(c)>0 β local minimum β valley.
"Positive = Valley."
6 The area of a circle is A=ΟrΒ². Then A''(r)=2Ο>0. Hence the curve is:
Second derivative is positive. Positive curvature means concave upward. Graph bends upward throughout.
A(r)=ΟrΒ² gives A''(r)=2Ο, which is always positive. Positive second derivative implies upward concavity. Thus the graph is concave up. The function is neither decreasing nor constant, and concave down would require a negative second derivative.
- Option A β Requires A''(r)<0.
- Option B β Area increases with radius.
- Option C β Function changes with r.
Used: Substitution
Application: Differentiate twice and inspect the sign.
Final Logic: Positive second derivative guarantees concave-up behavior.
"A'' positive β smile shape."
7 If profit changes show slope shifting from positive to negative, it indicates:
Positive slope means increasing profit. Negative slope means decreasing profit. Change from + to β indicates maximum.
The first derivative changes from positive to negative at a local maximum. Profit increases before the point and decreases afterward. Hence the point corresponds to maximum profit. Options B, C, and D do not match the derivative sign-change pattern described.
- Option B β Minimum requires β to + sign change.
- Option C β Profit value is not specified.
- Option D β Inflection concerns concavity changes.
Used: Elimination
Application: Use derivative sign-change rules.
Final Logic: + to β transition always signals a local maximum.
"Rise then fall = Maximum."
8 A function has exactly two turning points (one max, one min). The probability of selecting the extremum where f''(c)>0 is:
One maximum and one minimum exist. Only minimum has positive second derivative. Favorable outcomes = 1 out of 2.
Among two turning points, exactly one is a minimum. Since f''(c)>0 characterizes the minimum, there is one favorable choice among two extrema. Therefore probability = 1/2. The remaining options do not match the simple probability calculation.
- Option A β One favorable outcome exists.
- Option C β Not every extremum satisfies f''(c)>0.
- Option D β Probability cannot exceed 1.
Used: Substitution
Application: Count favorable and total outcomes.
Final Logic: 1 favorable Γ· 2 total = 1/2.
"One min, one max β half chance."
9 Let f(x)=xΒ². To confirm local minimum, evaluating f''(x) applies:
f'(x)=0 locates critical point. f''(x)>0 confirms minimum. This is the second derivative test.
For f(x)=xΒ², f'(0)=0 and f''(x)=2>0. The method that uses the second derivative to classify extrema is called the Second Derivative Test. Options A, B, and D refer to different concepts and are not the direct method described.
- Option A β Uses sign changes in f'.
- Option B β Integration is unrelated.
- Option D β Mean Value Theorem serves another purpose.
Used: Contextual/Tonal Matching
Application: Match the phrase "evaluating f''(x)" with the appropriate test.
Final Logic: Using the second derivative means Second Derivative Test.
"Check f'' β Second Test."
10 Using the first derivative instead of complicated second derivative in distance problems is preferred for:
First derivative often gives simpler calculations. Sign analysis is straightforward. Reduces computational effort.
The first derivative test frequently provides a quicker way to identify increasing/decreasing behavior and extrema. Therefore it is preferred for efficiency. It does not increase complexity, create inaccuracy, or cause obfuscation. Hence Option D best reflects the practical advantage.
- Option A β Simpler methods reduce complexity.
- Option B β Accuracy is not sacrificed.
- Option C β The purpose is clarity, not confusion.
Used: Contextual/Tonal Matching
Application: Select the practical benefit most consistent with optimization methods.
Final Logic: Simpler calculations make the method more efficient.
"First derivative = Faster route."
11 Calculate the local minimum value of the non-differentiable function f(x)=|xβ3|.
Absolute value graphs are V-shaped. Vertex occurs at x=3. Minimum function value equals 0.
For f(x)=|xβ3|, the expression inside the modulus becomes zero at x=3. Since absolute values are always non-negative, the smallest possible value is 0. Therefore the local as well as absolute minimum value is 0. Options A, C, and D are not minimum values of the function.
- Option A β 3 is the x-coordinate of the minimum point, not the minimum value.
- Option C β The function can attain values smaller than 4.
- Option D β Absolute value functions never produce negative outputs.
Used: Substitution
Application: Set the modulus expression equal to zero to locate the minimum value.
Final Logic: |xβ3| β₯ 0 and equals 0 at x=3.
"Absolute value is smallest at zero."
12 Assertion (A): x=0 is a critical point for f(x)=|x|.
Reason (R): f'(0) does not exist.
Critical points include points where derivative is undefined. f'(0) does not exist for |x|. Therefore x=0 is critical.
For f(x)=|x|, the left-hand derivative is β1 and the right-hand derivative is +1 at x=0, so f'(0) does not exist. A critical point is a point where f'(x)=0 or does not exist. Hence x=0 is a critical point, and the reason correctly explains the assertion.
- Option B β The reason is true.
- Option C β The assertion is also true.
- Option D β Both statements are actually true.
Used: Contextual/Tonal Matching
Application: Apply the definition of a critical point to the given function.
Final Logic: Undefined derivative at x=0 makes it a critical point.
"Corner point = Critical point."
13 Arrange the execution order to evaluate extrema using the 2nd derivative test:
1. Evaluate f''(c)
2. Find f'(x)
3. Determine critical point c
First compute the first derivative. Next locate critical points. Finally evaluate the second derivative.
The Second Derivative Test follows a fixed sequence. First find f'(x). Then solve f'(x)=0 (or identify critical points) to obtain c. Finally evaluate f''(c) to classify the critical point as a maximum, minimum, or inconclusive. Hence the correct order is 2, 3, 1.
- Option A β Critical point cannot be determined before finding f'.
- Option B β Second derivative cannot be evaluated first.
- Option D β The logical order of the test is reversed.
Used: Contextual/Tonal Matching
Application: Recall the procedural steps of the second derivative test.
Final Logic: f' β critical point β f''.
"FirstβCriticalβSecond."
14
Absolute extrema depend on function values. Endpoints and critical points are checked. Concavity tests are optional.
The Closed Interval Method requires evaluating the function at all critical points and interval endpoints. The greatest value gives the absolute maximum and the least gives the absolute minimum. Concavity tests may help understand the graph but are not mandatory. Therefore direct evaluation is sufficient.
- Option A β Second derivative tests are not always necessary.
- Option B β Minima can also be found by direct evaluation.
- Option C β Maxima likewise do not require concavity tests.
Used: Elimination
Application: Compare the passage statement with each option.
Final Logic: Closed interval extrema come from comparing values, not necessarily concavity.
"Evaluate, then compare."
15
Closed interval method applies. Check all critical points. Compare with endpoint values.
According to the passage and NCERT's Closed Interval Method, absolute extrema on a closed interval are obtained by evaluating the function at every critical point and both endpoints. Merely checking endpoints, roots, or inflection points can miss the actual maximum or minimum value.
- Option B β Interior critical points may contain extrema.
- Option C β Roots are unrelated to extrema in general.
- Option D β Inflection points are not necessarily maxima or minima.
Used: Contextual/Tonal Matching
Application: Directly apply the method stated in the passage.
Final Logic: Compare function values at critical points and endpoints.
"CEP = Critical, Endpoints, Pick."
16 For f(x)=xΒ³, the critical point at x=0 is a:
f'(0)=0 gives a critical point. Concavity changes around x=0. No local extremum exists.
For f(x)=xΒ³, f'(x)=3xΒ² and f''(x)=6x. The second derivative changes sign at x=0, indicating a change in concavity. Thus x=0 is a point of inflection. Since the first derivative does not change sign, it is neither a local maximum nor a local minimum.
- Option A β Not an absolute maximum.
- Option C β No local minimum occurs.
- Option D β No local maximum occurs.
Used: Substitution
Application: Compute derivatives and check sign behavior.
Final Logic: Concavity changes without an extremum.
"xΒ³ crosses, not peaks."
17 If f''(x)>0 for all real x, the graph is:
Positive second derivative indicates upward curvature. Holds for every real x. Entire graph remains concave up.
The sign of the second derivative determines concavity. If f''(x)>0 everywhere, the graph bends upward throughout its domain. This does not imply decreasing behavior or a horizontal line. Therefore the graph is always concave up.
- Option A β Concavity and monotonicity are different concepts.
- Option B β Requires negative second derivative.
- Option D β Horizontal lines have zero second derivative.
Used: Contextual/Tonal Matching
Application: Relate the sign of f'' directly to graph shape.
Final Logic: Positive f'' means concave upward.
"Positive second derivative = Smile."
18 A downward parabola has f''(x)<0. The vertex corresponds to:
Downward parabola is concave down. Vertex is highest point. Highest point is a maximum.
A downward-opening parabola has negative second derivative everywhere. Its vertex represents the highest point on the curve and hence a local as well as absolute maximum. It is not a minimum, corner, or inflection point because concavity does not change.
- Option A β Minima occur in upward parabolas.
- Option B β Parabolas are smooth curves.
- Option C β No concavity change occurs at the vertex.
Used: Elimination
Application: Identify the geometric meaning of a downward parabola.
Final Logic: Highest point of a concave-down parabola is a maximum.
"Upside-down U = Maximum."
19 An open-top box is made from a square sheet of side m by cutting squares of side x from corners. The volume is maximized when:
Volume function is optimized using derivatives. Critical point gives maximum volume. Standard NCERT result is x=m/6.
For a square sheet of side m, volume is V=x(mβ2x)Β². Differentiating and solving V'(x)=0 gives critical points x=m/6 and x=m/2. Since x=m/2 yields zero volume, the maximum volume occurs at x=m/6.
- Option B β Does not satisfy the maximizing condition.
- Option C β Produces zero volume.
- Option D β Gives smaller volume than the optimum case.
Used: Substitution
Application: Form the volume function and solve V'(x)=0.
Final Logic: Optimization yields x=m/6.
"Box maximum β one-sixth cut."
20 For f'(x)>0 for all x, the function is:
Positive derivative indicates rising function. Increase occurs throughout domain. Hence function is strictly increasing.
A positive first derivative everywhere means the function increases as x increases. Therefore the function is strictly increasing on its domain. A constant function would have derivative zero, while a decreasing function would require a negative derivative. The graph need not be a downward parabola.
- Option A β Shape alone does not guarantee f'(x)>0 everywhere.
- Option C β Constant functions have f'(x)=0.
- Option D β Decreasing functions have negative derivative.
Used: Contextual/Tonal Matching
Application: Use the standard interpretation of the first derivative.
Final Logic: f'(x)>0 everywhere β strictly increasing.
"Positive slope, always up."