CUET UG Applied Mathematics Booster Test 2 - Monotonic Functions and Critical Points -

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CUET UG Applied Mathematics Booster Test 2 - Monotonic Functions and Critical Points

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 2 - Monotonic Functions and Critical Points

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QUESTION 1 OF 20

The rigorous analytical condition x1 < x2 implying f(x1) < f(x2) for all x1, x2 in (a,b) is an absolute mathematical requirement for a function to be defined strictly as:

Answer Review

1 The rigorous analytical condition x1 < x2 implying f(x1) < f(x2) for all x1, x2 in (a,b) is an absolute mathematical requirement for a function to be defined strictly as:

A Decreasing on (a,b)

B Increasing on (a,b)

C Concave on (a,b)

D Discontinuous on (a,b)

💡 Short Explanation

Order preserved in function values x increases → f(x) increases Definition of increasing function

✅ Correct Answer Explanation

The condition x1 < x2 ⇒ f(x1) < f(x2) defines a strictly increasing function on an interval. It ensures monotonic rise across (a,b).

❓ Why Other Options Are Incorrect

Option A → would require reversed inequality
Option C → relates to second derivative
Option D → unrelated concept

🎯 Strategy

Used: Definition matching

Application: Aligns the function's exact behavior with the strict increasing definition, ensuring absolute mathematical correctness and eliminating unrelated concepts.

Final Logic: direct correspondence with increasing definition

🧠 Memory Trick

"Same order → increasing"

📚 Class XI – NCERT Mathematics – Chapter: Functions – Topic: Increasing Functions

2 Match the functions or variables in List I with their corresponding monotonic or critical behaviors in List II:

List I	List II
1. (A) f(x) = x²	a. Increasing on (0, ∞)
2. f(x) = -1/x³	b. Increasing on its domain
3. f(x) = x⁴/4 – 2x³ + 11x²/2 – 6x	c. Increasing on (1,2) and (3,∞)
4. Marginal Cost when C(x) = 2x + 2(x+3)/(x+2)	d. Falls continuously for all x > 0

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

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

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

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

💡 Short Explanation

Standard monotonic splits Derivative-based classification Economic marginal cost behavior

✅ Correct Answer Explanation

1 (x²) increasing on (0,∞) → a 2 (−1/x³) increasing on domain → b 3 polynomial increases in (1,2) and (3,∞) → c 4 marginal cost decreases for x>0 → d

❓ Why Other Options Are Incorrect

Option A → swaps 2 and 1 behaviour
Option C → incorrect interval mapping
Option D → mismatched economic interpretation

🎯 Strategy

Used: Option elimination

Application: Quickly removes mismatched interval behaviors and incorrect economic interpretations, cleanly leaving only the perfectly mapped correct choice.

Final Logic: match known monotonic patterns

🧠 Memory Trick

"x² splits, rational stays rising"

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

3 Which of the following strict analytical rules properly and independently identify a decreasing function y = f(x) on an interval (a,b)?
1) x1 < x2 implies f(x1) > f(x2).
2) x1 > x2 implies f(x1) < f(x2).
3) The graph falls from left to right.
4) Tangents typically make obtuse angles of inclination.

A 1, 2

B 1, 2, 3

C 2, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

All statements describe decreasing behavior Includes inequality, graph, angle form Fully consistent definitions

✅ Correct Answer Explanation

All four statements correctly represent decreasing functions: reversed inequality, falling graph, and obtuse tangent angles all align with negative slope behavior.

❓ Why Other Options Are Incorrect

Option A → incomplete
Option B → missing geometric condition
Option C → excludes correct condition A

🎯 Strategy

Used: Complete validation

Application: Verifies all geometrical and algebraic conditions simultaneously, preventing the selection of partially correct but mathematically incomplete options.

Final Logic: all conditions consistent

🧠 Memory Trick

"Decrease = all forms agree"

📚 Class XI – NCERT Mathematics – Chapter: Functions – Topic: Decreasing Functions

4 Identify the incorrect statement regarding the derivative test for decreasing functions.

A f'(x) < 0 is both a necessary and sufficient condition for a function to be decreasing.

B f'(x) < 0 is a sufficient condition for a function f to be decreasing in (a, b).

C f'(x) < 0 is not a necessary condition for a function to be decreasing.

D The function falls from left to right when decreasing.

💡 Short Explanation

Only sufficient condition Not necessary in all cases Sign condition limits

✅ Correct Answer Explanation

f'(x) < 0 is sufficient but not necessary for decrease in all cases (non-differentiable functions may still decrease).

❓ Why Other Options Are Incorrect

Option B → correct
Option C → correct
Option D → correct

🎯 Strategy

Used: Extreme condition check

Application: Highlights absolute wording like "necessary and sufficient", instantly exposing the mathematical falsehood in the restrictive statement.

Final Logic: "necessary and sufficient" makes statement false

🧠 Memory Trick

"Derivative test is sufficient, not always necessary"

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

5 A function like f(x) = -1/x³ is strictly increasing on (–∞, 0) and strictly increasing on (0, ∞). Since it strictly increases across its entire defined domain, it is analytically classified as a:

A Stationary

B Concave

C Monotonic

D Even

💡 Short Explanation

One-direction behavior No sign reversal Domain-wide increase

✅ Correct Answer Explanation

A function that is increasing over all subintervals of its domain is monotonic.

❓ Why Other Options Are Incorrect

Option A → constant behavior
Option B → curvature property
Option D → symmetry property

🎯 Strategy

Used: Concept classification

Application: Groups the unbroken, domain-wide positive slope under the overarching umbrella term "monotonic," eliminating overly narrow or unrelated definitions.

Final Logic: consistent increase → monotonic

🧠 Memory Trick

"Mono = one direction"

📚 Class XI – NCERT Mathematics – Chapter: Functions – Topic: Monotonic Functions

6 For the polynomial f(x) = x⁴/4 – 2x³ + 11x²/2 – 6x, the unbounded domain R (–∞, ∞) is structurally divided into how many strict open intervals by its critical points (1, 2, and 3) to evaluate monotonicity?

A 2 intervals

B 4 intervals

C 3 intervals

D 5 intervals

💡 Short Explanation

3 critical points divide line Intervals = n+1 rule 3 → 4 parts

✅ Correct Answer Explanation

Three critical points divide real line into four intervals for monotonicity analysis.

❓ Why Other Options Are Incorrect

Option A → too few
Option C → incorrect formula
Option D → overcounted

🎯 Strategy

Used: Mathematical pattern rule

Application: Reliably maps three critical points to exactly four structural intervals, bypassing unnecessary manual plotting or potential counting errors.

Final Logic: intervals = points + 1

🧠 Memory Trick

"Split rule: +1 interval"

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

7 The positive derivative test f'(x) > 0 for all x in (a,b) is a sufficient condition for a function to be increasing. Which function exemplifies being strictly increasing over R despite its derivative f'(0) explicitly being mathematically undefined?

A f(x) = x²

B f(x) = |x|

C f(x) = x³

D g(x) = x^(1/3)

💡 Short Explanation

Increasing everywhere Vertical tangent at 0 Still monotonic

✅ Correct Answer Explanation

x^(1/3) is strictly increasing over ℝ, even though derivative is undefined at 0.

❓ Why Other Options Are Incorrect

Option A → not increasing on full ℝ
Option B → not strictly increasing
Option C → derivative defined everywhere

🎯 Strategy

Used: Property recognition

Application: Quickly isolates the cube root function, which is famously increasing everywhere despite possessing an undefined vertical tangent at zero.

Final Logic: monotonic with singularity

🧠 Memory Trick

"Cube root always rises"

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

8 For the Total Cost function C(x) = 2x + 2(x+3)/(x+2), the derivative of Marginal Cost d(MC)/dx evaluates to -8/(x+2)³. For all x > 0, this derivative is strictly negative. Applying the negative derivative test confirms that the Marginal Cost:

A Falls continuously

B Rises continuously

C Constant

D Oscillates

💡 Short Explanation

Negative derivative Decreasing function Cost reduces

✅ Correct Answer Explanation

Since derivative is negative for all x>0, function decreases continuously.

❓ Why Other Options Are Incorrect

Option B → opposite sign
Option C → zero derivative needed
Option D → no oscillation

🎯 Strategy

Used: Sign test

Application: Confirms a strictly negative derivative algebraically, definitively proving the continuous downward trend of the marginal cost function.

Final Logic: negative slope → decreasing

🧠 Memory Trick

"Negative → falling graph"

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

9 A function like f(x) = x³ is generally considered an upward sloping increasing function over R. However, as an exception, its tangent is strictly parallel to the x-axis exactly at which critical point?

A 1

B −1

C 0

D 3

💡 Short Explanation

f'(x)=3x² Zero at x=0 Horizontal tangent

✅ Correct Answer Explanation

Derivative of x³ is 3x², which is zero only at x=0.

❓ Why Other Options Are Incorrect

Option A → derivative nonzero
Option B → same
Option D → same

🎯 Strategy

Used: Direct derivative evaluation

Application: Mathematically pinpoints the horizontal tangent precisely at the origin by simply isolating where the first derivative equals zero.

Final Logic: solve f'(x)=0

🧠 Memory Trick

"Cube → flat at origin"

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

10 Tangent of decreasing function makes:

A Obtuse

B Acute

C Right

D Zero

💡 Short Explanation

Negative slope Angle > 90° Downward trend

✅ Correct Answer Explanation

Decreasing function has negative slope, so tangent angle with x-axis is obtuse.

❓ Why Other Options Are Incorrect

Option B → positive slope
Option C → zero slope
Option D → invalid

🎯 Strategy

Used: Slope-angle relation

Application: Directly translates negative algebraic slopes into geometric obtuse angles, rapidly ruling out acute, right, or zero variations.

Final Logic: negative slope → obtuse angle

🧠 Memory Trick

"Downward = wide angle"

📚 Class XI – NCERT Mathematics – Chapter: Straight Lines – Topic: Slope and Angle

11 If c is a strict point of discontinuity for a function f(x) defined in an interval I, the derivative f'(c) is fundamentally not defined. According to critical point theory, the coordinate c is therefore analytically evaluated as:

A A point of local maximum

B A point of local minimum

C Always a critical point

D A stationary point

💡 Short Explanation

Critical points include undefined derivatives Discontinuity qualifies No smoothness required

✅ Correct Answer Explanation

A critical point is any interior point where f'(x)=0 or f'(x) is undefined. A discontinuity automatically makes derivative undefined, so it is a critical point.

❓ Why Other Options Are Incorrect

Option A → requires local behavior
Option B → requires minimum condition
Option D → requires derivative = 0, not undefined

🎯 Strategy

Used: Definition inclusion

Application: Forces the inclusion of discontinuities as critical points by strictly adhering to the fundamental undefined derivative rule.

Final Logic: undefined derivative ⇒ critical point

🧠 Memory Trick

"Undefined derivative = critical"

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

12 A critical point x=c where f'(c) = 0 is explicitly defined as a stationary point. Which of the following is non-stationary?

A x = 0

B x = 1/8

C x = 1

D x = -1

💡 Short Explanation

f'(0) undefined Not equal to zero Hence non-stationary

✅ Correct Answer Explanation

For f(x)=12x^(4/3)−6x^(1/3), derivative is undefined at x=0, so it is not stationary (stationary requires f'(x)=0).

❓ Why Other Options Are Incorrect

Option B → satisfies f'(x)=0
Option C → regular point
Option D → regular point

🎯 Strategy

Used: Condition filtering

Application: Strictly separates zero derivatives from undefined ones, correctly identifying x=0 as definitively non-stationary to eliminate confusion.

Final Logic: stationary requires derivative = 0

🧠 Memory Trick

"Stationary = zero slope only"

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

📄 Passage

A function may have a critical point where the derivative vanishes and it may not have any local extreme value at that point (or a local maximum or a local minimum value may not exist). For example, for the function h(x) = x³, h'(x) = 3x² and so h'(0) = 0. But '0' is neither a point of local maximum nor a point of local minimum, as the function is increasing on both sides of the point. If f'(x) does not change sign as x increases through 'c', then c is neither a point of local maximum nor a point of local minimum. Such a point is a point of inflexion. At this point, the curve possesses a tangent line and the concavity strictly changes.

According to the passage, x=0 for h(x)=x³ is:

A Local minimum

B Point of inflexion

C Local maximum

D Absolute extremum

💡 Short Explanation

No extrema at 0 Concavity changes Derivative zero

✅ Correct Answer Explanation

At x=0, h'(x)=0 but sign of derivative does not change; hence it is a point of inflexion.

❓ Why Other Options Are Incorrect

Option A → no minimum
Option C → no maximum
Option D → not extreme

🎯 Strategy

Used: Behavior classification

Application: Links a zero derivative with stable sign behaviour around the point, confidently classifying the coordinate as an inflexion.

Final Logic: zero slope + sign stability → inflexion

🧠 Memory Trick

"Flat but not extreme = inflexion"

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

📄 Passage

At an inflexion point, f'(x) as x increases through c:

A Does not change sign

B Reverses sign twice

C Changes positive to negative

D Changes negative to positive

💡 Short Explanation

No sign reversal Only concavity changes Slope behavior stable

✅ Correct Answer Explanation

At inflexion points, the first derivative does not change sign; only concavity (second derivative) changes.

❓ Why Other Options Are Incorrect

Option B → incorrect behavior
Option C → local maximum case
Option D → local minimum case

🎯 Strategy

Used: Sign analysis

Application: Focuses strictly on slope stability through the point, eliminating extremum choices by confirming the first derivative maintains its sign.

Final Logic: derivative sign remains same

🧠 Memory Trick

"Inflexion = no sign flip"

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

15 For the higher-order function f(x) = x⁴ – 8x³ + 22x² – 24x + 1, finding the stationary points by setting f'(x) = 0 mathematically yields the equation 4(x–1)(x–2)(x–3) = 0. The stationary points are exactly:

A 0,1, and 2

B 1,2, and 3

C −1,−2, and −3

D 2,4, and 6

💡 Short Explanation

f'(x)=0 roots Factorized form Direct solutions

✅ Correct Answer Explanation

Given f'(x)=4(x−1)(x−2)(x−3)=0 ⇒ x=1,2,3.

❓ Why Other Options Are Incorrect

Option A → incorrect roots
Option C → wrong sign values
Option D → unrelated

🎯 Strategy

Used: Factorization

Application: Cleanly breaks down the derived polynomial, instantly revealing the precise critical roots without needing complex algebraic manipulation.

Final Logic: roots of derivative

🧠 Memory Trick

"Shifted cubic roots = 1,2, and 3"

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

16 At the specific stationary points x=1, 2, 3 for the polynomial f(x) = x⁴ – 8x³ + 22x² – 24x + 1, the continuous curve structurally changes direction from increasing to decreasing or vice versa. These exact topological locations are geometrically defined as:

A Discontinuities

B Vertical tangents

C Turning points

D Cusps

💡 Short Explanation

f'(x)=0 Direction change Smooth curve

✅ Correct Answer Explanation

At stationary points, derivative is zero and curve changes direction, hence turning points.

❓ Why Other Options Are Incorrect

Option A → function continuous
Option B → slope infinite case
Option D → sharp corner

🎯 Strategy

Used: Geometric interpretation

Application: Visualizes the algebraic zero slope as a physical directional shift, confirming the location purely as a turning point.

Final Logic: derivative zero + smooth change

🧠 Memory Trick

"Stationary = turning"

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

17 Corner points occur where a graphical curve takes a sharp turn, rendering the function algebraically non-differentiable at that specific interior domain point. This lack of a defined derivative naturally classifies the corner point strictly as a:

A Critical point

B Inflexion point

C Stationary point

D Continuous limit

💡 Short Explanation

Derivative undefined Sharp corner Non-smooth point

✅ Correct Answer Explanation

Corner points have undefined derivative, hence they are critical points.

❓ Why Other Options Are Incorrect

Option B → requires concavity change
Option C → requires derivative zero
Option D → not classification

🎯 Strategy

Used: Definition recall

Application: Perfectly links geometric sharp turns to algebraic non-differentiability, satisfying the fundamental textbook condition for a critical point.

Final Logic: non-differentiable ⇒ critical

🧠 Memory Trick

"Corner = critical"

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

18 Which of the following analytical functions geometrically exhibits a vertical tangent at x=0, rendering the derivative f'(0) mathematically undefined, yet remains strictly monotonically increasing over R?

A f(x) = x³

B f(x) = |x| + 3

C h(x) = x²

D g(x) = x^(1/3)

💡 Short Explanation

Vertical tangent at 0 Monotonically increasing Derivative undefined

✅ Correct Answer Explanation

x^(1/3) increases everywhere but has infinite slope at x=0 (vertical tangent).

❓ Why Other Options Are Incorrect

Option A → smooth derivative
Option B → not strictly increasing
Option C → decreases on interval

🎯 Strategy

Used: Behavior-property match

Application: Accurately pairs an undefined center slope with overall upward algebraic movement, isolating the cube root function perfectly.

Final Logic: monotonic + vertical tangent

🧠 Memory Trick

"Cube root = steep center"

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Non-differentiable Points

19 Applying stringent interval analysis to the algebraic function f(x) = x⁴/4 – 2x³ + 11x²/2 – 6x, the function evaluates as strictly increasing in the union of which two specific open intervals?

A (−∞,1) and (2,3)

B (0,1) and (1,2)

C (1,2) and (3,∞)

D (−∞,2) and (2,∞)

💡 Short Explanation

Sign of derivative positive Interval splitting Standard analysis

✅ Correct Answer Explanation

From derivative sign analysis, function increases on (1,2) and (3,∞).

❓ Why Other Options Are Incorrect

Option A → incorrect intervals
Option B → partial
Option D → includes decreasing region

🎯 Strategy

Used: Sign chart

Application: Structurally maps out positive derivative regions, cleanly isolating the precise algebraic intervals where the function actively rises.

Final Logic: positive derivative intervals

🧠 Memory Trick

"Positive → rising segments"

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

20 When conducting a full critical point analysis on the rational function f(x) = 1/x³, we evaluate f'(x) = -3/x⁴. Because x=0 makes f'(x) strictly undefined, according to the mathematical guidelines provided in the text, x=0 is officially classified as:

A A critical point

B Stationary point

C Local maximum

D Local minimum

💡 Short Explanation

Derivative undefined at x=0 Not stationary Still critical

✅ Correct Answer Explanation

At x=0, function is undefined, so derivative undefined ⇒ critical point.

❓ Why Other Options Are Incorrect

Option B → derivative not zero
Option C → no extremum
Option D → no extremum

🎯 Strategy

Used: Definition application

Application: Rigorously flags the undefined derivative at x=0, definitively labeling it a critical point and eliminating non-applicable extremum choices.

Final Logic: undefined derivative ⇒ critical

🧠 Memory Trick

"No derivative = critical point"

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

CUET UG Applied Mathematics Booster Test 2 - Monotonic Functions and Critical Points

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