CUET UG Applied Mathematics Booster Test 1 - Monotonic Functions and Critical Points -

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CUET UG Applied Mathematics Booster Test 1 - 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 1 - Monotonic Functions and Critical Points

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QUESTION 1 OF 20

For a real function y = f(x) defined on an open interval (a,b), if x₁, x₂ ∈ (a,b) and x₁ > x₂ implies f(x₁) > f(x₂), the function is analytically:

Answer Review

1 For a real function y = f(x) defined on an open interval (a,b), if x₁, x₂ ∈ (a,b) and x₁ > x₂ implies f(x₁) > f(x₂), the function is analytically:

A Increasing

B Decreasing

C Stationary

D Undefined

💡 Short Explanation

Larger input gives larger output. This is the mathematical definition of a strictly increasing function. The function rises as x increases throughout the given interval.

✅ Correct Answer Explanation

A function is strictly increasing when every larger value of the independent variable corresponds to a larger function value. The given condition, x₁ > x₂ ⇒ f(x₁) > f(x₂), exactly matches this definition. Hence Option A is correct. Option B describes the opposite behaviour, while Options C and D do not satisfy the given analytical condition.

❓ Why Other Options Are Incorrect

Option B) Decreasing → A decreasing function satisfies x₁ > x₂ ⇒ f(x₁) < f(x₂), which is opposite to the given condition.
Option C) Stationary → A stationary point refers to a point where f′(x)=0, not the behaviour of the entire function over an interval.
Option D) Undefined → The condition clearly defines the function's monotonic behaviour, so the function is not undefined.

🎯 Strategy

Used: Elimination

Application: Compare the given inequality with the standard definitions of increasing and decreasing functions, then eliminate options that contradict the inequality.

Final Logic: Since larger x always gives larger f(x), the function is strictly increasing.

🧠 Memory Trick

"↑x ⇒ ↑f(x) = Increasing."

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Increasing and Decreasing Functions

2 Match the functions and their monotonic or critical point behaviors in List I with their corresponding domains or values in List II:

List I	List II
1. f(x) = x² interval of increase	a. (–∞, 0)
2. f(x) = x² interval of decrease	b. x = 0
3. f(x) = (x – 1)² + 2 critical point	c. (0, ∞)
4. f(x) = x³ inflexion point	d. x = 1

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

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

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

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

💡 Short Explanation

x² decreases on (−∞,0). x² increases on (0,∞). (x−1)²+2 has a critical point at x=1. x³ has an inflexion point at x=0.

✅ Correct Answer Explanation

For f(x)=x², the derivative is 2x, which is negative for x<0 and positive for x>0. Hence it decreases on (−∞,0) and increases on (0,∞). The function (x−1)²+2 has its stationary (critical) point at x=1. The function x³ changes concavity at x=0, giving its point of inflexion. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A) 1-a, 2-b, 3-c, 4-d→ Incorrect because x² increases on (0,∞), not (−∞,0), and the remaining matches are also misplaced.
Option C) 1-d, 2-c, 4-a, 4-b→ Incorrect because intervals and critical points are mismatched. The increasing interval and vertex location are assigned incorrectly.
Option D) 1-b, 2-a, 3-d, 4-c→ Although B and C are correct, A incorrectly assigns a point instead of an interval, and D gives the wrong inflexion location.

🎯 Strategy

Used: Option Grouping

Application: First determine each function's behaviour independently using derivatives, then compare the complete set of matches with the answer choices.

Final Logic: Only Option B correctly matches all four function properties.

🧠 Memory Trick

"x²: Left ↓, Right ↑; Shifted square → vertex; x³ → inflexion at 0."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Increasing and Decreasing Functions, Stationary Point and Point of Inflexion

3 Which of the following analytical conditions accurately signify a decreasing function \(f(x)\)on an interval \(\left(a\ ,\ b\right)\)?
1) \(x_{1}<x_{2}\)implies \(f(x_{1})>f(x_{2})\).
2) \(x_{1}>x_{2}\)implies \(f(x_{1})<f(x_{2})\).
3) \(f^{'}(x)<0\) for all \(x\) in \(\left(a\ ,\ b\right)\).
4) \(f^{'}(x)>0\) for all \(x\) in \(\left(a\ ,\ b\right)\).

A 1, 2, 4

B 1, 2

C 1, 2, 3

D 1, 2, 3, 4

💡 Short Explanation

A decreasing function gives smaller outputs for larger inputs. A and B are equivalent definitions. A negative derivative confirms decreasing behaviour throughout the interval.

✅ Correct Answer Explanation

A function is decreasing if larger values of \(x\) produce smaller values of \(f(x)\). Statements 1 and B express this definition in equivalent forms. Statement 3 is the derivative test: if \(f^{'}(x)<0\) throughout an interval, the function is decreasing there. 4 represents an increasing function, so Option C is correct.

❓ Why Other Options Are Incorrect

Option A) 1, 2, 4 → Incorrect because D states \(f^{'}(x)>0\), which indicates an increasing function, contradicting decreasing behaviour.
Option B) 1, 2 → Although A and B are correct, this option is incomplete because it omits 3, the standard derivative criterion for a decreasing function.
Option D) 1, 2, 3, 4 → Incorrect because 4 contradicts 1, 2 and 3 by describing increasing behaviour instead of decreasing behaviour.

🎯 Strategy

Used: Elimination

Application: Identify the derivative condition for decreasing functions and eliminate every option containing \(f^{'}(x)>0\).

Final Logic: Since 1, 2, and 3 are all correct while 4 is false, Option C is the only valid choice.

🧠 Memory Trick

"Negative derivative ⇒ Negative slope ⇒ Decreasing."

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

4 Identify the incorrect statement regarding decreasing interval conditions and graphs.

A \(x_{1}<x_{2}\)implies \(f(x_{1})>f(x_{2})\).

B The condition \(f^{'}(x)<0\) is sufficient for a function to be decreasing.

C The graph of a decreasing function falls from left to right.

D A decreasing function graph strictly rises from left to right.

💡 Short Explanation

A decreasing graph moves downward as \(x\) increases. Negative derivative indicates decreasing behaviour. Only Option D contradicts the definition of a decreasing function.

✅ Correct Answer Explanation

A decreasing function satisfies \(x_{1}<x_{2}\Rightarrow f(x_{1})>f(x_{2})\), and if \(f^{'}(x)<0\), it decreases on the interval. Such graphs fall from left to right. Therefore Options A, B, and C are correct statements. Option D incorrectly describes an increasing graph, making it the incorrect statement.

❓ Why Other Options Are Incorrect

Option A) \(x_{1}<x_{2}\)implies \(f(x_{1})>f(x_{2})\)→ This is the standard mathematical definition of a decreasing function.
Option B) The condition \(f^{'}(x)<0\) is sufficient for a function to be decreasing. → This is the first derivative test for monotonic decrease on an interval.
Option C) The graph of a decreasing function falls from left to right. → This correctly describes the geometric appearance of a decreasing graph.

🎯 Strategy

Used: Extreme Word Filter

Application: Compare each statement with the definition of decreasing functions. The statement claiming the graph "strictly rises" directly contradicts the concept.

Final Logic: A decreasing graph cannot rise from left to right; hence Option D is incorrect.

🧠 Memory Trick

"Decrease = Downward from left to right."

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

5 If a continuous function \(f(x)\)is strictly decreasing on \(\left(-\infty ,0\right)\)and increasing on \(\left(0\ ,\ \infty \right)\), such as \(f(x)=x^{2}\), we conclude that on the entire real line \(R=(-\infty ,\infty )\), the function is:

A Monotonic increasing

B Monotonic decreasing

C Neither increasing nor decreasing, hence not monotonic on \(R\)

D Entirely non-differentiable

💡 Short Explanation

A monotonic function must keep one trend throughout its domain. \(x^{2}\)decreases before zero and increases after zero. Hence it is not monotonic on the entire real line.

✅ Correct Answer Explanation

A function is monotonic only if it remains entirely increasing or entirely decreasing over its whole domain. The function \(x^{2}\)decreases on \(\left(-\infty ,0\right)\)and increases on \(\left(0\ ,\ \infty \right)\). Since its behaviour changes at \(x=0\), it is not monotonic on \(R\). Therefore Option C is correct. Options A, B, and D are inconsistent with this behaviour.

❓ Why Other Options Are Incorrect

Option A) Monotonic increasing → Incorrect because the function decreases on \(\left(-\infty ,0\right)\).
Option B) Monotonic decreasing → Incorrect because the function increases on \(\left(0\ ,\ \infty \right)\).
Option D) Entirely non-differentiable → Incorrect because \(x^{2}\)is differentiable for every real number; differentiability is unrelated to its failure to be monotonic on the entire domain.

🎯 Strategy

Used: Option Grouping

Application: Check whether the function keeps the same monotonic behaviour across the entire domain rather than only on individual intervals.

Final Logic: Since the function changes from decreasing to increasing at \(x=0\), it cannot be monotonic on all of \(R\).

🧠 Memory Trick

"Change of direction = Not monotonic."

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

6 To find the intervals where \(f(x)=\frac{x^{4}}{4}-2x^{3}+\frac{11x^{2}}{2}-6x\) is monotonically increasing or decreasing, we first locate critical points within the constraints. What are the critical points found by solving \(f^{'}(x)=0\)?

A 0, 1, 2

B 1, 2, 3

C –1, –2, –3

D 2, 3, 4

💡 Short Explanation

Differentiate the function. Factor \(f^{'}(x)=x^{3}-6x^{2}+11x-6=(x-1)(x-2)(x-3)\). Setting each factor equal to zero gives critical points at \(x=1, 2, 3\).

✅ Correct Answer Explanation

Differentiate the polynomial to obtain \(f^{'}(x)=x^{3}-6x^{2}+11x-6\). This factors as \(\left(x-1)(x-2)(x-3\right)\). Solving \(f^{'}(x)=0\) gives \(x=1, 2, 3\), which are the critical points used for monotonicity analysis. Therefore, Option B is correct. The remaining options contain values that are not roots of the derivative.

❓ Why Other Options Are Incorrect

Option A) 0, 1, 2 → Incorrect because \(x=0\) is not a root of \(f^{'}(x)\), while \(x=3\) is omitted.
Option C) –1, –2, –3 → Incorrect because none of these negative values satisfy \(f^{'}(x)=0\).
Option D) 2, 3, 4 → Incorrect because \(x=4\) is not a critical point, whereas \(x=1\) is missing.

🎯 Strategy

Used: Substitution

Application: Factor the derivative and substitute each option into \(f^{'}(x)\). Only one option makes the derivative zero for all listed values.

Final Logic: The derivative factors completely into three linear factors whose roots are 1, 2, and 3.

🧠 Memory Trick

"(1–2–3) are the roots of \(\left(x-1)(x-2)(x-3\right)\)."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Critical Points and Monotonicity

7 For \(f(x)=-\frac{1}{x^{3}}\), the derivative \(f^{'}(x)\)evaluates to \(\frac{3}{x^{4}}\). According to the positive derivative test, because \(f^{'}(x)>0\) for all \(x\) in its domain, the function is strictly:

A Increasing

B Decreasing

C Constant

D Concave downward

💡 Short Explanation

The derivative is \(3/x^{4}\). Since \(x^{4}>0\) for every \(x\neq 0\), \(f^{'}(x)\)is always positive. A positive derivative implies the function is increasing on its domain.

✅ Correct Answer Explanation

For \(f(x)=-x^{-3}\), differentiation gives \(f^{'}(x)=3/x^{4}\). Since \(x^{4}\)is always positive for \(x\neq 0\), the derivative remains positive throughout the domain. Hence the function is strictly increasing on each interval of its domain. Therefore, Option A is correct. Options B, C, and D do not follow from the first derivative test.

❓ Why Other Options Are Incorrect

Option B) Decreasing → Incorrect because a decreasing function requires \(f^{'}(x)<0\), but here the derivative is always positive.
Option C) Constant → Incorrect because a constant function has derivative zero everywhere in its domain.
Option D) Concave downward → Incorrect because concavity depends on the second derivative, not the sign of the first derivative.

🎯 Strategy

Used: Elimination

Application: Determine the sign of the derivative first. Eliminate every option inconsistent with a positive derivative.

Final Logic: Since \(f^{'}(x)>0\) for all \(x\neq 0\), the function is increasing on its domain.

🧠 Memory Trick

"Positive derivative ⇒ Increasing curve."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: First Derivative Test for Monotonicity

8 For \(f(x)=\frac{x^{4}}{4}-2x^{3}+\frac{11x^{2}}{2}-6x\), the derivative is \(f^{'}(x)=(x-1)(x-2)(x-3)\). In the interval \(\left(1\ ,\ 2\right)\), \(f^{'}(x)\)is positive. In the interval \(\left(2\ ,\ 3\right)\), \(f^{'}(x)\)evaluates as:

A Positive, so \(f\) is increasing

B Zero, so \(f\) is stationary

C Negative, so \(f\) is decreasing

D Undefined

💡 Short Explanation

In \(\left(2\ ,\ 3\right)\), \(\left(x\ −\ 1\right)\)and \(\left(x\ −\ 2\right)\)are positive. \(\left(x\ −\ 3\right)\)is negative. Their product is negative, so the function decreases.

✅ Correct Answer Explanation

For \(2<x<3\), the factors satisfy \((x-1)>0\), \((x-2)>0\), and \((x-3)<0\). Their product is therefore negative, making \(f^{'}(x)<0\). A negative derivative indicates the function decreases throughout the interval. Hence Option C is correct. The derivative is neither zero nor undefined within the open interval.

❓ Why Other Options Are Incorrect

Option A) Positive, so \(f\) is increasing → Incorrect because the derivative is negative, not positive, on \(\left(2\ ,\ 3\right)\).
Option B) Zero, so \(f\) is stationary → Incorrect because \(f^{'}(x)=0\) only at \(x=2\) and \(x=3\), not between them.
Option D) Undefined → Incorrect because the derivative is a polynomial expression and is defined for every real value.

🎯 Strategy

Used: Substitution

Application: Choose any test point, such as \(x=2.5\), determine the sign of each factor, and evaluate the sign of their product.

Final Logic: Two positive factors and one negative factor produce a negative derivative, so the function decreases

🧠 Memory Trick

"++− = Negative ⇒ Decreasing."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Sign of First Derivative and Monotonicity

9 When a function is strictly increasing in its domain, the tangents drawn to the upward sloping curve generally make what type of angle of inclination?

A Obtuse angles

B Acute angles

C Right angles

D Reflex angles

💡 Short Explanation

An increasing function has a positive slope. Positive slope corresponds to an angle of inclination between \(0^{\circ }\)and \({90}^{\circ }\). Hence, tangents make acute angles with the positive x-axis.

✅ Correct Answer Explanation

The slope of a tangent is \(m=tan⁡\theta\), where \(\theta\) is the angle of inclination. For an increasing function, the slope is positive, so \(tan⁡\theta >0\). Therefore, the angle lies between \(0^{\circ }\)and \({90}^{\circ }\), making it an acute angle. Thus, Option B is correct. The remaining options correspond to different slope conditions.

❓ Why Other Options Are Incorrect

Option A) Obtuse angles → Obtuse angles (\({90}^{\circ }<\theta <{180}^{\circ }\)) have negative tangent values, representing decreasing functions.
Option C) Right angles → A right angle corresponds to a vertical tangent with undefined slope, not the general case of an increasing function.
Option D) Reflex angles → Reflex angles are not used to represent the standard angle of inclination of a tangent line in coordinate geometry.

🎯 Strategy

Used: Elimination

Application: Relate the sign of the slope to the angle of inclination. Eliminate options corresponding to negative or undefined slopes.

Final Logic: Positive slope implies an acute angle; therefore, Option B is correct.

🧠 Memory Trick

"Positive slope → Acute angle."

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

10 When a function is monotonically decreasing in its domain, the tangents drawn to the downward sloping curve generally make what type of geometric angle of inclination?

A Acute angles

B Right angles

C Obtuse angles

D Zero degree angles

💡 Short Explanation

A decreasing function has a negative slope. Negative slope corresponds to an angle between \({90}^{\circ }\)and \({180}^{\circ }\). Therefore, tangent lines make obtuse angles with the positive x-axis.

✅ Correct Answer Explanation

For a decreasing function, the tangent has a negative slope. Since \(m=tan⁡\theta\), a negative tangent value corresponds to an angle of inclination between \({90}^{\circ }\)and \({180}^{\circ }\). Such an angle is obtuse. Therefore, Option C is correct. Acute angles indicate positive slopes, while right and zero-degree angles do not represent decreasing functions.

❓ Why Other Options Are Incorrect

Option A) Acute angles → Acute angles have positive tangent values, representing increasing rather than decreasing functions.
Option B) Right angles → A right angle indicates a vertical tangent with undefined slope, not the usual tangent to a decreasing curve.
Option D) Zero degree angles → A zero-degree angle gives a horizontal tangent with zero slope, indicating a stationary point rather than a decreasing function.

🎯 Strategy

Used: Elimination

Application: Use the relationship between slope and angle of inclination. Remove options that represent positive, zero, or undefined slopes.

Final Logic: A negative slope always corresponds to an obtuse angle; hence Option C is correct.

🧠 Memory Trick

"Negative slope → Obtuse angle."

📚 Class XI – NCERT Mathematics – Chapter: Straight Lines – Topic: Angle of Inclination of a Line

11 A continuous function \(f(x)\)is defined in interval \(I\) with an interior point \(c\). If \(f(x)\)is continuous at \(x=c\) but \(f^{'}(c)\)is algebraically not defined, then \(c\) is explicitly classified as:

A A critical point

B An absolute minimum

C An absolute maximum

D A stationary point

💡 Short Explanation

A critical point occurs where \(f^{'}(x)=0\) or \(f^{'}(x)\)does not exist. Continuity is maintained. Since \(f^{'}(c)\)is undefined, \(c\) is a critical point.

✅ Correct Answer Explanation

A critical point is any point in the domain where the derivative is either zero or does not exist, provided the function itself is defined there. Since \(f(x)\)is continuous at \(c\) but \(f^{'}(c)\)is undefined, \(c\) satisfies the definition of a critical point. Therefore, Option A is correct. The information does not guarantee a maximum, minimum, or stationary point.

❓ Why Other Options Are Incorrect

Option B) An absolute minimum → Incorrect because an undefined derivative alone does not imply the function attains its least value.
Option C) An absolute maximum → Incorrect because continuity and an undefined derivative do not guarantee the greatest function value.
Option D) A stationary point → Incorrect because a stationary point requires \(f^{'}(c)=0\), whereas here the derivative is not defined.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Recall the formal definition of a critical point and compare it with the given condition of an undefined derivative.

Final Logic: Since the derivative does not exist at a point where the function is defined, the point is critical, making Option A correct.

🧠 Memory Trick

"Critical = Derivative zero or undefined."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Critical Points and Stationary Points

12 Which of the following statements correctly distinguishes the geometric point \(x=0\) for the specific function \(f(x)=x^{2/3}\)?

A It is a critical point but not a stationary point because \(f^{'}(0)\)is not defined.

B It is a stationary point but not a critical point.

C It is both a critical point and a stationary point.

D It is neither a critical nor a stationary point.

💡 Short Explanation

A critical point occurs where the derivative is zero or undefined. Here, \(f^{'}(0)\)does not exist. Since the derivative is not zero, it is not a stationary point.

✅ Correct Answer Explanation

For \(f(x)=x^{2/3}\), the derivative is \(f^{'}(x)=\frac{2}{3}x^{-1/3}\), which is undefined at \(x=0\). Hence, \(x=0\) is a critical point because the function exists but its derivative does not. A stationary point requires \(f^{'}(x)=0\), which is not satisfied. Therefore, Option A is correct, while the remaining options contradict these definitions.

❓ Why Other Options Are Incorrect

Option B) It is a stationary point but not a critical point. → Incorrect because a stationary point requires \(f^{'}(0)=0\), whereas the derivative is undefined.
Option C) It is both a critical point and a stationary point. → Incorrect because although it is a critical point, it is not stationary since the derivative is not zero.
Option D) It is neither a critical nor a stationary point. → Incorrect because an undefined derivative at a point where the function exists makes it a critical point.

🎯 Strategy

Used: Elimination

Application: Recall the definitions of critical and stationary points, then eliminate options that confuse an undefined derivative with a zero derivative.

Final Logic: Since \(f^{'}(0)\)is undefined but \(f(0)\)exists, the point is critical but not stationary.

🧠 Memory Trick

"Critical = Zero or Undefined; Stationary = Zero only."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Critical and Stationary Points

📄 Passage

A function may have a critical point where the derivative vanishes or is strictly undefined. The curve \(f(x)=∣x∣+3\) has a critical point at \(x=0\) because 0 is a point of discontinuity or \(f^{'}(0)\)is not defined. The graph takes a sharp turn at \(\left(0\ ,\ 3\right)\), creating a corner. Another example is the graph of the function \(f(x)=x^{2/3}\), where \(f^{'}(0)\)is not defined. In this case, 0 is a critical point, and at \(\left(0\ ,\ 0\right)\), there is a cusp in the graph. Lastly, a function like \(g(x)=x^{1/3}\)has a vertical tangent at the point \(\left(0\ ,\ 0\right)\), which is the y-axis itself, and its derivative \(g^{'}(0)\)is not defined.

According to the passage, why is \(x=0\) mathematically considered a critical point for the absolute value function \(f(x)=∣x∣+3\)?

A Because \(f^{'}(0)=0\) and the tangent is horizontal.

B Because \(f^{'}(0)\)is not defined and there is a geometric corner.

C Because it represents a smooth turning point.

D Because the curve becomes parallel to the x-axis.

💡 Short Explanation

The graph has a sharp corner at \(x=0\). Left and right derivatives differ. Therefore, \(f^{'}(0)\)is undefined, making it a critical point.

✅ Correct Answer Explanation

The graph of \(f(x)=∣x∣+3\) has a corner at \(\left(0\ ,\ 3\right)\). At this point, the left-hand and right-hand derivatives are unequal, so \(f^{'}(0)\)does not exist. A point where the function is defined but the derivative is undefined is a critical point. Hence, Option B is correct. The remaining options incorrectly describe the geometry or derivative.

❓ Why Other Options Are Incorrect

Option A) Because \(f^{'}(0)=0\) and the tangent is horizontal. → Incorrect because the derivative is undefined, not zero.
Option C) Because it represents a smooth turning point. → Incorrect because a corner is not a smooth turning point.
Option D) Because the curve becomes parallel to the x-axis. → Incorrect because there is no horizontal tangent at the corner.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Identify the exact statement in the passage explaining why \(x=0\) is critical and match it with the options.

Final Logic: The passage explicitly states that the derivative is not defined due to a corner, making Option B correct.

🧠 Memory Trick

"Corner ⇒ Derivative Undefined ⇒ Critical Point."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Non-Differentiable Critical Points

📄 Passage

Based on the passage, for the function \(f(x)=x^{2/3}\), the critical point located at \(\left(0\ ,\ 0\right)\)geometrically represents a:

A Cusp

B Point of inflexion

C Smooth minimum

D Smooth maximum

💡 Short Explanation

The graph of \(x^{2/3}\)forms a sharp pointed curve. Its derivative is undefined at the origin. This pointed shape is called a cusp.

✅ Correct Answer Explanation

For \(f(x)=x^{2/3}\), the derivative is undefined at \(x=0\), and the graph has a sharp pointed shape known as a cusp. This is a standard example of a non-differentiable point in calculus. Therefore, Option A is correct. The graph is neither a smooth turning point nor an inflexion point.

❓ Why Other Options Are Incorrect

Option B) Point of inflexion → Incorrect because the graph forms a cusp rather than merely changing concavity with a well-defined tangent.
Option C) Smooth minimum → Incorrect because the point is not smooth; the derivative is undefined.
Option D) Smooth maximum → Incorrect because the graph has neither a smooth maximum nor a differentiable turning point.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Use the passage directly, which explicitly identifies the geometric nature of the point at \(\left(0\ ,\ 0\right)\).

Final Logic: The passage clearly states that \(\left(0\ ,\ 0\right)\)is a cusp, making Option A the correct answer.

🧠 Memory Trick

"Two-thirds → Cusp; One-third → Vertical Tangent."

15 For the function \(f(x)=x^{3}-\frac{3}{2}x^{2}-18x+1\), finding the stationary points requires setting the derivative equal to zero. What are these exact \(x\)-values?

A \(x=1\) and \(x=-6\)

B \(x=2\) and \(x=-3\)

C \(x=3\) and \(x=-2\)

D \(x=0\) and \(x=3\)

💡 Short Explanation

Differentiate the function. Solve \(f^{'}(x)=3x^{2}-3x-18=0\). Factoring gives \(3(x-3)(x+2)=0\), so the stationary points occur at \(x=3\) and \(x=-2\).

✅ Correct Answer Explanation

Stationary points occur where the derivative equals zero. Differentiating gives \(f^{'}(x)=3x^{2}-3x-18=3(x-3)(x+2)\). Solving \(f^{'}(x)=0\) gives \(x=3\) and \(x=-2\). Therefore, Option C is correct. The remaining options contain values that do not satisfy the derivative equation.

❓ Why Other Options Are Incorrect

Option A) \(x=1\) and \(x=-6\)→ Incorrect because substituting these values into \(3x^{2}-3x-18\) does not give zero.
Option B) \(x=2\) and \(x=-3\)→ Incorrect because neither value is a root of the derivative.
Option D) \(x=0\) and \(x=3\)→ Incorrect because \(x=0\) is not a stationary point, although \(x=3\) is.

🎯 Strategy

Used: Substitution

Application: Differentiate the function, factor the quadratic, and verify the roots by substitution into the derivative.

Final Logic: Only \(x=3\) and \(x=-2\) satisfy \(f^{'}(x)=0\), making Option C correct.

🧠 Memory Trick

"Factor first: \(3(x-3)(x+2)\)."

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

16 At the evaluated stationary points \(x=3\) and \(x=-2\) for \(f(x)=x^{3}-\frac{3}{2}x^{2}-18x+1\), the curve momentarily rests and takes a smooth turn. These turning points exclusively represent:

A Points where the tangent is perpendicular to the x-axis

B Points where the tangent is parallel to the x-axis

C Points of absolute discontinuity

D Points where the derivative is undefined

💡 Short Explanation

At stationary points, \(f^{'}(x)=0\). Zero slope means a horizontal tangent. A horizontal tangent is parallel to the x-axis and indicates a smooth turning point.

✅ Correct Answer Explanation

A stationary point is characterized by \(f^{'}(x)=0\). Since the slope of the tangent is zero, the tangent line is horizontal, making it parallel to the x-axis. Therefore, Option B is correct. A perpendicular tangent has undefined slope, while discontinuity and undefined derivatives are not properties of stationary points.

❓ Why Other Options Are Incorrect

Option A) Points where the tangent is perpendicular to the x-axis → Incorrect because a perpendicular (vertical) tangent has an undefined slope, not zero.
Option C) Points of absolute discontinuity → Incorrect because stationary points occur on continuous, differentiable curves.
Option D) Points where the derivative is undefined → Incorrect because stationary points require the derivative to be zero, not undefined.

🎯 Strategy

Used: Elimination

Application: Recall the definition of a stationary point and remove options involving undefined derivatives or discontinuity.

Final Logic: Since \(f^{'}(x)=0\), the tangent is horizontal and parallel to the x-axis; therefore Option B is correct.

🧠 Memory Trick

"Stationary = Zero slope = Horizontal tangent."

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

17 For the continuous function \(g(x)=x^{1/3}\), the geometric point \(\left(0\ ,\ 0\right)\)features a vertical tangent. Because the tangent line exists and the concavity of the curve strictly changes at this point, it is classified as a:

A Point of inflexion

B Point of local minimum

C Point of local maximum

D Cusp

💡 Short Explanation

The graph has a vertical tangent at the origin. Its concavity changes across \(x=0\). Hence, \(\left(0\ ,\ 0\right)\)is a point of inflexion, not an extremum or cusp.

✅ Correct Answer Explanation

For \(g(x)=x^{1/3}\), the graph has a vertical tangent at the origin, and its concavity changes from one side of the point to the other. A change in concavity is the defining property of a point of inflexion. Therefore, Option A is correct. The point is neither a maximum, minimum, nor a cusp.

❓ Why Other Options Are Incorrect

Option B) Point of local minimum → Incorrect because the function continues to increase through the origin and does not attain a minimum.
Option C) Point of local maximum → Incorrect because there is no highest value near the origin.
Option D) Cusp → Incorrect because a cusp has two meeting branches with sharp opposing tangents, whereas \(x^{1/3}\)has a vertical tangent, not a cusp.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Identify the defining property stated in the question—change in concavity—and match it with the appropriate geometric concept.

Final Logic: Since concavity changes at the point, it is a point of inflexion, making Option A correct.

🧠 Memory Trick

"One-third → Vertical tangent + Inflexion."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Point of Inflexion

18 A formal point of inflexion dynamically occurs when a continuous curve possesses a tangent line and simultaneously experiences a structural change in its:

A Independent variable

B Concavity

C Absolute extremum

D Vertical asymptote

💡 Short Explanation

A point of inflexion is identified by a change in concavity. The curve changes from concave upward to downward or vice versa. Hence, concavity is the defining feature.

✅ Correct Answer Explanation

A point of inflexion is a point where a continuous curve changes its concavity. The graph changes from concave upward to concave downward or vice versa. The existence of a tangent alone is insufficient; the essential condition is the change in concavity. Therefore, Option B is correct. The remaining options do not define an inflexion point.

❓ Why Other Options Are Incorrect

Option A) Independent variable → Incorrect because the independent variable does not undergo any structural change; only the curve's shape changes.
Option C) Absolute extremum → Incorrect because an inflexion point need not be a maximum or minimum.
Option D) Vertical asymptote → Incorrect because a vertical asymptote is unrelated to the definition of an inflexion point.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Recall the formal definition of a point of inflexion and identify the property that changes at that point.

Final Logic: A point of inflexion is characterized by a change in concavity, making Option B correct.

🧠 Memory Trick

"Inflexion = Flip in Concavity."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Point of Inflexion

19 For the polynomial \(f(x)=\frac{x^{4}}{4}-2x^{3}+\frac{11x^{2}}{2}-6x\), the algebraic sign of \(f^{'}(x)\)is strictly less than 0 in the interval \(\left(-\infty ,1\right)\). Thus, the function is:

A Decreasing over \(\left(-\infty ,1\right)\)

B Increasing over \(\left(-\infty ,1\right)\)

C Constant over \(\left(-\infty ,1\right)\)

D Undefined over \(\left(-\infty ,1\right)\)

💡 Short Explanation

A negative derivative indicates decreasing behaviour. Since \(f^{'}(x)<0\) throughout \(\left(-\infty ,1\right)\), the function continuously decreases over the entire interval.

✅ Correct Answer Explanation

The first derivative test states that if \(f^{'}(x)<0\) throughout an interval, the function is strictly decreasing on that interval. Since the derivative is negative for every \(x\in (-\infty ,1)\), the graph continuously falls as \(x\) increases. Therefore, Option A is correct. The remaining options contradict the derivative test.

❓ Why Other Options Are Incorrect

Option B) Increasing over \(\left(-\infty ,1\right)\)→ Incorrect because an increasing function requires \(f^{'}(x)>0\).
Option C) Constant over \(\left(-\infty ,1\right)\)→ Incorrect because a constant function has derivative equal to zero throughout the interval.
Option D) Undefined over \(\left(-\infty ,1\right)\)→ Incorrect because both the polynomial and its derivative are defined for all real numbers.

🎯 Strategy

Used: Elimination

Application: Apply the first derivative test and eliminate every option inconsistent with a negative derivative.

Final Logic: Since \(f^{'}(x)<0\), the function is decreasing; therefore Option A is correct.

🧠 Memory Trick

"Negative derivative → Negative slope → Decreasing."

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

20 Given \(f(x)=12x^{4/3}-6x^{1/3}\)on \(\left[-1,1\right]\), setting \(f^{'}(x)=0\) yields \(x=\frac{1}{8}\). Additionally, \(f^{'}(x)\)is not defined at \(x=0\). Therefore, executing critical point analysis, the critical points in the open interval \(\left(-1,1\right)\)are precisely:

A 1 and -1

B 1/4 and 0

C 1/8 and 0

D 1/8 and 1/2

💡 Short Explanation

Critical points occur where the derivative is zero or undefined. Here, \(f^{'}(x)=0\) at \(x=\frac{1}{8}\). Also, \(f^{'}(0)\)is undefined, giving two critical points.

✅ Correct Answer Explanation

A critical point occurs where the derivative is zero or does not exist, provided the function is defined. Here, solving \(f^{'}(x)=0\) gives \(x=\frac{1}{8}\), while \(f^{'}(0)\)is undefined. Both points lie in the open interval \(\left(-1,1\right)\). Hence, the critical points are \(\frac{1}{8}\)and \(0\), making Option C correct.

❓ Why Other Options Are Incorrect

Option A) 1 and -1 → Incorrect because these are interval endpoints and are not critical points in the open interval.
Option B) 1/4 and 0 → Incorrect because \(x=\frac{1}{4}\)does not satisfy \(f^{'}(x)=0\).
Option D) 1/8 and 1/2 → Incorrect because \(x=\frac{1}{2}\)is not a point where the derivative is zero or undefined.

🎯 Strategy

Used: Option Grouping

Application: Identify all points where the derivative is zero or undefined, then compare the complete set with the options.

Final Logic: The only pair containing both \(x=\frac{1}{8}\)and \(x=0\) is Option C.

🧠 Memory Trick

"Critical = Zero + Undefined."

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

CUET UG Applied Mathematics Booster Test 1 - Monotonic Functions and Critical Points

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