CUET UG Mathematics Booster Test 2 - Increasing and Decreasing Functions -

🚀

CUET UG Mathematics Booster Test 2 - Increasing and Decreasing Functions

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 2 - Increasing and Decreasing Functions

1 / 20

QUESTION 1 OF 20

If \(f(x)\)is strictly increasing on \(\left[a\ ,\ b\right]\)and \(f(x)>0\) on \(\left[a\ ,\ b\right]\), then \(\int_{a}^{b}\,f(x) dx\) is:

Answer Review

1 If \(f(x)\)is strictly increasing on \(\left[a\ ,\ b\right]\)and \(f(x)>0\) on \(\left[a\ ,\ b\right]\), then \(\int_{a}^{b}\,f(x) dx\) is:

A Strictly positive

B Zero

C Strictly negative

D Cannot be determined

💡 Short Explanation

Function values are positive throughout. Integral represents signed area. Positive function gives positive area.

✅ Correct Answer Explanation

Since \(f(x)>0\) for every \(x\in [a,b]\), the graph lies above the x-axis. Therefore the definite integral represents a positive area. The fact that the function is strictly increasing strengthens positivity but is not essential. Hence \(\int_{a}^{b}\,f(x) dx>0\), making Option A correct.

❓ Why Other Options Are Incorrect

Option B → Zero integral requires no net area.
Option C → Negative integral occurs when function is below the x-axis.
Option D → Positivity of the function fully determines the sign.

🎯 Strategy

Used: Elimination

Application: Use the geometric meaning of definite integrals.

Final Logic: Positive function ⇒ positive area ⇒ positive integral.

🧠 Memory Trick

Above x-axis ⇒ Positive Integral

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Geometrical Interpretation of Definite Integral

2 A function is strictly decreasing on \(\left[a\ ,\ c\right]\)and strictly increasing on \(\left[c\ ,\ b\right]\). If a point is chosen uniformly from \(\left[a\ ,\ b\right]\), the probability that the slope is negative lies in:

A \(\left[c\ ,\ b\right]\)

B \(\left[a\ ,\ b\right]\)

C \(\left[a\ ,\ c\right]\)

D \(\left(a\ ,\ c\right)\)

💡 Short Explanation

Negative slope means decreasing function. Function decreases before c. Open interval avoids endpoint ambiguity.

✅ Correct Answer Explanation

A strictly decreasing function has \(f^{'}(x)<0\). Since the function is strictly decreasing on \(\left[a\ ,\ c\right]\), the derivative is negative throughout the interior interval \(\left(a\ ,\ c\right)\). At \(c\), derivative behavior may change. Hence the interval where the slope is certainly negative is \(\left(a\ ,\ c\right)\), making Option D correct.

❓ Why Other Options Are Incorrect

Option A → Function is increasing there.
Option B → Includes both increasing and decreasing regions.
Option C → Includes endpoint c where sign change may occur.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Relate decreasing behavior to derivative sign.

Final Logic: Decreasing interval ⇒ negative slope interval.

🧠 Memory Trick

Decrease Before c

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

3 Assertion (A): For an increasing function \(f\), if \(x_{1}<x_{2}\), then \(f(x_{1})\leq f(x_{2})\).
Reason (R): The definition of an increasing (non-strict) function allows constant values over intervals.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Increasing functions preserve order. Equality may occur in non-strict increase. Reason explains the inequality sign.

✅ Correct Answer Explanation

For an increasing function, \(x_{1}<x_{2}\Rightarrow f(x_{1})\leq f(x_{2})\) The non-strict inequality is used because the function may remain constant on parts of the interval. Thus both Assertion and Reason are true, and the Reason correctly explains the Assertion. Hence Option C is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Compare definition and explanation.

Final Logic: Non-strict increase allows equality.

🧠 Memory Trick

Increasing ⇒ ≤

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

4 Let \(\vec{r}(t)=(x(t),y(t))\). If \(y(t)\)is strictly decreasing, then the dot product of velocity \(\vec{v}(t)\)with unit vector \(\hat{j}\)is:

A Positive

B Negative

C Zero

D Variable

💡 Short Explanation

Velocity component along y-axis is \(dy/dt\). Decreasing y means negative derivative. Dot product equals y-component.

✅ Correct Answer Explanation

The velocity vector is \(\vec{v}(t)=\left(\frac{dx}{dt}\ ,\ \frac{dy}{dt}\right)\) and \(\vec{v}⋅\hat{j}=\frac{dy}{dt}\) Since \(y(t)\)is strictly decreasing, \(\frac{dy}{dt}<0\). Therefore the dot product is negative. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Would indicate increasing y.
Option C → Represents constant y.
Option D → The sign is determined as negative.

🎯 Strategy

Used: Substitution

Application: Evaluate the dot product directly.

Final Logic: \(\vec{v}⋅\hat{j}=dy/dt<0\).

🧠 Memory Trick

j-component = dy/dt

📚 Class XII – NCERT Mathematics – Chapter: Vector Algebra – Topic: Components of Vectors

5 Match the Following For \(f(x)=C\):

List I	List II
1. \(f^{'}(x)\)	a. Horizontal line
2. \(df/dx\)	b. Neither strictly increasing nor strictly decreasing
3. Graph type	c. 0
4. Monotonicity	d. 0

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

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

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

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

💡 Short Explanation

Constant function has zero derivative. Graph is horizontal. Not strictly monotonic.

✅ Correct Answer Explanation

For \(f(x)=C\), \(f^{'}(x)=0,\frac{df}{dx}=0\) Its graph is a horizontal line. Since function values do not increase or decrease strictly, it is neither strictly increasing nor strictly decreasing. Thus the correct matching is Option A.

❓ Why Other Options Are Incorrect

Option B → Matches derivatives incorrectly.
Option C → Assigns wrong graph type.
Option D → Incorrectly matches monotonicity.

🎯 Strategy

Used: Option Grouping

Application: Match each property of a constant function.

Final Logic: Constant ⇒ zero derivative ⇒ horizontal graph.

🧠 Memory Trick

Constant = Flat Line

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Constant Function

6 A quantity increases at constant rate \(k\). If \(f(0)=0\) and \(f(2)=10\), then \(k\) is:

A 2

B 3

C 5

D 10

💡 Short Explanation

Constant rate means linear growth. Rate = change ÷ time. Apply directly.

✅ Correct Answer Explanation

Since the quantity increases at a constant rate, \(k=\frac{f(2)-f(0)}{2-0}k=\frac{10-0}{2}=5\) Therefore the constant rate is 5, making Option C correct.

❓ Why Other Options Are Incorrect

Option A → Underestimates the rate.
Option B → Incorrect calculation.
Option D → Equals total change, not rate.

🎯 Strategy

Used: Substitution

Application: Use rate = change/time.

Final Logic: \(10/2=5\).

🧠 Memory Trick

Rate = Rise ÷ Time

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

7 Steps to prove a function is increasing at a point:
1. Conclude \(f^{'}(x_{0})>0\)
2. Evaluate the sign of derivative at \(x_{0}\)
3. Differentiate \(f(x)\)
4. Identify the point \(x_{0}\)

A 1, 2, 3, 4

B 4, 2, 3, 1

C 3, 1, 4, 2

D 4, 3, 2, 1

💡 Short Explanation

Identify the point. Differentiate the function. Check derivative sign and conclude.

✅ Correct Answer Explanation

The logical procedure is: 1. Identify the point \(x_{0}\). 2. Differentiate the function. 3. Evaluate the derivative sign at \(x_{0}\). 4. Conclude whether \(f^{'}(x_{0})>0\). Hence the correct sequence is 4, 3, 2, 1, corresponding to Option D.

❓ Why Other Options Are Incorrect

Option A → Starts with the conclusion.
Option B → Checks sign before differentiation.
Option C → Uses an illogical order.

🎯 Strategy

Used: Elimination

Application: Follow the standard mathematical workflow.

Final Logic: Point → Derivative → Sign → Conclusion.

🧠 Memory Trick

Point–Diff–Check–Conclude

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

8 If a function decreases at a point \(x_{0}\), the graph near \(x_{0}\)shows:

A Local valley

B Downward slope

C Local peak

D Horizontal tangent

💡 Short Explanation

Decreasing means falling graph. Tangent slope is negative. Curve descends left to right.

✅ Correct Answer Explanation

A decreasing function has negative slope locally. Therefore near \(x_{0}\), the graph moves downward as x increases. This is represented by a downward slope. Option B correctly describes the graphical behavior.

❓ Why Other Options Are Incorrect

Option A → Valley indicates a minimum point.
Option C → Peak indicates a maximum point.
Option D → Horizontal tangent has zero slope.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match graphical description with monotonicity.

Final Logic: Decrease ⇒ downward slope.

🧠 Memory Trick

Decrease = Downhill

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

9 If \(f^{'}(x)<0\) for all real x, then the function is:

A Decreasing

B Increasing

C Constant

D Undefined

💡 Short Explanation

Negative derivative means negative slope. Function falls throughout domain. Standard derivative test.

✅ Correct Answer Explanation

If \(f^{'}(x)<0\) for every x, the tangent slope is always negative. By the First Derivative Test, the function decreases throughout its domain. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Requires positive derivative.
Option C → Requires zero derivative.
Option D → Function is perfectly defined.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Use derivative sign directly.

Final Logic: Negative derivative ⇒ decreasing function.

🧠 Memory Trick

Negative ⇒ Decreasing

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

10 If \(f^{'}(x)<0\), then:
I. Function is decreasing
II. Tangent makes an obtuse angle with positive x-axis
III. Function is constant

A I, II, and III

B I and III only

C I and II only

D II and III only

💡 Short Explanation

Negative derivative means decreasing. Negative slope corresponds to an obtuse inclination angle. Constant functions have zero derivative.

✅ Correct Answer Explanation

If \(f^{'}(x)<0\), the function is decreasing and the tangent has a negative slope. A line with negative slope makes an obtuse angle with the positive x-axis. Therefore Statements I and II are true. Statement III is false because a constant function has derivative zero.

❓ Why Other Options Are Incorrect

Option A → Includes false Statement III.
Option B → Omits Statement II.
Option D → Includes false Statement III.

🎯 Strategy

Used: Option Grouping

Application: Test each statement independently.

Final Logic: Only I and II follow from \(f^{'}(x)<0\).

🧠 Memory Trick

Negative Slope = Obtuse Angle

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Geometrical Meaning of Derivative

11 Which of the following is NOT required for applying the First Derivative Test on an interval I?

A f is continuous on I

B f is twice differentiable on I

C f is differentiable on I

D The sign of \(f^{'}(x)\)must be checked

💡 Short Explanation

First Derivative Test uses first derivative only. Continuity and differentiability are required. Second derivative is not necessary.

✅ Correct Answer Explanation

The First Derivative Test requires continuity on the interval, differentiability where applicable, and examination of the sign of \(f^{'}(x)\). It does not require the function to be twice differentiable. Twice differentiability is associated with the Second Derivative Test. Hence Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A → Continuity is a standard condition.
Option C → Differentiability is required to evaluate \(f^{'}(x)\).
Option D → Sign analysis of \(f^{'}(x)\)is the core of the test.

🎯 Strategy

Used: Odd One Out

Application: Identify the condition unrelated to the First Derivative Test.

Final Logic: First Derivative Test does not need a second derivative.

🧠 Memory Trick

First Test → First Derivative Only

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

12 Let \(f^{'}(x)=-2\) on \(\left[0\ ,\ 3\right]\). Find the area of the region where f is decreasing, bounded by \(x=0\), \(x=3\), the curve \(y=f^{'}(x)\), and the x-axis.

A 3

B 4

C 5

D 6

💡 Short Explanation

Derivative graph is \(y=-2\). Area forms a rectangle. Area = length × height.

✅ Correct Answer Explanation

Since \(f^{'}(x)=-2\), the graph of the derivative is a horizontal line below the x-axis. The required area is \(Area=3\times ∣-2∣=6\) where 3 is the interval length and 2 is the vertical distance from the x-axis. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Uses incorrect dimensions.
Option B → Does not equal rectangle area.
Option C → Miscalculates the product.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Compute geometric area directly.

Final Logic: Area = \(3\times 2=6\).

🧠 Memory Trick

Rectangle Area = Base × Height

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Area Under Curves

13 The cubic function \(f(x)=x^{3}-3x^{2}+4x\) is strictly decreasing in:

A \(\left(0\ ,\ 2\right)\)

B \(\left(-\infty ,0\right)\)

C \(\left(2\ ,\ \infty \right)\)

D None

💡 Short Explanation

Find the derivative. Check whether it becomes negative. Determine decreasing intervals.

✅ Correct Answer Explanation

\(f^{'}(x)=3x^{2}-6x+4=3(x-1)^{2}+1\) Since \(\left(x-1)^{2}\geq 0\right.\), \(f^{'}(x)>0\) for all real x. Thus the function is strictly increasing everywhere and never decreasing. Therefore Option D is correct. The provided answer A is incorrect.

❓ Why Other Options Are Incorrect

Option A → Derivative remains positive on \(\left(0\ ,\ 2\right)\).
Option B → Function is increasing there.
Option C → Function is also increasing there.

🎯 Strategy

Used: Substitution

Application: Analyze the sign of the derivative.

Final Logic: Positive derivative everywhere ⇒ no decreasing interval.

🧠 Memory Trick

Square + Positive Constant > 0

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

14 For \(f(x)=x^{3}-3x^{2}+4x\), the sum of its critical points is:

A 1

B 3

C 2

D 0

💡 Short Explanation

Critical points occur where derivative is zero. Check discriminant of derivative. Determine whether critical points exist.

✅ Correct Answer Explanation

\(f^{'}(x)=3x^{2}-6x+4\) The discriminant is \((-6)^{2}-4(3)(4)=36-48=-12\) Since the discriminant is negative, \(f^{'}(x)\)has no real roots. Hence there are no real critical points. The sum of real critical points is therefore 0. Option D is correct.

❓ Why Other Options Are Incorrect

Option A → No real critical point exists.
Option B → Based on incorrect roots.
Option C → Assumes two real critical points.

🎯 Strategy

Used: Substitution

Application: Examine roots of the derivative.

Final Logic: No real critical points ⇒ sum = 0.

🧠 Memory Trick

Negative Discriminant ⇒ No Real Critical Points

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

15 In which interval is \(f(x)=cos⁡x\) strictly decreasing?

A \(\left(-\pi ,0\right)\)

B \(\left(0\ ,\ \pi \right)\)

C \(\left(\pi \ ,\ 2\pi \right)\)

D \(\left(0\ ,\ 2\pi \right)\)

💡 Short Explanation

Differentiate cos x. Check sign of \(-sin⁡x\). Negative derivative implies decrease.

✅ Correct Answer Explanation

\(f^{'}(x)=-sin⁡x\) On \(\left(0\ ,\ \pi \right)\), \(sin⁡x>0\), so \(f^{'}(x)<0\). Hence \(\cos\,x\) decreases throughout this interval. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Derivative is positive there.
Option C → Function increases there.
Option D → Contains both increasing and decreasing parts.

🎯 Strategy

Used: Elimination

Application: Analyze derivative sign interval-wise.

Final Logic: \(-sin⁡x<0\) on \(\left(0\ ,\ \pi \right)\).

🧠 Memory Trick

cos Falls from 0 to π

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

16 For \(f(x)=sin⁡(2x)\), the function is strictly decreasing in:

A \(\left(0\ ,\ \pi /2\right)\)

B \(\left(\pi /2\ ,\ \pi \right)\)

C \(\left(\pi \ ,\ 3\pi /2\right)\)

D \(\left(\pi /4\ ,\ 3\pi /4\right)\)

💡 Short Explanation

Differentiate \(sin⁡(2x)\). Determine where derivative is negative. Match interval.

✅ Correct Answer Explanation

\(f^{'}(x)=2cos⁡(2x)\) The function decreases when \(cos⁡(2x)<0\) which occurs for \(\frac{\pi }{2}<2x<\frac{3\pi }{2}\) Dividing by 2, \(\frac{\pi }{4}<x<\frac{3\pi }{4}\) Thus Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Contains both positive and negative derivative regions.
Option B → Only part of a decreasing interval.
Option C → Includes increasing portions.

🎯 Strategy

Used: Substitution

Application: Differentiate and solve the inequality.

Final Logic: \(2cos⁡(2x)<0\).

🧠 Memory Trick

sin(2x) ↓ on (π/4, 3π/4)

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

17 For \(f(x)=x^{2}-4x\), the critical point is:

A 2

B 0

C -2

D 4

💡 Short Explanation

Differentiate the function. Set derivative equal to zero. Solve for x.

✅ Correct Answer Explanation

\(f^{'}(x)=2x-4\) Setting \(2x-4=0\) gives \(x=2\) Therefore the critical point is 2, making Option A correct.

❓ Why Other Options Are Incorrect

Option B → Derivative equals −4.
Option C → Not a solution.
Option D → Derivative equals 4.

🎯 Strategy

Used: Substitution

Application: Solve \(f^{'}(x)=0\).

Final Logic: Critical point occurs at x = 2.

🧠 Memory Trick

2x − 4 = 0 ⇒ x = 2

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

18 For \(f(x)=x^{2}-4x\), the function is increasing in:

A \(\left(-\infty ,2\right)\)

B \(\left(2\ ,\ \infty \right)\)

C \(\left(-\infty ,\infty \right)\)

D None

💡 Short Explanation

Check derivative sign. Positive derivative implies increase. Determine interval.

✅ Correct Answer Explanation

\(f^{'}(x)=2x-4\) For \(x>2\), \(2x-4>0\) Hence the function is increasing on \(\left(2\ ,\ \infty \right)\). Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Derivative is negative there.
Option C → Function is not increasing everywhere.
Option D → An increasing interval exists.

🎯 Strategy

Used: Substitution

Application: Analyze derivative sign.

Final Logic: \(f^{'}(x)>0\) for \(x>2\).

🧠 Memory Trick

Right of 2 ⇒ Increasing

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Sign Chart Method

📄 Passage

Consider \(f(x)=x^{3}-3x^{2}+4x\). Then

What is the sign of \(f^{'}(x)\)in the final interval?

A Negative

B Zero

C Positive

D Undefined

💡 Short Explanation

Analyze derivative expression. Derivative is always positive. Final interval therefore has positive sign.

✅ Correct Answer Explanation

\(f^{'}(x)=3x^{2}-6x+4=3(x-1)^{2}+1\) which is always positive. Therefore every interval, including the final interval, has positive derivative. Option C is correct. The passage statement about sign alternation is mathematically incorrect.

❓ Why Other Options Are Incorrect

Option A → Derivative never becomes negative.
Option B → Derivative never equals zero.
Option D → Derivative is defined everywhere.

🎯 Strategy

Used: Substitution

Application: Simplify derivative and inspect sign.

Final Logic: \(3(x-1)^{2}+1>0\).

🧠 Memory Trick

Positive Square Form

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

📄 Passage

Consider \(f(x)=x^{3}-3x^{2}+4x\). Then

The total number of turning points in the open interval is:

A 0

B 1

C 3

D 2

💡 Short Explanation

Turning points require critical points. Derivative never becomes zero. No turning points exist.

✅ Correct Answer Explanation

Since \(f^{'}(x)=3x^{2}-6x+4=3(x-1)^{2}+1>0\) for all x, the derivative never vanishes and never changes sign. Therefore the function is strictly increasing everywhere and has no turning points. The provided answer D is incorrect. The correct answer is Option A.

❓ Why Other Options Are Incorrect

Option B → Requires one sign change.
Option C → Impossible for this cubic.
Option D → Requires two critical points.

🎯 Strategy

Used: Substitution

Application: Check whether the derivative has real zeros.

Final Logic: No critical points ⇒ no turning points.

🧠 Memory Trick

No Critical Point = No Turning Point

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

CUET UG Mathematics Booster Test 2 - Increasing and Decreasing Functions

Test Complete!

Answer Review