CUET UG Mathematics Booster Test 1 - Introduction and Basic Concepts -

🚀

UG Mathematics Booster Test 1 - Introduction and Basic Concepts

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 1 - Introduction and Basic Concepts

1 / 20

QUESTION 1 OF 20

The concept of a mathematical relation is generalized from which of the following real-world applications?

Answer Review

1 The concept of a mathematical relation is generalized from which of the following real-world applications?

A Physics principles of energy transfer

B English language recognizable connections between objects

C Geometric axioms

D Chemical atomic bonding

💡 Short Explanation

Relation idea comes from everyday language Mathematical relations express connections NCERT introduces relation concept similarly

✅ Correct Answer Explanation

The mathematical idea of relation originates from recognizable connections between objects in ordinary English language usage. NCERT explains that mathematical relations abstract real-life associations into ordered pairs and subsets of Cartesian products. Hence Option B is correct. Physics, geometry, and chemistry are unrelated to the conceptual origin of relations.

❓ Why Other Options Are Incorrect

Option A → Energy transfer belongs to physics concepts, not set-theoretic relation origin.
Option C → Geometric axioms define geometry foundations, not relations.
Option D → Atomic bonding is a chemistry concept unrelated to mathematical relation definitions.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Match the conceptual origin described in NCERT with real-world language usage.

Final Logic:

Relations mathematically model recognizable connections between objects.

🧠 Memory Trick

"Relation = Connection"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Meaning of Relation

2 Let A = {1, 2}. A subset of A × A is chosen at random to form a relation. What is the probability that the chosen relation is the universal relation?

A 1/4

B 1/8

C 1/16

D 1/2

💡 Short Explanation

A × A has 4 elements Total relations = 2⁴ = 16 Only one universal relation exists

✅ Correct Answer Explanation

Since A = {1,2}, the Cartesian product A × A contains 4 ordered pairs. Total possible relations are subsets of A × A, equal to 2⁴ = 16. The universal relation contains all ordered pairs and is unique. Therefore probability = 1/16. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Assumes incorrect total number of relations.
Option B → Uses 8 total relations instead of 16.
Option D → Universal relation is only one among all possible subsets.

🎯 Strategy

Used: Substitution

Application:

Substitute number of elements into relation-count formula 2^(n²).

Final Logic:

Only one universal relation exists among 16 possible relations.

🧠 Memory Trick

"Universal = Full set only once"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Number of Relations

3 Roster form lists elements explicitly. Which of the sets below correctly represent the relation R = {(x,y) : x+y=5, x,y ∈ N}?
I. {(1,4), (2,3)}
II. {(3,2), (4,1)}
III. {(0,5)}

A I only

B II only

C I and III

D I and II

💡 Short Explanation

Natural numbers exclude 0 here Pairs must satisfy x+y=5 Both I and II satisfy condition

✅ Correct Answer Explanation

For x,y ∈ N and x+y=5, valid ordered pairs include (1,4), (2,3), (3,2), and (4,1). Thus Statements I and II are correct. Statement III contains 0, which is not taken as a natural number in standard NCERT convention. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores valid ordered pairs in Statement II.
Option B → Ignores valid ordered pairs in Statement I.
Option C → Includes Statement III, which violates natural number condition.

🎯 Strategy

Used: Elimination

Application:

Check each ordered pair against the condition x+y=5 and natural number restriction.

Final Logic:

Only Statements I and II fully satisfy the relation.

🧠 Memory Trick

"Sum 5, no zero"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Roster Form

4 Which is an incorrect characteristic of the set-builder form of a relation?

A It always lists every single ordered pair explicitly without conditions.

B It defines the mathematical rule connecting the components of ordered pairs.

C It uses a colon or vertical bar to mean "such that".

D It is particularly useful for representing infinite sets.

💡 Short Explanation

Set-builder uses rules, not full listing Symbols mean "such that" Useful for infinite relations

✅ Correct Answer Explanation

Set-builder form describes a relation using a condition or rule instead of explicitly listing all ordered pairs. Hence Option A is incorrect. Options B and C correctly describe notation and purpose. Option D is also correct because infinite sets are conveniently represented using conditions rather than exhaustive listing.

❓ Why Other Options Are Incorrect

Option B → Correct because set-builder form specifies defining conditions.
Option C → Correct since ":" or "|" means "such that."
Option D → Infinite sets are efficiently represented through rules.

🎯 Strategy

Used: Extreme Word Filter

Application:

The word "always" signals an overly broad and incorrect statement.

Final Logic:

Explicit listing belongs to roster form, not set-builder form.

🧠 Memory Trick

"Builder uses rules, not lists"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Set-builder Form

5 Match the given relations to their corresponding domains (assuming natural numbers):

List I	List II
1. {(1,2), (3,4)}	a. {1, 2}
2. {(x,y) : y = x, x < 3}	b. {2}
3. {(2,1), (2,3)}	c. {1, 3}
4. y = x², for x ∈ {4}	d. {4}

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

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

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

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

💡 Short Explanation

Domain contains first elements Extract x-values from relations Match domains carefully

✅ Correct Answer Explanation

For Relation 1, domain = {1,3} → c. For Relation 2, x < 3 in natural numbers gives {1,2} → a. For Relation 3, only first element is 2 → b. For Relation 4, x belongs to {4} → d. Thus Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrectly assigns domain of Relation 1.
Option C → Misidentifies domain of Relation 2.
Option D → Completely mismatches domains for Relations 1 and 4.

🎯 Strategy

Used: Option Grouping

Application:

Identify first coordinates systematically and match them with domain sets.

Final Logic:

Domain always consists of all first components.

🧠 Memory Trick

"Domain = Left side values"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Domain

6 A chemical mixture problem defines a relation between time (x) and concentration percentage (y):
R = {(1, 10), (2, 20), (3, 30)}. If the range values represent proportions out of 100, what is the average of the range elements?

A 10

B 30

C 20

D 60

💡 Short Explanation

Range values are 10, 20, 30 Average = sum ÷ number Result equals 20

✅ Correct Answer Explanation

The range of the relation consists of second coordinates: 10, 20, and 30. Their average is: (10+20+30)/3 = 60/3 = 20. Hence Option C is correct. Other options either represent totals or individual values instead of the arithmetic mean.

❓ Why Other Options Are Incorrect

Option A → Represents smallest range value only.
Option B → Represents largest range value only.
Option D → Equals total sum, not average.

🎯 Strategy

Used: Substitution

Application:

Extract range values and apply average formula directly.

Final Logic:

Average = (10+20+30)/3 = 20.

🧠 Memory Trick

"Add outputs, divide count"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Range

7 Assertion (A): The relation R = {(1,2), (1,3)} is a valid function.
Reason (R): In a function, an element in the domain cannot map to multiple distinct elements in the codomain.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Input 1 has two outputs Functions allow unique image only Reason correctly states function rule

✅ Correct Answer Explanation

The relation {(1,2),(1,3)} is not a function because the element 1 maps to two different outputs. Therefore Assertion is false. The Reason is true because a function requires each domain element to have exactly one image in the codomain. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Reason is mathematically correct.
Option B → Assertion is false since uniqueness condition fails.
Option C → Assertion itself is incorrect, so explanation cannot hold.

🎯 Strategy

Used: Elimination

Application:

Check whether any input corresponds to more than one output.

Final Logic:

One input with two outputs violates function definition.

🧠 Memory Trick

"One input → one output"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Definition of Function

8 Given f(x) = x² for x = 1, 2, 3, 4, 5. The sequential output values are 1, 4, 9, 16, 25. Calculate the first three 3-period moving averages. What is the lowest moving average among them?

A 14/3

B 29/3

C 15/3

D 50/3

💡 Short Explanation

Moving averages use consecutive triples Averages are 14/3, 29/3, and 50/3 Lowest is 14/3

✅ Correct Answer Explanation

The outputs are 1, 4, 9, 16, 25. First 3-period average = (1+4+9)/3 = 14/3. Second = (4+9+16)/3 = 29/3. Third = (9+16+25)/3 = 50/3. The lowest moving average is 14/3. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Represents the second moving average.
Option C → Incorrect arithmetic calculation.
Option D → Represents the third moving average.

🎯 Strategy

Used: Substitution

Application:

Compute consecutive averages carefully using grouped outputs.

Final Logic:

Lowest among computed moving averages is 14/3.

🧠 Memory Trick

"First window gives smallest average"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Real Valued Functions

9 What geometric boundary is formed by the graph of the relation |x| + |y| = 1?

A Circle

B Square

C Triangle

D Parabola

💡 Short Explanation

Absolute value creates straight segments Boundary joins four vertices Resulting figure is a square

✅ Correct Answer Explanation

The equation |x| + |y| = 1 forms four linear boundaries in different quadrants. The graph connects points (1,0), (0,1), (-1,0), and (0,-1), producing a square rotated by 45°. Therefore Option B is correct. It is neither circular nor parabolic in shape.

❓ Why Other Options Are Incorrect

Option A → Circle follows x²+y²=r², not absolute value equation.
Option C → Figure has four vertices, not three.
Option D → Parabola arises from quadratic equations, not modulus sums.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret modulus equation geometrically by examining quadrant-wise linear behavior.

Final Logic:

Four straight boundary segments form a square.

🧠 Memory Trick

"|x|+|y| gives diamond square"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Graphs of Relations

10 Arrange the analytical steps to visually verify if a mapping diagram from set X to set Y represents a function:
1. Check if any domain element has no arrow originating from it.
2. Identify the domain set X.
3. Conclude it is a function if each element in X has exactly one arrow.
4. Check if any domain element has more than one arrow.

A 2, 1, 3, 4

B 1, 2, 4, 3

C 2, 1, 4, 3

D 4, 3, 2, 1

💡 Short Explanation

Start with identifying domain Verify arrow conditions carefully Conclude function at end

✅ Correct Answer Explanation

To verify a function visually, first identify the domain set X. Next check whether any domain element lacks an arrow. Then verify that no domain element has multiple arrows. Finally conclude whether each element has exactly one image. Thus the correct order is 2 → 1 → 4 → 3, making Option C correct.

❓ Why Other Options Are Incorrect

Option A → Conclusion cannot be reached before checking multiple arrows.
Option B → Domain identification should come first logically.
Option D → Conclusion must always appear after verification steps.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Arrange steps according to logical mathematical verification flow.

Final Logic:

Identify → verify arrows → conclude function status.

🧠 Memory Trick

"Check domain, then arrows"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Mapping Diagrams

📄 Passage

A relation R in a set A is called reflexive, if (a, a) ∈ R, for every a ∈ A; symmetric, if (a1, a2) ∈ R implies that (a2, a1) ∈ R; and transitive, if (a1, a2) ∈ R and (a2, a3) ∈ R implies that (a1, a3) ∈ R. An equivalence relation must satisfy all three.

Based on the passage, the family relation R = {(a, b): a is the sibling of b} on the set of all children in a family is:

A Reflexive only

B Transitive only

C Asymmetric

D Symmetric

💡 Short Explanation

Sibling relation works mutually If a is sibling of b, reverse also true Relation is symmetric in nature

✅ Correct Answer Explanation

The sibling relation is symmetric because if a is a sibling of b, then b is also a sibling of a. Hence Option D is correct. It is not reflexive since a person is not their own sibling. It is also not asymmetric because symmetry clearly exists.

❓ Why Other Options Are Incorrect

Option A → Sibling relation is not reflexive because no person is sibling of themselves.
Option B → Sibling relation alone does not establish transitivity strictly in relation theory.
Option C → Asymmetric relations cannot be symmetric simultaneously.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret family relation properties using the passage definitions directly.

Final Logic:

Sibling relation satisfies mutual connection, hence symmetric.

🧠 Memory Trick

"Sibling works both ways"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Symmetric Relation

📄 Passage

Based on the passage, the numerical relation "x is less than or equal to y" on Real numbers is:

A Reflexive and transitive

B Symmetric only

C Reflexive and symmetric

D Transitive and symmetric

💡 Short Explanation

x ≤ x always true Order relation preserves transitivity Reverse condition need not hold

✅ Correct Answer Explanation

The relation x ≤ y is reflexive because every number satisfies x ≤ x. It is transitive because x ≤ y and y ≤ z imply x ≤ z. However, it is not symmetric since x ≤ y does not imply y ≤ x unless x = y. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Relation is not symmetric in general.
Option C → Symmetry fails whenever x < y.
Option D → Symmetry is not satisfied for order relations.

🎯 Strategy

Used: Elimination

Application:

Check each property individually using standard inequality examples.

Final Logic:

≤ is reflexive and transitive but not symmetric.

🧠 Memory Trick

"Less-than keeps order forward only"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Reflexive and Transitive Relations

13 An ordered pair (x, y) can represent a 2D vector. If the relation maps any vector v to a new vector 2v, what is the resulting image of the vector represented by (3, -1)?

A (3, -2)

B (6, -2)

C (1.5, -0.5)

D (-6, 2)

💡 Short Explanation

Multiply each component by 2 Vector scaling doubles coordinates Result becomes (6, -2)

✅ Correct Answer Explanation

The transformation v → 2v doubles every component of the vector. For vector (3,−1), multiplying each coordinate by 2 gives (6,−2). Therefore Option B is correct. Other options either partially scale coordinates or incorrectly change signs.

❓ Why Other Options Are Incorrect

Option A → Only y-coordinate is doubled incorrectly.
Option C → Represents halving instead of doubling.
Option D → Changes sign unnecessarily while scaling.

🎯 Strategy

Used: Substitution

Application:

Apply scalar multiplication directly to vector coordinates.

Final Logic:

2(3,−1) = (6,−2).

🧠 Memory Trick

"Scalar multiplies every coordinate"

📚 Class XI – NCERT Mathematics – Chapter: Introduction to Three Dimensional Geometry – Topic: Vector Representation

14 If the relation notation aRb mathematically signifies that b = 4a³, what is the area under the curve formed by this relation evaluated from a = 0 to a = 1?

A 2

B 3

C 1

D 4

💡 Short Explanation

Integrate 4a³ from 0 to 1 Antiderivative becomes a⁴ Final value equals 1

✅ Correct Answer Explanation

The required area is: \(\int_{0}^{1}\,4a^{3} da\) Integrating gives a⁴ evaluated from 0 to 1: 1⁴ − 0⁴ = 1. Hence Option C is correct. Other options arise from incorrect integration rules or coefficient handling.

❓ Why Other Options Are Incorrect

Option A → Incorrect evaluation after integration.
Option B → Results from arithmetic error.
Option D → Equals coefficient only, not integral value.

🎯 Strategy

Used: Substitution

Application:

Replace b with 4a³ and apply definite integration directly.

Final Logic:

∫₀¹4a³da = [a⁴]₀¹ = 1.

🧠 Memory Trick

"Power +1 cancels coefficient"

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Definite Integrals

15 A universal relation on the set of real numbers restricted from 0 to 2 is the Cartesian product ×. What is the area of this region in the coordinate plane?

A 2

B 8

C 6

D 4

💡 Short Explanation

Cartesian product forms square region Side length equals 2 Area = 2 × 2 = 4

✅ Correct Answer Explanation

The universal relation on [0,2] consists of all ordered pairs (x,y) where both coordinates lie between 0 and 2. This forms a square region in the coordinate plane with side length 2. Therefore area = 2 × 2 = 4. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Represents only one dimension length.
Option B → Incorrect multiplication beyond square area.
Option C → Does not correspond to Cartesian square geometry.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Interpret Cartesian product geometrically as a rectangular region.

Final Logic:

[0,2] × [0,2] forms square of area 4.

🧠 Memory Trick

"Cartesian square → side²"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Cartesian Product

16 Match the function f: R → R with its conceptual type:

List I	List II
1. f(x) = x	a. One-one and onto
2. f(x) = x²	b. Many-one, not onto
3. f(x) = 3	c. Many-one (Constant)
4. f(x) = e^x	d. One-one, not onto

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

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

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

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

💡 Short Explanation

Identity function is bijective x² is many-one and not onto e^x is one-one but not onto R

✅ Correct Answer Explanation

f(x)=x is both one-one and onto → a. f(x)=x² maps x and −x together and misses negatives → b. f(x)=3 is constant and many-one → c. f(x)=e^x is injective but its range excludes negative reals → d. Thus Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrectly classifies exponential function as onto R.
Option C → x² is not one-one and onto over R.
Option D → Constant function is not onto over real numbers.

🎯 Strategy

Used: Option Grouping

Application:

Compare injective and surjective properties of standard functions.

Final Logic:

Each function matches its textbook mapping behavior.

🧠 Memory Trick

"Identity bijective, square many-one"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Types of Functions

17 Assertion (A): In general function compositions, fog is not equal to gof.
Reason (R): Composition of functions is always a commutative operation.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Function composition order matters Composition is generally non-commutative Reason contradicts actual property

✅ Correct Answer Explanation

In general, fog ≠ gof because the order of applying functions affects outputs. Hence Assertion is true. The Reason is false because function composition is not commutative in general. Therefore Option B is correct according to standard properties of composite functions.

❓ Why Other Options Are Incorrect

Option A → Assertion is mathematically correct.
Option C → Reason itself is false and cannot explain Assertion.
Option D → Composition usually depends on order, so Assertion is true.

🎯 Strategy

Used: Extreme Word Filter

Application:

The word "always" signals incorrect universal generalization.

Final Logic:

Composition is generally non-commutative.

🧠 Memory Trick

"Order matters in composition"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Composition of Functions

18 Which of the following conditions must hold for a function f to have a valid inverse?
I. f must be injective.
II. f must be a polynomial.
III. f must be surjective.

A I only

B II and III

C I and III

D III only

💡 Short Explanation

Inverse requires bijection Bijection means injective and surjective Polynomial condition unnecessary

✅ Correct Answer Explanation

A function has a valid inverse only if it is bijective. Bijective means both injective (one-one) and surjective (onto). Thus Statements I and III are necessary. Statement II is not required because invertible functions need not be polynomials. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Injectivity alone does not guarantee inverse over codomain.
Option B → Polynomial form is unrelated to invertibility requirement.
Option D → Surjectivity alone cannot ensure uniqueness of inverse.

🎯 Strategy

Used: Elimination

Application:

Use the theorem: invertible ⇔ bijective.

Final Logic:

Inverse exists only for one-one and onto functions.

🧠 Memory Trick

"Inverse needs bijection"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Invertible Functions

19 Arrange the analytical process of formalizing a real-life relation into mathematical notation:
1. Define the sets involved (e.g., set of people A).
2. Represent the set as R subset of A × A.
3. Identify the real-life link (e.g., "is a friend of").
4. Write out the specific ordered pairs (x, y) fulfilling the link.

A 1, 2, 3, 4

B 3, 4, 1, 2

C 2, 1, 4, 3

D 3, 1, 4, 2

💡 Short Explanation

Begin with real-life connection Define involved sets next Then form ordered pairs and relation

✅ Correct Answer Explanation

To formalize a relation, first identify the real-life link. Then define the relevant sets involved. Next list ordered pairs satisfying the relation. Finally express the relation formally as a subset of Cartesian product. Hence the correct order is 3 → 1 → 4 → 2, making Option D correct.

❓ Why Other Options Are Incorrect

Option A → Formal notation appears before identifying the relation meaning.
Option B → Sets should be defined before writing ordered pairs.
Option C → Cartesian representation cannot precede relation formation steps.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Arrange steps according to natural mathematical modeling process.

Final Logic:

Identify relation → define set → form pairs → formalize subset.

🧠 Memory Trick

"Link → Set → Pairs → Relation"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Real-life Relations

20 Which is an incorrect statement about abstract equivalence relations?

A They divide a set into overlapping, non-disjoint subsets.

B They divide a set into mutually disjoint equivalence classes.

C The union of all equivalence classes is the original set.

D Elements in the same equivalence class are related to each other.

💡 Short Explanation

Equivalence classes are disjoint Their union forms original set Related elements belong together

✅ Correct Answer Explanation

Equivalence relations partition a set into mutually disjoint equivalence classes. Therefore Option A is incorrect because equivalence classes never overlap. Options B, C, and D correctly describe partition properties and relation behavior within the same equivalence class.

❓ Why Other Options Are Incorrect

Option B → Correct property of equivalence class partitioning.
Option C → Entire set is covered by equivalence classes together.
Option D → Members of same class satisfy the equivalence relation.

🎯 Strategy

Used: Odd One Out

Application:

Identify the statement contradicting partition properties of equivalence relations.

Final Logic:

Equivalence classes are always disjoint, never overlapping.

🧠 Memory Trick

"Equivalence classes never overlap"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Equivalence Relations

UG Mathematics Booster Test 1 - Introduction and Basic Concepts

Test Complete!

Answer Review