CUET UG Mathematics Booster Test 2 - Introduction and Basic Concepts -

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UG Mathematics Booster Test 2 - Introduction and Basic Concepts

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 2 - Introduction and Basic Concepts

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QUESTION 1 OF 20

Arrange the following relations from most restrictive (containing the fewest ordered pairs) to least restrictive on the set A = {1, 2, 3}:
1. Universal relation
2. Empty relation
3. Identity relation
4. R = {(1,2), (2,1), (1,1), (2,2)}

Answer Review

1 Arrange the following relations from most restrictive (containing the fewest ordered pairs) to least restrictive on the set A = {1, 2, 3}:
1. Universal relation
2. Empty relation
3. Identity relation
4. R = {(1,2), (2,1), (1,1), (2,2)}

A 1, 4, 3, 2

B 2, 4, 3, 1

C 2, 3, 4, 1

D 3, 2, 1, 4

💡 Short Explanation

Empty relation has 0 pairs Identity relation has 3 pairs Universal relation has maximum pairs

✅ Correct Answer Explanation

For A = {1,2,3}, universal relation contains 9 ordered pairs, identity relation contains 3 pairs, and Relation 4 contains 4 pairs. Empty relation contains no ordered pairs. Therefore ordering from fewest to most pairs is: Empty relation → Identity relation → Relation 4 → Universal relation. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Starts with universal relation, which has maximum ordered pairs.
Option B → Relation 4 has more pairs than identity relation.
Option D → Identity relation cannot come before empty relation.

🎯 Strategy

Used: Option Grouping

Application:

Count ordered pairs in each relation and arrange numerically.

Final Logic:

0 < 3 < 4 < 9 ordered pairs.

🧠 Memory Trick

"Empty < Identity < Partial < Universal"

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

2 Set A has exactly 3 elements. A relation on A is picked at random from all possible relations. What is the probability that the picked relation contains the element (1,1)?

A 1/512

B 1/8

C 1/4

D 1/2

💡 Short Explanation

A × A has 9 ordered pairs Half of all subsets contain (1,1) Probability equals 1/2

✅ Correct Answer Explanation

A set with 3 elements gives 3×3 = 9 ordered pairs in A × A. Total relations are 2⁹. For exactly half of these relations, the pair (1,1) is included. Thus favorable relations = 2⁸. Hence probability = 2⁸/2⁹ = 1/2. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Uses total relations incorrectly as probability.
Option B → Assumes only one ordered pair selection process.
Option C → Incorrect subset-counting logic.

🎯 Strategy

Used: Elimination

Application:

Recognize that inclusion/exclusion of one fixed element splits subsets equally.

Final Logic:

Any specific ordered pair appears in exactly half the relations.

🧠 Memory Trick

"One pair appears in half the subsets"

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

3 Which is incorrect about converting the relation R = {(x, y) : x² + y² = 25, x, y ∈ Z} to roster form?

A It contains the ordered pair (5, 5).

B It contains a total of 12 elements.

C It contains both (3, 4) and (-3, 4).

D It contains both (0, 5) and (5, 0).

💡 Short Explanation

(5,5) does not satisfy equation 5² + 5² = 50 Valid pairs satisfy total 25 only

✅ Correct Answer Explanation

For the relation x² + y² = 25, the pair (5,5) gives 25 + 25 = 50, not 25. Hence Option A is incorrect. Valid integer solutions include (±3,±4), (±4,±3), (±5,0), and (0,±5), totaling 12 ordered pairs. Therefore other options are correct.

❓ Why Other Options Are Incorrect

Option B → Correct because total integer solutions equal 12 ordered pairs.
Option C → Both ordered pairs satisfy 3² + 4² = 25.
Option D → Both pairs satisfy 0² + 5² = 25.

🎯 Strategy

Used: Substitution

Application:

Substitute coordinates into the equation x²+y²=25 directly.

Final Logic:

(5,5) violates the defining equation.

🧠 Memory Trick

"Check equation before accepting pair"

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

4 Which of these set-builder notations define a proper equivalence relation on the set of integers?
I. R = {(a, b) : a - b is an integer}
II. R = {(a, b) : a + b is even}
III. R = {(a, b) : a < b}

A I only

B I and II

C II and III

D I, II, and III

💡 Short Explanation

Equivalence needs reflexive, symmetric, transitive I and II satisfy all properties Less-than is not symmetric

✅ Correct Answer Explanation

Statement I is always true for integers, so it forms an equivalence relation. Statement II groups integers by parity and satisfies reflexive, symmetric, and transitive properties. Statement III fails reflexivity and symmetry because a < a is false and reverse order need not hold. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores Statement II, which is also equivalence relation.
Option C → Statement III is not an equivalence relation.
Option D → Less-than relation violates equivalence properties.

🎯 Strategy

Used: Elimination

Application:

Test each relation against reflexive, symmetric, and transitive conditions.

Final Logic:

Only Statements I and II satisfy all equivalence requirements.

🧠 Memory Trick

"Equivalence = RST (Reflexive, Symmetric, Transitive)"

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

5 Match the real-valued function to its maximum possible domain in R:

List I	List II
1. f(x) = 1/x	a. x ≥ 0
2. f(x) = √x	b. x > 0
3. f(x) = 1/(x² - 1)	c. R - {0}
4. f(x) = ln(x)	d. R - {-1, 1}

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

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

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

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

💡 Short Explanation

Denominator cannot be zero Square root requires nonnegative input Logarithm requires positive input

✅ Correct Answer Explanation

For 1/x, x ≠ 0 → c. For √x, x ≥ 0 → a. For 1/(x²−1), denominator cannot vanish, so x ≠ ±1 → d. For ln(x), x > 0 → b. Thus matching is 1-c, 2-a, 3-d, 4-b, giving Option C.

❓ Why Other Options Are Incorrect

Option A → Logarithm domain is not all reals except zero.
Option B → √x allows 0, so x ≥ 0 is correct domain.
Option D → 1/x does not require x ≥ 0 only.

🎯 Strategy

Used: Elimination

Application:

Apply standard domain restrictions for roots, logarithms, and rational functions.

Final Logic:

Avoid zero denominators and invalid logarithm/root inputs.

🧠 Memory Trick

"Root ≥0, Log >0"

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

6 An EMI financial function maps loan tenure in years (x) to total interest percentage (y). If the function evaluates to y = 5x + 2, and the domain is {1, 2, 3, 4}, what is the maximum value in the range?

A 12

B 17

C 7

D 22

💡 Short Explanation

Substitute domain values in function Largest x gives largest y Maximum value equals 22

✅ Correct Answer Explanation

Evaluate y = 5x + 2 for x ∈ {1,2,3,4}: 7, 12, 17, 22. The largest value in the range is 22. Hence Option D is correct. Other options correspond to smaller outputs obtained from smaller domain values.

❓ Why Other Options Are Incorrect

Option A → Corresponds to x = 2 only.
Option B → Corresponds to x = 3 only.
Option C → Corresponds to x = 1 only.

🎯 Strategy

Used: Substitution

Application:

Substitute all domain values and compare outputs.

Final Logic:

Largest input gives maximum output here.

🧠 Memory Trick

"Check all outputs for range"

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

7 Assertion (A): A relation where one x-value maps to two different y-values is a valid mathematical function.
Reason (R): Functions inherently allow multiple outputs for a single input to model complex trajectories.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Functions require unique output One input cannot have two outputs Reason contradicts function definition

✅ Correct Answer Explanation

A function assigns exactly one output to each input. Therefore a relation where one x-value maps to multiple y-values is not a function, making Assertion false. The Reason is also false because functions never allow multiple outputs for a single input. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Assertion is false, not true.
Option C → Both statements contradict the definition of function.
Option D → Reason is also mathematically incorrect.

🎯 Strategy

Used: Extreme Word Filter

Application:

Check whether the statements violate the uniqueness rule of functions.

Final Logic:

Functions always assign exactly one image to each input.

🧠 Memory Trick

"One input, one output only"

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

8 A real valued function generates sequence values 2, 6, 12, 20. The first difference between consecutive values generates a new analytical sequence. What is the 2-period moving average of the first two values of this new sequence?

A 8

B 5

C 6

D 4

💡 Short Explanation

First differences are 4, 6, 8 First two values are 4 and 6 Average equals 5

✅ Correct Answer Explanation

The first differences are: 6−2 = 4, 12−6 = 6, 20−12 = 8. The first two values of the new sequence are 4 and 6. Their 2-period moving average is (4+6)/2 = 5. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Sum of first two differences, not average.
Option C → Second difference value only.
Option D → First difference value only.

🎯 Strategy

Used: Substitution

Application:

Compute difference sequence first, then apply moving average formula.

Final Logic:

(4+6)/2 = 5.

🧠 Memory Trick

"Difference first, average next"

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

9 The graph of relation R is defined by the inequality x² + y² < 1. This means R analytically represents:

A A standard function mapping.

B The strict boundary of a unit circle.

C The interior region of a unit circle.

D The exterior region of a unit circle.

💡 Short Explanation

x²+y²=1 gives unit circle boundary Strict inequality means inside region Boundary excluded from graph

✅ Correct Answer Explanation

The equation x²+y²=1 represents the unit circle boundary. Since the inequality is x²+y²<1, all points strictly inside the circle are included while boundary points are excluded. Therefore Option C is correct. Options involving only boundary or exterior regions are incorrect.

❓ Why Other Options Are Incorrect

Option A → Relation describes a region, not a standard function.
Option B → Boundary requires equality, not strict inequality.
Option D → Exterior region corresponds to x²+y²>1.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret inequality signs geometrically in coordinate plane.

Final Logic:

"Less than" indicates interior of the circle.

🧠 Memory Trick

"< means inside circle"

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

10 Match the mapping diagram scenarios to their rigorous relation properties:

List I	List II
1. Every x points to exactly one distinct y, and all y are covered.	a. Not a function
2. Every x points to exactly one distinct y, but some y are not covered.	b. Many-one function
3. Multiple x elements point to the same y.	c. Bijective function
4. One x element points to multiple y elements.	d. Injective but not surjective

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

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

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

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

💡 Short Explanation

Bijective means one-one and onto Injective may miss some outputs Multiple outputs violate function rule

✅ Correct Answer Explanation

Statement 1 describes bijection → c. Statement 2 is injective but not surjective → d. Statement 3 describes many-one mapping → b. Statement 4 violates function definition because one input has multiple outputs → a. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrectly identifies bijection as not a function.
Option B → Many-one functions still remain valid functions.
Option C → Statement 2 is injective, not many-one.

🎯 Strategy

Used: Option Grouping

Application:

Match standard mapping behaviors with formal function classifications.

Final Logic:

One-one and onto → bijection; multiple outputs → not a function.

🧠 Memory Trick

"Bijective = one-one + onto"

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

11 For the family relation defined as "x is a descendant of y":
I. It is transitive.
II. It is not reflexive.
III. It is symmetric.

A I and II

B I only

C II and III

D I, II, and III

💡 Short Explanation

Descendant relation is transitive Nobody is own descendant Relation is not symmetric

✅ Correct Answer Explanation

The descendant relation is transitive because if x is a descendant of y and y is a descendant of z, then x is a descendant of z. It is not reflexive since no person is their own descendant. It is also not symmetric because descendant direction cannot reverse. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Ignores the correct non-reflexive property.
Option C → Symmetry does not hold for descendant relations.
Option D → Descendant relation is definitely not symmetric.

🎯 Strategy

Used: Elimination

Application:

Test descendant relation against standard relation properties individually.

Final Logic:

Transitive and non-reflexive are valid; symmetry fails.

🧠 Memory Trick

"Descendants move downward only"

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

12 The relation mapping is defined by the natural exponential function y = e^x. Find the integral of this function from 0 to ln(2).

A 2

B 1

C e

D 0

💡 Short Explanation

Integral of e^x is e^x Apply upper and lower limits Result equals 2 − 1 = 1

✅ Correct Answer Explanation

The required integral is: \(\int_{0}^{ln⁡(2)}\,e^{x} dx\) Since ∫e^x dx = e^x, evaluate between limits: e^(ln2) − e^0 = 2 − 1 = 1. Therefore Option B is correct. Other options arise from incomplete evaluation or misunderstanding exponential integration.

❓ Why Other Options Are Incorrect

Option A → Equals upper-limit exponential value only.
Option C → Incorrectly assumes upper limit equals 1.
Option D → Positive exponential function cannot give zero area here.

🎯 Strategy

Used: Substitution

Application:

Use standard exponential integration and substitute limits directly.

Final Logic:

e^(ln2) − 1 = 1.

🧠 Memory Trick

"Integral of e^x stays e^x"

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

13 An ordered pair (a,b) is a vector in 2D. Under relation R, vector v maps to vector w such that w is the orthogonal projection of v onto the x-axis. What does the pair (4, 5) map to?

A (0, 5)

B (5, 4)

C (4, 0)

D (0, 0)

💡 Short Explanation

Projection on x-axis removes y-coordinate x-coordinate remains unchanged Result becomes (4,0)

✅ Correct Answer Explanation

Orthogonal projection onto the x-axis keeps the x-coordinate unchanged while making the y-coordinate zero. Therefore vector (4,5) maps to (4,0). Hence Option C is correct. Other options either preserve the wrong coordinate or incorrectly interchange components.

❓ Why Other Options Are Incorrect

Option A → Represents projection onto y-axis instead.
Option B → Coordinates are incorrectly interchanged.
Option D → Zero vector occurs only when x-coordinate is also zero.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret geometric meaning of orthogonal projection directly.

Final Logic:

Projection onto x-axis keeps x and removes y.

🧠 Memory Trick

"x-axis projection kills y"

📚 Class XI – NCERT Mathematics – Chapter: Introduction to Three Dimensional Geometry – Topic: Coordinate Vectors

14 If aRb represents the set of points where a ≥ 0, b ≥ 0, and a + b ≤ 2, what is the area of the polygon formed by this restricted relation?

A 1

B 4

C 3

D 2

💡 Short Explanation

Region forms right triangle Base and height equal 2 Area = ½ × 2 × 2 = 2

✅ Correct Answer Explanation

The inequalities a ≥ 0, b ≥ 0, and a+b ≤ 2 form a triangular region bounded by coordinate axes and line a+b=2. The triangle has base 2 and height 2. Therefore area = (1/2)×2×2 = 2. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Uses incorrect dimensions for triangle.
Option B → Equals rectangle area, not triangular region.
Option C → Incorrect geometric calculation.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Interpret inequalities geometrically and apply area formula.

Final Logic:

Right triangle area equals 2 square units.

🧠 Memory Trick

"Half base × height"

📚 Class XI – NCERT Mathematics – Chapter: Linear Inequalities – Topic: Graphical Regions

📄 Passage

A function f: X → Y is defined to be invertible, if there exists a function g: Y → X such that gof = Ix and fog = Iy. The function g is called the inverse of f and is denoted by f⁻¹. Furthermore, if f is invertible, then f must necessarily be both one-one and onto.

Although the passage defines function invertibility, an inverse relation R⁻¹ always exists by simply reversing pairs. If R is a bijective function, what kind of relation is its inverse R⁻¹?

A A bijective function

B An empty relation

C A many-one function

D Not a function

💡 Short Explanation

Bijective functions are invertible Inverse reverses ordered pairs uniquely Inverse remains bijective function

✅ Correct Answer Explanation

A bijective function is both one-one and onto, so its inverse exists as a valid function. Reversing ordered pairs preserves uniqueness and surjectivity. Therefore the inverse relation R⁻¹ is also bijective. Hence Option A is correct. Other options contradict properties of invertible functions.

❓ Why Other Options Are Incorrect

Option B → Inverse of bijection cannot be empty.
Option C → Bijective inverse preserves one-one property.
Option D → Inverse of bijective function is always a function.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use invertibility conditions stated directly in the passage.

Final Logic:

Inverse of bijection remains bijection.

🧠 Memory Trick

"Bijection ↔ Invertible"

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

📄 Passage

Based on the provided passage, if a function is NOT onto (surjective), can it be invertible?

A Yes, if it is one-one.

B No, it must be both one-one and onto.

C Yes, if it is many-one.

D No, only continuous functions are invertible.

💡 Short Explanation

Invertible functions must be bijective Onto property is compulsory Non-surjective functions lack inverse

✅ Correct Answer Explanation

The passage explicitly states that invertible functions must be both one-one and onto. If a function is not onto, some codomain elements have no preimage, making inverse impossible. Therefore Option B is correct. Other options ignore the surjectivity requirement or add irrelevant continuity conditions.

❓ Why Other Options Are Incorrect

Option A → One-one alone is insufficient for invertibility.
Option C → Many-one functions cannot possess inverses.
Option D → Continuity is unrelated to basic invertibility condition here.

🎯 Strategy

Used: Elimination

Application:

Use the bijection condition directly from the passage.

Final Logic:

Invertibility requires both injective and surjective properties.

🧠 Memory Trick

"Inverse needs onto too"

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

17 Which analytical statement about the composition gof is incorrect?

A Its domain is the domain of f.

B Its codomain is the codomain of g.

C It is mathematically defined even if the range of f does not intersect the domain of g.

D gof(x) means applying function f first, then applying g to the result.

💡 Short Explanation

Composition needs compatible mappings Range of f must fit domain of g Otherwise composition undefined

✅ Correct Answer Explanation

For composition gof to exist, outputs of f must belong to the domain of g. If the range of f does not intersect the domain of g, composition cannot be defined. Therefore Option C is incorrect. Other options correctly describe properties and operational meaning of composite functions.

❓ Why Other Options Are Incorrect

Option A → Domain of gof originates from domain of f where composition exists.
Option B → Outputs of gof belong to codomain of g.
Option D → Function f acts first, then g on the result.

🎯 Strategy

Used: Elimination

Application:

Check whether composition requirements satisfy function compatibility.

Final Logic:

Composition needs range(f) ⊆ domain(g).

🧠 Memory Trick

"Output of f enters g"

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

18 Arrange the algebraic steps to find the inverse of the function f(x) = 2x + 3:
1. Replace y with x, and x with f⁻¹(x).
2. Set y = 2x + 3.
3. The final inverse is f⁻¹(x) = (x - 3)/2.
4. Solve for x algebraically in terms of y: x = (y - 3)/2.

A 2, 1, 4, 3

B 1, 2, 3, 4

C 4, 3, 2, 1

D 2, 4, 1, 3

💡 Short Explanation

Begin with y = f(x) Solve algebraically for x Interchange variables finally

✅ Correct Answer Explanation

To find inverse: first write y = 2x+3. Then solve for x: x = (y−3)/2. Next interchange variables to express inverse notation. Finally write: f⁻¹(x) = (x−3)/2. Hence correct order is 2 → 4 → 1 → 3, making Option D correct.

❓ Why Other Options Are Incorrect

Option A → Variable replacement occurs before solving algebraically.
Option B → Final inverse cannot appear before solving equation.
Option C → Starts from intermediate step without defining original function.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Follow standard inverse-function derivation procedure step-by-step.

Final Logic:

Define → solve → interchange → conclude inverse.

🧠 Memory Trick

"Write, solve, swap, finish"

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

19 In a network of computers, relation R represents "is connected to". If it strictly acts as an equivalence relation, what does a resulting equivalence class represent?

A A fully isolated sub-network where all computers are connected to each other.

B The central server connecting all nodes together.

C A completely broken connection.

D A single cable connecting two distinct computers.

💡 Short Explanation

Equivalence classes partition sets All members are mutually related Connected subgroup forms one class

✅ Correct Answer Explanation

Equivalence classes contain elements mutually related under an equivalence relation. Therefore a class here represents a subgroup of computers all connected within that network component. Hence Option A is correct. Other options describe specific devices or incomplete structures rather than equivalence classes.

❓ Why Other Options Are Incorrect

Option B → Equivalence classes are not defined by a central server.
Option C → Broken connections imply absence of relation.
Option D → One cable alone does not represent an entire equivalence class.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret equivalence classes through practical network grouping.

Final Logic:

Equivalence classes form mutually connected groups.

🧠 Memory Trick

"Same class = all connected"

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

20 Assertion (A): The relation "modulus of difference is a multiple of 4" on the set of integers is a valid equivalence relation.
Reason (R): Any relation mathematically based on the divisibility of a difference is only transitive, not reflexive.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Congruence modulo 4 is equivalence relation Difference divisibility gives reflexivity too Reason incorrectly denies reflexive property

✅ Correct Answer Explanation

The relation |a−b| divisible by 4 is equivalent to congruence modulo 4, which is reflexive, symmetric, and transitive. Hence Assertion is true. The Reason is false because divisibility-based difference relations are also reflexive since a−a=0 is divisible by 4. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Assertion correctly describes equivalence relation.
Option C → Reason is mathematically incorrect.
Option D → Divisibility-based congruence relations are valid equivalence relations.

🎯 Strategy

Used: Elimination

Application:

Check reflexive, symmetric, and transitive conditions systematically.

Final Logic:

Congruence modulo n always forms equivalence relation.

🧠 Memory Trick

"Difference divisible ⇒ congruence"

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

UG Mathematics Booster Test 2 - Introduction and Basic Concepts

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