CUET UG Categorized PYQ Mathematics Unit 1

📚

CUET UG Categorized PYQ Mathematics Unit 1

Mathematics

⏱️ 30 Minutes

❓ 12 Questions

📅 2022,2023&2024

✍️ Long Answers

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

CUET UG Categorized PYQ Mathematics Unit 1

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QUESTION 1 OF 12

Match List I with List II.

List I	List II
A. R = {(x, y): x and y are students of the same school}	I. Symmetric
B. R = {(L₁, L₂): L₁ ⊥ L₂, L₁, L₂ ∈ L}, where L is a set of all lines	II. One-one
C. A function f: ℝ → ℝ defined by f(x) = 2 - 3x	III. Bijective
D. A function f: ℝ⁺ → ℝ defined by f(x) = 1 + x²	IV. Equivalence

Answer Review

1 Match List I with List II.

List I	List II
A. R = {(x, y): x and y are students of the same school}	I. Symmetric
B. R = {(L₁, L₂): L₁ ⊥ L₂, L₁, L₂ ∈ L}, where L is a set of all lines	II. One-one
C. A function f: ℝ → ℝ defined by f(x) = 2 - 3x	III. Bijective
D. A function f: ℝ⁺ → ℝ defined by f(x) = 1 + x²	IV. Equivalence

A (A) - (I), (B) - (IV), (C) - (II), (D) - (III)

B (A) - (IV), (B) - (I), (C) - (III), (D) - (II)

C (A) - (I), (B) - (IV), (C) - (III), (D) - (II)

D (A) - (IV), (B) - (I), (C) - (II), (D) - (III)

💡 Short Explanation

Same-school relation is reflexive, symmetric and transitive. Therefore, it is an equivalence relation. Perpendicularity of lines is symmetric. f(x) = 2 - 3x is bijective from ℝ to ℝ. f(x) = 1 + x² is one-one on ℝ⁺ because it is strictly increasing.

✅ Correct Answer Explanation

❓ Why Other Options Are Incorrect

Option A) This is incorrect because the same-school relation is equivalence, not only symmetric, and f(x) = 1 + x² is not bijective from ℝ⁺ to ℝ.
Option C) This is incorrect because (A) should match equivalence and (B) should match symmetric.
Option D) This is incorrect because f(x) = 2 - 3x is bijective and f(x) = 1 + x² is one-one; C and D are interchanged.

🎯 Strategy

Used: Concept Matching

Identify the exact property of each relation or function.
Same group relation usually indicates equivalence.
A non-zero linear function over ℝ generally gives a bijective function.
Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Types of Relations and Functions

2 The relation R in the set A = {1, 2, 3, 4} is given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then R is: (PYQ 2023 Shift 1)

A Reflexive and symmetric but not transitive

B Reflexive and transitive but not symmetric

C Symmetric and transitive but not reflexive

D An equivalence relation

💡 Short Explanation

The relation contains all identity pairs, so it is reflexive. It contains (1, 2), but the reverse pair (2, 1) is absent. Therefore, it is not symmetric. The required linked pairs satisfy transitivity. Hence, it is reflexive and transitive but not symmetric.

✅ Correct Answer Explanation

❓ Why Other Options Are Incorrect

Option A) This is incorrect because the relation is not symmetric.
Option C) This is incorrect because the relation is reflexive and not symmetric.
Option D) This is incorrect because an equivalence relation must be reflexive, symmetric and transitive; here symmetry fails.

🎯 Strategy

Used: Relation Property Test

Check all identity pairs for reflexivity.
Check reverse pairs for symmetry.
Check linked pairs for transitivity.
Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Types of Relations

3 If the given functions are f(x) = 27x³ and g(x) = x¹ᐟ³, then (g ∘ f)(x) is: (PYQ 2023 Shift 1)

A x

B 2x

C 3x

D 0

💡 Short Explanation

Composite function (g ∘ f)(x) means g(f(x)). Substitute f(x) = 27x³ into g(x). So, g(f(x)) = g(27x³). Cube root of 27x³ is 3x.

✅ Correct Answer Explanation

❓ Why Other Options Are Incorrect

Option A) This is incorrect because it ignores the cube root of 27.
Option B) This is incorrect because the cube root of 27 is 3, not 2.
Option D) This is incorrect because the composite function does not become zero.

🎯 Strategy

Used: Function Composition

Write (g ∘ f)(x) as g(f(x)).
Substitute the inner function first.
Simplify using cube root rules.
Final Answer → Option C.

🧠 Memory Trick

📚 → Section/Theme: Composition of Functions

4 Set A has 4 elements and set B has 6 elements. Then the number of injective mappings that can be defined from A to B is: (PYQ 2023 Shift 1)

A 360

B 15

C 24

D 1296

💡 Short Explanation

Injective mapping means different elements of A must go to different elements of B. First element of A has 6 choices. Second element has 5 choices. Third element has 4 choices. Fourth element has 3 choices. Total = 6 × 5 × 4 × 3 = 360.

✅ Correct Answer Explanation

❓ Why Other Options Are Incorrect

Option B) This is incorrect because 15 = ⁶C₄, which counts selections, not mappings.
Option C) This is incorrect because 24 = 4!, which ignores the 6 available elements in set B.
Option D) This is incorrect because 1296 = 6⁴ counts all functions, not only injective mappings.

🎯 Strategy

Used: Injective Function Counting

Use permutation because distinct images are required.
Apply mPn where m = 6 and n = 4.
Multiply decreasing choices: 6 × 5 × 4 × 3.
Final Answer → Option A.

🧠 Memory Trick

📚 → Section/Theme: One-One Functions

5 For real numbers a and b, define aRb if b - a + √5 is an irrational number. Then the relation R is: (PYQ 2023 Shift 2)

A Symmetric

B Reflexive

C Transitive

D Equivalence

💡 Short Explanation

For reflexive relation, aRa must be true for every real number a. Put b = a in the given condition. Then b - a + √5 = a - a + √5 = √5. Since √5 is irrational, aRa is true for all real numbers a. Hence, the relation is reflexive.

✅ Correct Answer Explanation

❓ Why Other Options Are Incorrect

Option A) This is incorrect because aRb does not always imply bRa.
Option C) This is incorrect because the irrational condition is not always preserved through three elements.
Option D) This is incorrect because an equivalence relation must be reflexive, symmetric and transitive, but this relation is not symmetric and not transitive.

🎯 Strategy

Used: Relation Property Test

For reflexive property, replace both variables with the same element.
The expression becomes √5, which is irrational.
Eliminate equivalence because symmetry and transitivity are not fully satisfied.
Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Types of Relations

6 The domain of the function f(x) = log(x² - 4) is: (PYQ 2023 Shift 2)

A [2, ∞)

B (-∞, -2) ∪ (2, ∞)

C (2, ∞)

D (-∞, -2] ∪ (-2, -1)

💡 Short Explanation

Logarithm is defined only when its argument is positive. So, x² - 4 > 0. Factorise: (x - 2)(x + 2) > 0. This is true when x < -2 or x > 2. Therefore, the domain is (-∞, -2) ∪ (2, ∞).

✅ Correct Answer Explanation

❓ Why Other Options Are Incorrect

Option A) This is incorrect because it includes x = 2 where log(0) is undefined, and it misses the valid interval (-∞, -2).
Option C) This is incorrect because it omits the valid negative interval (-∞, -2).
Option D) This is incorrect because x = -2 is not allowed and the interval (-2, -1) does not satisfy x² - 4 > 0.

🎯 Strategy

Used: Log Domain Rule

For log expression, make the inside quantity greater than zero.
Solve the quadratic inequality x² - 4 > 0.
Exclude points where the log argument becomes zero.
Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Domain of Logarithmic Functions

7 Let f: ℝ → ℝ be defined by f(x) = x² for every x ∈ ℝ. Then f is: (PYQ 2023 Shift 2)

A One-one and onto

B One-one and not onto

C Neither one-one nor onto

D Onto and not one-one

💡 Short Explanation

f(x) = x² gives the same output for opposite inputs. For example, f(2) = 4 and f(-2) = 4. Hence, it is not one-one. The range is [0, ∞), not all real numbers. Hence, it is not onto ℝ.

✅ Correct Answer Explanation

❓ Why Other Options Are Incorrect

Option A) This is incorrect because f(x) = x² is neither one-one nor onto from ℝ to ℝ.
Option B) This is incorrect because the function is not one-one.
Option D) This is incorrect because the function is not onto ℝ.

🎯 Strategy

Used: Counterexample Method

Use two different inputs such as 2 and -2 to disprove one-one.
Use negative real numbers to disprove onto.
Compare range with codomain.
Final Answer → Option C.

🧠 Memory Trick

📚 → Section/Theme: One-One and Onto Functions

8 Let f(x) = |x| and g(x) = [x], where [x] denotes the greatest integer function. Then the value of (f ∘ g)(-9/2) - (g ∘ f)(-9/2) is: (PYQ 2023 Shift 3)

A -1

B 1

C -9

D 9

💡 Short Explanation

Apply the inner function first in each composition. g(-9/2) = [-4.5] = -5. So, f(g(-9/2)) = |-5| = 5. f(-9/2) = 9/2. So, g(f(-9/2)) = [4.5] = 4. Required value = 5 - 4 = 1.

✅ Correct Answer Explanation

❓ Why Other Options Are Incorrect

Option A) This is incorrect because the order of subtraction gives 5 - 4, not 4 - 5.
Option C) This is incorrect because the compositions give small integer outputs 5 and 4, not -9.
Option D) This is incorrect because the absolute value and greatest integer calculations give 1, not 9.

🎯 Strategy

Used: Composition Evaluation

Evaluate the inner function first.
Use the greatest integer rule carefully for negative numbers.
Subtract the second composition from the first.
Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Composite Functions

9 Let R be a relation on the set of integers ℤ such that R = {(a, b): a = 2ᵏb, a, b, k ∈ ℤ}. Then R is: (PYQ 2023 Shift 3)

A Reflexive but not symmetric and transitive

B Symmetric and reflexive but not transitive

C Equivalence relation

D Reflexive and transitive but not symmetric

💡 Short Explanation

For reflexive property, take k = 0, so a = 2⁰a = a. For symmetric property, if a = 2ᵏb, then b = 2⁻ᵏa. For transitive property, exponents add. So, the relation is reflexive, symmetric and transitive. Hence, it is an equivalence relation.

✅ Correct Answer Explanation

❓ Why Other Options Are Incorrect

Option A) This is incorrect because the relation is symmetric and transitive also.
Option B) This is incorrect because the relation is transitive.
Option D) This is incorrect because the relation is symmetric also.

🎯 Strategy

Used: Property Verification

Check reflexive property using k = 0.
Check symmetric property using the exponent -k.
Check transitive property by adding exponents.
Final Answer → Option C.

🧠 Memory Trick

📚 → Section/Theme: Equivalence Relations

10 The function f: ℝ → ℝ, f(x) = x² is: (PYQ 2023 Shift 3)

A Injective but not surjective

B Surjective but not injective

C Injective as well as surjective

D Neither injective nor surjective

💡 Short Explanation

f(x) = x² gives the same value for x and -x. For example, f(1) = f(-1) = 1. Therefore, it is not injective. The function never gives negative real numbers as outputs. Therefore, it is not surjective onto ℝ.

✅ Correct Answer Explanation

❓ Why Other Options Are Incorrect

Option A) This is incorrect because f(x) = x² is not injective.
Option B) This is incorrect because f(x) = x² is not surjective onto ℝ.
Option C) This is incorrect because the function is neither injective nor surjective.

🎯 Strategy

Used: Mapping Test

Use f(1) and f(-1) to test injectivity.
Compare the range [0, ∞) with the codomain ℝ.
Eliminate options claiming injective or surjective behavior.
Final Answer → Option D.

🧠 Memory Trick

📚 → Section/Theme: Injective and Surjective Functions

11 If the function f: ℕ → ℕ is defined as
\(f(n)=\left\{\begin{pmatrix}n-1, & if n is even\\ n+1, & if n is odd\end{pmatrix}\right.\)then:
(A) f is injective
(B) f is into
(C) f is surjective
(D) f is invertible
Choose the correct answer from the options given below: (PYQ 2024 Shift 1)

A (B) only

B (A), (B) and (D) only

C (A) and (C) only

D (A), (C) and (D) only

💡 Short Explanation

The function swaps even and odd natural numbers. Each input has a unique output, so it is injective. Every natural number is covered, so it is surjective. Since it is both injective and surjective, it is invertible.

✅ Correct Answer Explanation

❓ Why Other Options Are Incorrect

Option A) This is incorrect because it includes only "into" and ignores injective, surjective and invertible properties.
Option B) This is incorrect because it wrongly includes "into" and excludes surjective.
Option C) This is incorrect because it misses invertibility, which follows from bijectivity.

🎯 Strategy

Used: Function Mapping Check

Write the first few values of the function.
Observe that the function creates perfect even-odd pairing.
Bijective functions are invertible.
Final Answer → Option D.

🧠 Memory Trick

📚 → Section/Theme: Functions and Invertible Functions

12 Let R be the relation over the set A of all straight lines in a plane such that l₁ R l₂ ⟺ l₁ is parallel to l₂. Then R is: (PYQ 2024 Shift 1)

A Symmetric

B Equivalence relation

C Transitive

D Reflexive

💡 Short Explanation

Every line is parallel to itself, so the relation is reflexive. If l₁ is parallel to l₂, then l₂ is parallel to l₁, so it is symmetric. If l₁ ∥ l₂ and l₂ ∥ l₃, then l₁ ∥ l₃, so it is transitive. Therefore, the relation is an equivalence relation.

✅ Correct Answer Explanation

❓ Why Other Options Are Incorrect

Option A) This is incomplete because symmetric is only one property; the relation satisfies all three equivalence properties.
Option C) This is incomplete because transitive is only one property; the relation is also reflexive and symmetric.
Option D) This is incomplete because reflexive is only one property; the relation is also symmetric and transitive.

🎯 Strategy

Used: RST Property Check

Check reflexive property: a line is parallel to itself.
Check symmetric property: parallelism works both ways.
Check transitive property: parallelism is preserved through a third line.
Final Answer → Option B.

🧠 Memory Trick

📚 → Section/Theme: Equivalence Relations

CUET UG Categorized PYQ Mathematics Unit 1

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