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100+ Free PSEB Class 12 Maths Practice Questions

Punjab School Education Board (PSEB) Senior Secondary Mathematics (Class 12) practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: PSEB Class 12 Maths Exam

~3 hours

Typical PSEB Senior Secondary Mathematics theory exam duration

PSEB Class 12 theory sitting logistics for Mathematics

Code 028

PSEB subject code for Senior Secondary Mathematics

Punjab School Education Board scheme/date-sheet practice

80 + 20

Theory 80 marks + Internal Assessment 20 marks = 100

PSEB Scheme of Studies Class XII

33% pass

Separate and aggregate pass floor framing under current PSEB Senior Secondary rules — confirm year circular

PSEB scheme of studies / result rules; verify official circular

NCERT Class 12

Syllabus themes track NCERT Class 12 Mathematics chapter map

PSEB Class 12 Mathematics syllabus alignment

English MCQ adaptation

This free local bank is not the official PSEB board paper

OpenExamPrep practice policy

PSEB Senior Secondary Mathematics is the Punjab Class 12 board Maths paper (code 028, Theory 80 + INA 20 = 100, typically 3 hours, 33% separate and aggregate pass). Syllabus tracks NCERT Class 12 Maths. This free 2026 bank is an English MCQ study adaptation with heavy calculus weight — not an official paper simulation.

Sample PSEB Class 12 Maths Practice Questions

Try these sample questions to test your PSEB Class 12 Maths exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1On A = {1, 2, 3}, let R = {(1,1), (2,2), (3,3), (1,2), (2,1)}. Which statement is correct?
A.R is an equivalence relation
B.R is reflexive and symmetric but not transitive
C.R is transitive but not reflexive
D.R is symmetric but not reflexive
Explanation: Diagonals make R reflexive. (1,2) and (2,1) are both in R, so R is symmetric. Non-trivial chains close with pairs already present, so R is transitive. Hence R is an equivalence relation (classes {1,2} and {3}).
2Let f: ℝ → ℝ be given by f(x) = 2x + 7. Then f is:
A.Onto but not one-one
B.One-one and onto
C.Neither one-one nor onto
D.One-one but not onto
Explanation: f(x₁)=f(x₂) ⇒ 2x₁+7=2x₂+7 ⇒ x₁=x₂, so one-one. For any y∈ℝ, x=(y−7)/2 works, so onto. Thus bijective.
3If f(x) = x² + 1 and g(x) = √x (x ≥ 0), then (g∘f)(3) equals:
A.√8
B.4
C.√10
D.10
Explanation: f(3)=10; (g∘f)(3)=g(10)=√10.
4Let f: {1,2,3} → {a,b,c} with f(1)=a, f(2)=b, f(3)=a. Then f is:
A.Neither one-one nor onto
B.One-one but not onto
C.Onto but not one-one
D.Bijective
Explanation: f(1)=f(3)=a with 1≠3 ⇒ not one-one. Image {a,b} misses c ⇒ not onto.
5If f(x) = 3x − 4 is invertible, f⁻¹(5) equals:
A.1
B.3
C.5/3
D.−1
Explanation: f⁻¹(y)=(y+4)/3, so f⁻¹(5)=3. Check: f(3)=5.
6On ℝ, define a * b = a + b − ab. The identity element for * is:
A.0
B.−1
C.2
D.1
Explanation: Need a*e=a: a+e−ae=a ⇒ e(1−a)=0 for all a ⇒ e=0. Check: a*0=a.
7If A = {1, 2, 3, 4} and R = {(1,2), (2,3), (1,3), (3,4)} on A, how many ordered pairs must be added at minimum to make R transitive?
A.3
B.4
C.2
D.1
Explanation: (2,3)+(3,4) force (2,4); (1,3)+(3,4) force (1,4). Minimum add 2 pairs.
8Let f(x) = x³ − 3x + 2. Counting multiplicity, the number of real roots of f(x)=0 is:
A.0
B.1
C.2
D.3
Explanation: f(1)=0. Factor: (x−1)²(x+2)=0. Roots 1 (mult. 2) and −2 (mult. 1); total multiplicity 3.
9The principal value of sin⁻¹(1/2) is:
A.π/6
B.π/3
C.π/4
D.π/2
Explanation: sin(π/6)=1/2 and π/6∈[−π/2,π/2].
10The principal value of cos⁻¹(−1/2) is:
A.π/3
B.π/6
C.−π/3
D.2π/3
Explanation: Principal range [0,π]; cos(2π/3)=−1/2.

About the PSEB Class 12 Maths Practice Questions

Verified exam format metadata for Punjab School Education Board (PSEB) Senior Secondary Mathematics (Class 12) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.