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100+ Free Karnataka II PUC Mathematics (KSEAB) Practice Questions

Karnataka School Examination and Assessment Board (KSEAB) II PUC Class 12 Mathematics (Subject Code 35) practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: Karnataka II PUC Mathematics (KSEAB) Exam

Code 35

Official II PUC Mathematics subject code

KSEAB / DPUE II PUC blueprint 2025–26

80/120

Marks to earn vs marks in the question framework

KSEAB II PUC Mathematics weightage framework 2025–26

3 h 15 m

Typical II PUC exam duration including reading time

KSEAB II PUC exam logistics / date sheet reporting

~30% / ~33%

Commonly reported per-subject minimum and overall aggregate pass range

KSEAB pass criteria reporting (confirm current circular)

80+20 pattern

Commonly described theory + internal split (confirm year notice)

KSEAB II PUC assessment reporting

Mixed paper

Official format is objective + subjective, not pure MCQ

KSEAB II PUC Mathematics blueprint Parts A–E

English MCQ adaptation

This free local bank is not the official mixed paper format

OpenExamPrep practice policy

KSEAB II PUC Mathematics (code 35) is a Class 12 board subject with a mixed objective + subjective paper (~3 h 15 min) under a 120-mark framework earning 80 theory marks plus ~20 internal. Pass is commonly ~30% per subject and ~33% aggregate as notified. Fees are paid through colleges. This free 2026 bank is an English MCQ study adaptation — not an official paper simulation.

Sample Karnataka II PUC Mathematics (KSEAB) Practice Questions

Try these sample questions to test your Karnataka II PUC Mathematics (KSEAB) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Let R be a relation on the set A = {1, 2, 3} defined by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}. Then R is:
A.reflexive only
B.symmetric only
C.reflexive and symmetric but not transitive
D.an equivalence relation
Explanation: R contains (1,1), (2,2), (3,3), so it is reflexive. Both (1,2) and (2,1) are present, so it is symmetric. For transitivity, the only non-diagonal pairs involve 1 and 2: (1,2) with (2,1) yields (1,1)∈R and (2,1) with (1,2) yields (2,2)∈R. Thus R is an equivalence relation partitioning A into {1,2} and {3}.
2The number of binary relations on a set with 3 elements is:
A.9
B.27
C.512
D.256
Explanation: A set with 3 elements has 3×3 = 9 ordered pairs in A×A. Each pair may either belong to a relation or not, so there are 2^9 = 512 binary relations.
3Let f: R → R be defined by f(x) = x² + 1. Then f is:
A.one-one and onto
B.one-one but not onto
C.onto but not one-one
D.neither one-one nor onto
Explanation: f(1) = f(−1) = 2, so f is not one-one. Range is [1, ∞), not all of R, so f is not onto R. Hence neither one-one nor onto.
4If f(x) = 2x − 3 and g(x) = x² + 1, then (f ∘ g)(2) equals:
A.3
B.5
C.7
D.11
Explanation: g(2) = 4 + 1 = 5. Then f(g(2)) = f(5) = 2·5 − 3 = 7.
5Let A = {1, 2, 3} and B = {a, b}. The number of onto functions from A to B is:
A.6
B.8
C.9
D.2
Explanation: Total functions: 2³ = 8. Exactly 2 constant functions are not onto. So onto functions = 8 − 2 = 6. Equivalently, 2! · S(3,2) = 2 · 3 = 6.
6If f: R → R is defined by f(x) = 3x + 2, then f⁻¹(x) is:
A.(x − 2)/3
B.(x + 2)/3
C.3x − 2
D.x/3 − 2
Explanation: Set y = 3x + 2 ⇒ x = (y − 2)/3. Replacing y by x gives f⁻¹(x) = (x − 2)/3.
7A relation R on Z defined by aRb iff a − b is divisible by 5 is:
A.reflexive only
B.symmetric only
C.an equivalence relation
D.neither reflexive nor symmetric
Explanation: a − a = 0 is divisible by 5 (reflexive). If 5 | (a−b) then 5 | (b−a) (symmetric). If 5|(a−b) and 5|(b−c) then 5|(a−c) (transitive). Hence R is an equivalence relation (congruence modulo 5).
8If f(x) = sin x and g(x) = x², then (g ∘ f)(π/6) equals:
A.1/4
B.1/2
C.√3/2
D.π²/36
Explanation: f(π/6) = sin(π/6) = 1/2. Then g(f(π/6)) = (1/2)² = 1/4.
9The principal value of sin⁻¹(1/2) is:
A.π/6
B.π/3
C.π/2
D.5π/6
Explanation: The principal range of sin⁻¹ is [−π/2, π/2]. Since sin(π/6) = 1/2 and π/6 lies in that range, sin⁻¹(1/2) = π/6.
10The domain of the function f(x) = cos⁻¹(2x − 1) is:
A.[0, 1]
B.[−1, 1]
C.[0, 2]
D.[−1, 0]
Explanation: cos⁻¹ is defined for arguments in [−1, 1], so −1 ≤ 2x − 1 ≤ 1. Adding 1: 0 ≤ 2x ≤ 2, so 0 ≤ x ≤ 1.

About the Karnataka II PUC Mathematics (KSEAB) Practice Questions

Verified exam format metadata for Karnataka School Examination and Assessment Board (KSEAB) II PUC Class 12 Mathematics (Subject Code 35) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.