CBSE Class 12 Mathematics 2026 Main Paper 65/1/2 · Q345 Marksrelations_and_functions
CBSE Class 12 Mathematics 2026 Main Paper 65/1/2 · Question 34 (Relations And Functions)
A relation R is defined on Z, the set of integers, as R = {(x, y) : |x − y| is divisible by a prime number ‘p’, x, y ∈ Z}. Check whether R is an equivalence relation or not. OR A function f : R − {3/5} → R − {3/5} is defined as f(x) = (3x+2)/(5x−3). Show that f is one-one and onto.
Alternative 1 5 Marks
A relation R is defined on Z, the set of integers, as R = {(x, y) : |x − y| is divisible by a prime number ‘p’, x, y ∈ Z}. Check whether R is an equivalence relation or not.
Answer: R is an equivalence relation.
- Reflexive: For any x, |x−x|=0 divisible by p, so (x,x)∈R.
- Symmetric: If |x−y| divisible by p, then |y−x|=|x−y| divisible by p, so (y,x)∈R. Transitive: If |x−y| and |y−z| divisible by p, then (x−y) and (y−z) divisible by p ⇒ (x−z) = (x−y)+(y−z) divisible by p ⇒ |x−z| divisible by p, so (x,z)∈R. Hence R is equivalence.
- R is an equivalence relation.
Alternative 2 5 Marks
A function f : R − {3/5} → R − {3/5} is defined as f(x) = (3x+2)/(5x−3). Show that f is one-one and onto.
Answer: f is one-one and onto.
- One-one: Let f(x₁)=f(x₂) ⇒ (3x₁+2)/(5x₁−3) = (3x₂+2)/(5x₂−3). Cross multiply and simplify to get x₁=x₂.
- Onto: For any y in codomain, solve y = (3x+2)/(5x−3) ⇒ x = (3y+2)/(5y−3), which is in domain (since y≠3/5). Thus f is bijective.
- f is one-one and onto.