CBSE Class 12 Mathematics 2026 Supplementary Paper 65/7/1 · Q263 Marksrelation_domain
CBSE Class 12 Mathematics 2026 Supplementary Paper 65/7/1 · Question 26 (Relations And Functions)
On positive natural numbers, \(R=\{(x,y):x+2y=25\}\). Find its domain and range and decide whether it is an equivalence relation. OR Determine whether \(f:\mathbb R\to\mathbb R\), \(f(x)=2x^3-7\), is bijective.
Alternative 1 3 Marks
On positive natural numbers, \(R=\{(x,y):x+2y=25\}\). Find its domain and range and decide whether it is an equivalence relation.
Answer: Domain {1,3,…,23}; range {1,2,…,12}; not an equivalence relation.
- Since x=25−2y≥1 and y≥1, y ranges from 1 through 12.
- The corresponding x-values are the odd numbers from 23 down to 1. Reflexivity fails, for example 1+2·1≠25.
- Hence the domain is {1,3,…,23}, the range {1,…,12}, and R is not an equivalence relation.
Alternative 2 3 Marks
Determine whether \(f:\mathbb R\to\mathbb R\), \(f(x)=2x^3-7\), is bijective.
Answer: Bijective.
- If f(u)=f(v), then u³=v³, giving u=v.
- For any real t, x=∛((t+7)/2) is a real preimage, and substitution gives f(x)=t.
- Thus f is both one-one and onto.