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Relations and Functions
Previous Year Questions (2022-2024)
Year 2022
- Check if f(x) = x2 is injective and surjective for f: R -> R.
Show Solution
Not injective: f(1)=1, f(-1)=1\nNot surjective: no x gives f(x)<0\nf is neither injective nor surjective. - Prove that R = {(a,b): a-b is divisible by 3} on Z is an equivalence relation.
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Reflexive: a-a=0 divisible by 3.\nSymmetric: if a-b divisible by 3 then b-a also divisible by 3.\nTransitive: a-b=3k, b-c=3m => a-c=3(k+m) divisible by 3.\nHence equivalence relation.
Year 2023
- Check if f(x) = x2 is injective and surjective for f: R -> R.
Show Solution
Not injective: f(1)=1, f(-1)=1\nNot surjective: no x gives f(x)<0\nf is neither injective nor surjective. - Prove that R = {(a,b): a-b is divisible by 3} on Z is an equivalence relation.
Show Solution
Reflexive: a-a=0 divisible by 3.\nSymmetric: if a-b divisible by 3 then b-a also divisible by 3.\nTransitive: a-b=3k, b-c=3m => a-c=3(k+m) divisible by 3.\nHence equivalence relation.
Year 2024
- Check if f(x) = x2 is injective and surjective for f: R -> R.
Show Solution
Not injective: f(1)=1, f(-1)=1\nNot surjective: no x gives f(x)<0\nf is neither injective nor surjective. - Prove that R = {(a,b): a-b is divisible by 3} on Z is an equivalence relation.
Show Solution
Reflexive: a-a=0 divisible by 3.\nSymmetric: if a-b divisible by 3 then b-a also divisible by 3.\nTransitive: a-b=3k, b-c=3m => a-c=3(k+m) divisible by 3.\nHence equivalence relation.