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For what values of k the relation R:{(k2 , 5), (3k, 6)} is not a function?

1 Answer

4 votes

Answer:


k=0 and
k=3.

Explanation:

To this relation be a function, the horizontal coordinates can't be equal. So, let's find the value of
k that makes those coordinates equal.


k^(2)=3k\\ k^(2)-3k=0\\k(k-3)=0\\

Using the null factor property, we have


k=0\\k-3=0 \implies k=3

Therefore, the given relation is not a function when
k=0 and
k=3.

User MIPB
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