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junior college 2 | H1 Maths
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junior college 2 chevron_right H1 Maths chevron_right Singapore

Hi, can anyone show me the solution for the last part? I do not know how to get the answer 13

Date Posted: 3 months ago
Views: 101
3 months ago
Since the first function is mod it is going to be positive integer. So we need points that can make (k-6) negative . Which is any thing less than or equal to 5.
For k <=0 (k-6) * positive non zero integer is going to less than or equal to -6.

So that's 11 points. And since the function will evaluate to 0 at 1 and 6, between 2-5 just put the numbers in the function you will get two more at k=3 and 4