Data Sufficiency (DS) | Re: If n ≠ 2, is n^2-1/n-2 ≥8 ?

\(\frac{n^2-1}{n-2}\) ≥8

Whenn≤2
n^2-8n+15≤0
n=[3,5]
Not possible

Whenn≥2
n^2-8n+15≥0
n=[2,3]∪[5,∞ ]

Statement 1
n is odd number
When n>1,\(\frac{n^2-1}{n-2}\) ≥8
When n≤1,\(\frac{n^2-1}{n-2}\) ≤8

Statement 2- n>1
ifn=[1,2)∪(3,5), \(\frac{n^2-1}{n-2}\) ≤8
ifn=[2,3]∪[5,∞), \(\frac{n^2-1}{n-2}\)≥8

Combining both equations
n is odd number and greater than 1
\(\frac{n^2-1}{n-2}\) ≥8 is always true.
Sufficient.





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