CAT 2019 Slot 2 — QA Question 8

Given : $\frac{(n^2+7n+12)}{(n^2-n-12)}$

$\frac{(n^2+7n+12)}{(n^2-n-12)}$ = $\frac{(n^2+4n+3n+12)}{(n^2-4n+3n-12)}$ = $\frac{(n+4)(n+3)}{(n-4)(n+3)}$

Now, n cannot be equal to -3, since denominator cannot be 0

⇒ $\frac{(n^2+7n+12)}{(n^2-n-12)}$ = $\frac{(n+4)}{(n-4)}$ = $\frac{(n-4+8)}{(n-4)}$ = $1 + \frac{8}{(n-4)}$

For $\frac{(n^2+7n+12)}{(n^2-n-12)}$ to be an integer, 8/(n-4) should also be an integer.

Largest value of n – 4 = 8

∴ largest possible value of n = 12.

Hence, option (d).