Discussion

Question:

What is the largest positive integer such that $\frac{(n^{2} + 7 n + 12)}{(n^{2} - n - 12)}$ is also positive integer?

Explanation:

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

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

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

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

For $\frac{(n^{2} + 7 n + 12)}{(n^{2} - n - 12)}$ 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).

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