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

- $8$
- $12$
- $16$
- $6$

1 vote

$\cfrac{n^{2}+7n+12}{n^{2}-n-12} = \cfrac{n^{2}+3n+4n+12}{n^{2}+3n-4n-12}$

$= \cfrac{n(n+3)+4(n+3)}{n(n+3)-4(n+3)}$$\require{cancel}$

$= \cfrac{\cancel{(n+3)}(n+4)}{\cancel{(n+3)}(n-4)} = \cfrac{(n+4)}{(n-4)}$

$\therefore \cfrac{n^{2}+7n+12}{n^{2}-n-12} = \cfrac{(n+4)}{(n-4)}$

Now let’s substitute $n$ values given in options

- $ n = 8 $

$\cfrac{(n+4)}{(n-4)} = \cfrac{(8+4)}{(8-4)} =\cfrac{(12)}{(4)} = 3$ (Positive Integer)

- $ n = 12 $

$\cfrac{(n+4)}{(n-4)} = \cfrac{(12+4)}{(12-4)} =\cfrac{(16)}{(8)} = 2$ (Positive Integer)

- $ n = 16 $

$\cfrac{(n+4)}{(n-4)} = \cfrac{(16+4)}{(16-4)} =\cfrac{(20)}{(12)} \sim 1.67$ (Not an integer)

- $ n = 6 $

$\cfrac{(n+4)}{(n-4)} = \cfrac{(6+4)}{(6-4)} =\cfrac{(10)}{(2)} = 5$ (Positive Integer)

Options A, B, D have produced the result as a positive integer. In question, given that we need to find “largest positive integer $n$ such that $\cfrac{n^{2}+7n+12}{n^{2}-n-12}$ is also a positive integer.” *n*

Hence correct answer is Option B.

Note: Option D would have been correct when they ask the question as “For what value of $n$ the **result** of the following expression $\cfrac{n^{2}+7n+12}{n^{2}-n-12}$ is maximum?”