Question

If $\sqrt{9-{(n-2)}^2}$ is a real number, then the number of integral values of $n$ is

• A. $3$
• B. $5$
• C. $7$
• D. Infinitely many

Answer 1

The correct option is C $7$

Since, $\sqrt{9-{(n-2)}^2}$ is a real number.
$\Rightarrow 9-{(n-2)}^2\ge 0$
$\Rightarrow (n-2{)}^2\le 9$
$\Rightarrow n^2-4n+4\le 9$
$\Rightarrow n^2-4n-5\le 0$
$\Rightarrow n^2-5n+n-5\le 0$
$\Rightarrow (n+1)(n-5)\le 0$
Two cases arise:
$\text{Case:I}$
$(n+1)\le 0\text{and}(n-5)\ge 0$
$\Rightarrow n\le -1\text{and}n\ge 5$
No such $n$ exists
$\text{Case:II}$
$(n+1)\ge 0\text{and}(n-5)\le 0$
$\Rightarrow n\ge -1\text{and}n\le 5$
$\Rightarrow -1\le n\le 5$
The integrals values of $n$ from $-1$ to $5$ are $-1,0,1,2,3,4,5$
Hence, $7$ is the correct answer.