Question

If $\frac{\sqrt{21-4a-a^2}}{a+1}\le 1,$ then complete set of values of $a$, is equal to -

• A. $a\in [-7,-1)\cup [2,3]$
• B. $a\in [-7,-1)$
• C. $a\in [-7,-1)\cup (-1,2]$
• D. none of these

Answer 1

The correct option is A $a\in [-7,-1)\cup [2,3]$

Case (i) $a+1<0\text{and}21-4a-a^2\ge 0$
$a\in [-7,-1)$
Case (ii) $\begin{array}{c}a+1>0\text{and}21-4a-a^2\ge 0\\ \text{and}(21-4a-a^2)\le (a+1{)}^2\end{array}\}$
$a\in [2,3]$
So $a\in [-7,-1)\cup [2,3]$