Question

The equation $|x+1||x-1|=a^2-2a-3$ can have real solutions for x if a belongs to

• A. $(-\infty ,-1]\cup [3,\infty )$
• B. $[1-\sqrt{5},1+\sqrt{5}]$
• C. $[1-\sqrt{5},-1]\cup [3,1+\sqrt{5}]$
• D. None of the above

Answer 1

The correct option is A $(-\infty ,-1]\cup [3,\infty )$

$\begin{array}{cc}& |x+1||x-1|=a^2-2a-3\\ \Rightarrow & |x^2-1|=a^2-2a-3\\ \Rightarrow & x^2-1=\pm (a^2-2a-3)\\ but& a^2-2a-3\ge 0\\ \therefore & x^2-1=a^2-2a-3\\ \Rightarrow & x^2=a^2-2a-2,-a^2+2a+4\ge 0\end{array}$
For real solutions,
$a^2-2a-2\ge 0and-a^2+2a+4\ge 0$
Thus we have $aϵ(-\infty ,-1]\cup [3,\infty )$