Question

If $(x^2-9)\sqrt{x^2-1}<0$, then what are the possible values of $x$?

• A. $x\in (-3,-2]\cup [2,3)$
• B. $x\in (-3,-1)\cup (1,3)$
• C. $x\in (-3,3)$
• D. $x\in (-\infty ,-1]\cup [1,\infty )$

Answer 1

The correct option is B

$x\in (-3,-1)\cup (1,3)$

Given $(x^2-9)\sqrt{x^2-1}<0$

We must have $x^2-1\ge 0$ [The entry inside a square root is non-negative.]

$\Rightarrow (x+1)(x-1)\ge 0$
But x can't be $\pm 1$ as per given inequality in the question
$\Rightarrow x<-1orx>1$ ------------(1)

Also as $\sqrt{x^2-1}$ is always positive,

($x^2-9)\sqrt{x^2-1}<0$

$\Rightarrow (x^2-9)<0$

$\Rightarrow (x+3)(x-3)<0$
$\Rightarrow -3<x<3$ --------(2)

From (1) & (2)

$x\in (-3,-1)\cup (1,3)$