Question

$\text{If}(x^2-4)\sqrt{x^2-1}<0$ then $x$ will lie in the interval

• A. $(1,2)$
• B. $(-2,-1)$
• C. $[-1,1]$
• D. $(-2,2)$

Answer 1

Answer: (A), (B)

The correct options are
A $(1,2)$
B $(-2,-1)$

Given $(x^2-4)\sqrt{x^2-1}<0$
for inequality to define we must have $x^2-1>0$
$\text{or}(x-1)(x+1)>0$
$\text{or}x<-1\text{or}x>1\cdots \left(1\right)$

$\text{also}(x^2-4)\sqrt{x^2-1}<0$
as we know that square root is always non negative quantity
$\therefore x^2-4<0\Rightarrow (x-2)(x+2)<0$

$\Rightarrow -2<x<2\cdots \left(2\right)$
From equation $\left(1\right)\text{and}\left(2\right)$, we have
$x\in (-2,-1)\cup (1,2)$