Question

If $\frac{(x^2-1)(x+2)(x+1{)}^2}{(x-2)}<0$ , then x lies in the interval

• A. $(-2,-1)\cup (1,2)$
• B. $(-\infty ,-2)\cup (2,\infty )$
• C. $(-2,-1)\cup (2,\infty )$
• D. $(-2,-1)\cup (1,\infty )$

Answer 1

The correct option is A $(-2,-1)\cup (1,2)$

$\frac{(x^2-1)(x+2)(x+1{)}^2}{(x-2)}<0$
$\frac{(x^2-1)(x^2-4)(x+1{)}^2}{(x-2{)}^2}<0$
$(x^2-1)(x^2-4)<0$
$1<x^2<4$
$-2<x<-1or1<x<2$