Question

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

• A. (-2, -1) ∪ (1, 2)
• B. (-∞, -2) ∪ (2, ∞)
• C. (-2, -1) ∪ (2, ∞)
• D. (-2, -1) ∪ (1, ∞)

Answer 1

The correct option is A (-2, -1) ∪ (1, 2)

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