Question

If $\frac{(2-x^2){(x-3)}^3}{(x+1)(x^2-3x-4)}\ge 0$, then $x$ belong to

• A. $[-\sqrt{2},0)\cup (-1,\sqrt{2}]\cup [3,4)$
• B. $[-\sqrt{2},0)\cup (-1,\sqrt{2}]$
• C. $[-\sqrt{2},-1)\cup (-1,\sqrt{2}]$
• D. $[-\sqrt{2},-1)\cup (-1,\sqrt{2}]\cup [3,4)$

Answer 1

Answer: (D)

$[-\sqrt{2},-1)\cup (-1,\sqrt{2}]\cup [3,4)$

$\frac{(2-x^2)(x-3{)}^3}{(x+1)(x^2-3x-4)}\ge 0$

$\Rightarrow x^2-3x-4$

$=(x-4)(x+1)$

$\Rightarrow x=4,-1$

$x=\pm \sqrt{2},3,-1,4,-1$

$\Rightarrow \frac{(2-x^2)(x-3{)}^3}{(x+1{)}^2(x-4)}\ge 0$

$\Rightarrow \frac{(x^2-2)(x-3{)}^3}{(x+1{)}^2(x-4)}\le 0$

$x\ne -1,4$

$\therefore x\in [-\sqrt{2},-1)\cup (-1,\sqrt{2}]\cup [3,4)$.