Question

If $|x^2-2x-8|+|x^2+x-2|=3|x+2|$, then the set of all real values of x is

• A. $[1,4]\cup -2$
• B. $[1,4]$
• C. $[-2,1]\cup [4,\infty ]$
• D. $[-\infty ,-2]\cup [1,4]$

Answer 1

Answer: (A)

$[1,4]\cup -2$

We have $|x^2-2x-8|+|x^2+x-2|=3|x+2|$

Above is possible when

$x^2-2x-8\le 0$ and $x^2+x-2\ge 0$

or

$x^2-2x-8\ge 0$ and $x^2+x-2\le 0$

For Case 1 when $x^2-2x-8<0$ and $x^2+x-2>0$

$(x-4)(x+2)\le 0$ and $(x+2)(x-1)\ge 0$

$x\in [-2,4]$ and $x\in R-(-2,1)$

$\therefore x\in [1,4]\cup \{-2\}$

For Case 2 when $x^2-2x-8\ge 0$ and $x^2+x-2\le 0$

$(x-4)(x+2)\ge 0$ and $(x+2)(x-1)\le 0$

$x\in R-(-2,4)$ and $x\in [-2,1]$

So, for the case 2, $x\in \varphi$

$\therefore x\in [1,4]\cup \{-2\}$