Question

The set \[ \left\{x \in \mathbb{R} \,/\, \frac{\sqrt{|x|^2-2|x|-8}} {\log(2-x-x^2)} \text{ is a real number} \right\} \] is equal to:

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

Hint:

For expressions involving logarithms and square roots, always check: \[ \text{(i) quantity inside square root } \geq 0 \] and \[ \text{(ii) quantity inside logarithm } \gt 0 \] simultaneously.

Answer 1

The Correct Option is B

Step 1: Understand what makes the expression real.
The expression $\dfrac{\sqrt{|x|^2-2|x|-8}}{\log(2-x-x^2)}$ is real only when the square root is real and the logarithm is defined (and non-zero). We will collect every condition.
Step 2: Handle the square root.
We need $|x|^2-2|x|-8 \ge 0$. Since $|x|^2=x^2$, put $y=|x|\ge 0$, giving $y^2-2y-8\ge 0$, that is $(y-4)(y+2)\ge 0$.
Step 3: Solve for $|x|$.
Because $y\ge 0$, the only valid part is $y\ge 4$, so $|x|\ge 4$. This means $x\le -4$ or $x\ge 4$.
Step 4: Handle the logarithm.
For $\log(2-x-x^2)$ to exist we need $2-x-x^2>0$, i.e. $x^2+x-2<0$, i.e. $(x+2)(x-1)<0$, giving $-2<x<1$.
Step 5: Combine both conditions.
The square root forces $x\le -4$ or $x\ge 4$, while the logarithm forces $-2<x<1$. These two sets have no common point.
Step 6: Conclude.
Since the intersection is empty, no real $x$ satisfies all requirements, so the set is the empty set $\phi$.
\[ \boxed{\phi} \]