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:
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:
Show 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.
Updated On: Jun 22, 2026
- \((-\infty,-4] \cup [4,\infty)\)
- \(\phi\)
- \((-1,2)\)
- \((-\infty,-4] \cup (-1,2) \cup [4,\infty)\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Condition for the square root to be real.
We require \[ |x|^2-2|x|-8 \geq 0 \] Since \[ |x|^2=x^2, \] the inequality becomes \[ x^2-2|x|-8 \geq 0 \] Let \[ y=|x| \] Then, \[ y^2-2y-8 \geq 0 \] Factorizing, \[ (y-4)(y+2)\geq 0 \] Since \(y=|x|\geq 0\), we get \[ y\geq 4 \] Hence, \[ |x|\geq 4 \] Therefore, \[ x\leq -4 \quad \text{or} \quad x\geq 4 \]
Step 2: Condition for the logarithm to exist.
For \[ \log(2-x-x^2) \] to be defined, we need \[ 2-x-x^2\gt 0 \] That is, \[ x^2+x-2\lt 0 \] Factorizing, \[ (x+2)(x-1)\lt 0 \] Thus, \[ -2\lt x\lt 1 \]
Step 3: Denominator should not be zero.
Also, \[ \log(2-x-x^2)\neq 0 \] Since \[ \log 1=0, \] we require \[ 2-x-x^2 \neq 1 \] So, \[ x^2+x-1\neq 0 \] However, this condition is secondary because first both domain conditions must overlap.
Step 4: Find the common set.
From the square root condition: \[ x\in (-\infty,-4]\cup [4,\infty) \] From the logarithm condition: \[ x\in (-2,1) \] There is no common value satisfying both conditions simultaneously.
Hence, the required set is \[ \phi \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\phi} \]
We require \[ |x|^2-2|x|-8 \geq 0 \] Since \[ |x|^2=x^2, \] the inequality becomes \[ x^2-2|x|-8 \geq 0 \] Let \[ y=|x| \] Then, \[ y^2-2y-8 \geq 0 \] Factorizing, \[ (y-4)(y+2)\geq 0 \] Since \(y=|x|\geq 0\), we get \[ y\geq 4 \] Hence, \[ |x|\geq 4 \] Therefore, \[ x\leq -4 \quad \text{or} \quad x\geq 4 \]
Step 2: Condition for the logarithm to exist.
For \[ \log(2-x-x^2) \] to be defined, we need \[ 2-x-x^2\gt 0 \] That is, \[ x^2+x-2\lt 0 \] Factorizing, \[ (x+2)(x-1)\lt 0 \] Thus, \[ -2\lt x\lt 1 \]
Step 3: Denominator should not be zero.
Also, \[ \log(2-x-x^2)\neq 0 \] Since \[ \log 1=0, \] we require \[ 2-x-x^2 \neq 1 \] So, \[ x^2+x-1\neq 0 \] However, this condition is secondary because first both domain conditions must overlap.
Step 4: Find the common set.
From the square root condition: \[ x\in (-\infty,-4]\cup [4,\infty) \] From the logarithm condition: \[ x\in (-2,1) \] There is no common value satisfying both conditions simultaneously.
Hence, the required set is \[ \phi \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\phi} \]
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