Question

The domain of the real valued function: \[ f(x) \;=\; \sqrt[3]{\frac{x \,-\, 2}{2x^2 \,-\, 7x \,+\,5}} \;+\; \log\bigl(x^2 \,-\, x \,-\, 2\bigr). \]

• A. \(\displaystyle (-\infty,\,-1)\;\cup\;\Bigl(2,\,\tfrac{5}{2}\Bigr)\;\cup\;\Bigl(\tfrac{5}{2},\,\infty\Bigr)\)
• B. \(\displaystyle \mathbb{R}\setminus\Bigl\{1,\,\tfrac{5}{2}\Bigr\}\)
• C. \(\displaystyle (-\infty,\,-1)\;\cup\;(2,\,\infty)\)
• D. \(\displaystyle (-1,\,2)\)

Hint:

For a real logarithm, its argument must be strictly positive.
- A cube root has no sign restriction on its argument, but you must exclude values that make any denominator zero.

Answer 1

The Correct Option is A

Step 1: Condition from the logarithm.
For \(\log\bigl(x^2 - x - 2\bigr)\) to be valid (real and defined), we need: \[ x^2 - x - 2 \;>\; 0 \;\;\Longrightarrow\;\;(x-2)(x+1) \;>\; 0, \] which is true for: Thus, from the logarithm part alone, \[ x \;\in\; (-\infty,\,-1)\;\cup\;(2,\,\infty). \] Step 2: Condition from the cube root’s denominator.
Inside the cube root, we have \(\displaystyle \frac{x-2}{2x^2 -7x+5}\). The numerator can be any real number, but we must avoid a zero denominator: \[ 2x^2 - 7x + 5 \;\neq\; 0. \] Factoring, \[ 2x^2 - 7x + 5 = (x-1)(2x-5), \] so exclude \[ x = 1 \quad \text{and}\quad x = \tfrac{5}{2}. \] Step 3: Final domain.
We intersect \(\;(-\infty,-1)\cup(2,\infty)\;\) with the restriction that \(\;x \neq 1,\; x \neq \tfrac{5}{2}.\) Note that \(x=1\) is already not in \((2,\infty)\), so the only relevant exclusion in \((2,\infty)\) is \(x=\tfrac{5}{2}\). Hence, the combined domain is: \[ \boxed{(-\infty,\,-1)\;\cup\;\bigl(2,\tfrac{5}{2}\bigr)\;\cup\;\bigl(\tfrac{5}{2},\,\infty\bigr).} \]