Question:
\( \{ x \in \mathbb{R} : \frac{2x - 1}{x^{3} + 4x^{2} + 3x} \in \mathbb{R} \} \) equals
\( \{ x \in \mathbb{R} : \frac{2x - 1}{x^{3} + 4x^{2} + 3x} \in \mathbb{R} \} \) equals
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In rational functions, exclude values where the denominator is zero.
Updated On: Apr 10, 2026
- $R-\{0\}$
- $R-\{0, 1, 3\}$
- $R-\{0, -1, -3\}$
- $R-\{0, -1, -3, +\frac{1}{2}\}$
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The Correct Option is C
Solution and Explanation
Step 1: Domain Condition
The expression is a real number if the denominator is not equal to zero.
Step 2: Factoring the Denominator
$x^{3} + 4x^{2} + 3x = x(x^{2} + 4x + 3) = x(x + 3)(x + 1)$.
Step 3: Finding Zeros
The denominator is zero when $x = 0$, $x = -3$, or $x = -1$.
Step 4: Conclusion
Therefore, $x$ can be any real number except $\{0, -1, -3\}$.
Final Answer: (c)
The expression is a real number if the denominator is not equal to zero.
Step 2: Factoring the Denominator
$x^{3} + 4x^{2} + 3x = x(x^{2} + 4x + 3) = x(x + 3)(x + 1)$.
Step 3: Finding Zeros
The denominator is zero when $x = 0$, $x = -3$, or $x = -1$.
Step 4: Conclusion
Therefore, $x$ can be any real number except $\{0, -1, -3\}$.
Final Answer: (c)
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