Question

Let \( f : R \setminus \left\{ -\frac{1}{2} \right\} \to R \) and \( g : R \setminus \left\{ -\frac{5}{2} \right\} \to R \) be defined as \( f(x) = \frac{2x + 3}{2x + 1} \) and \( g(x) = \frac{|x| + 1}{2x + 5} \). Then the domain of the function \( f(g(x)) \) is:

• A. \( R \setminus \left\{ -\frac{5}{2} \right\} \)
• B. \( R \)
• C. \( R \setminus \left\{ -\frac{7}{4} \right\} \)
• D. \( R \setminus \left\{ -\frac{5}{2}, -\frac{7}{4} \right\} \)

Answer 1

The Correct Option is A

For the function composition \(f \circ g(x)\) to be defined, \(g(x)\) must be in the domain of \(f\). Given:  
\(f(g(x)) = \frac{2g(x) + 3}{2g(x) + 1}.\)

The domain of \(f\) excludes \(x = -\frac{3}{2}\). Therefore, \(2g(x) + 1 \neq 0\). Solving:  
\(2\left(\frac{|x| + 1}{2x + 5}\right) + 1 = 0.\)

Simplifying:  
\(\frac{2(|x| + 1)}{2x + 5} + 1 = 0 \implies |x| + 1 = -\frac{2x + 5}{2}.\)

This yields no valid solutions in \(\mathbb{R}\). Therefore, the domain is:  
\(\mathbb{R} \setminus \left\{\frac{-5}{2}\right\}.\)

Thus, the correct answer is: \( R \setminus \left\{ -\frac{5}{2} \right\} \)