Let `f:R-{-1/2}toR` and `g:R-{-5/2}toR` be defined as `f(x)=(2x+3)/(2x+1)` `g(x)=(abs(x)+1)/(2x+5)` then the domain of the function f(g(x) is:

Options

A. `R`
B. `R-{-7/4}`
C. `R-{-7/4,-5/2}`
D. `R-{-5/2}`

Answer

To find the domain of the function \( f(g(x)) \), we need to analyze the functions \( f(x) \) and \( g(x) \) given by:

\[ f(x) = \frac{2x + 3}{2x + 1} \]
\[ g(x) = \frac{|x| + 1}{2x + 5} \]

### Step 1: Determine the domain of \( g(x) \)

The function \( g(x) \) is defined for all real numbers except where the denominator is zero. Therefore, we need to find when:

\[ 2x + 5 = 0 \]

Solving for \( x \):

\[ 2x = -5 \implies x = -\frac{5}{2} \]

Thus, the domain of \( g(x) \) is all real numbers except \( x = -\frac{5}{2} \):

\[ \text{Domain of } g(x) = \mathbb{R} \setminus \left\{-\frac{5}{2}\right\} \]

### Step 2: Determine the range of \( g(x) \)

Next, we need to find the range of \( g(x) \) to see what values \( g(x) \) can take. The function \( g(x) \) can potentially take any real value except for values that make \( f(g(x)) \) undefined.

To find the values that make \( f(x) \) undefined, we set the denominator of \( f(x) \) to zero:

\[ 2g(x) + 1 = 0 \]

Solving for \( g(x) \):

\[ 2g(x) = -1 \implies g(x) = -\frac{1}{2} \]

Thus, \( f(g(x)) \) is undefined when \( g(x) = -\frac{1}{2} \).

### Step 3: Solve for when \( g(x) = -\frac{1}{2} \)

Now we need to find \( x \) such that:

\[ \frac{|x| + 1}{2x + 5} = -\frac{1}{2} \]

Cross-multiplying gives:

\[ 2(|x| + 1) = -1(2x + 5) \]

This simplifies to:

\[ 2|x| + 2 = -2x - 5 \]

Rearranging terms:

\[ 2|x| + 2 + 2x + 5 = 0 \implies 2|x| + 2x + 7 = 0 \]

### Case 1: \( x \geq 0 \)

If \( x \geq 0 \), then \( |x| = x \):

\[ 2x + 2x + 7 = 0 \implies 4x + 7 = 0 \implies x = -\frac{7}{4} \]

This value is not in the domain since \( x \) must be non-negative.

### Case 2: \( x < 0 \)

If \( x < 0 \), then \( |x| = -x \):

\[ 2(-x) + 2x + 7 = 0 \implies -2x + 2x + 7 = 0 \implies 7 = 0 \]

This case does not yield any valid solutions.

### Step 4: Combine the restrictions

From the analysis, we find that:

1. \( g(x) \) is defined for all \( x \) except \( -\frac{5}{2} \).
2. \( f(g(x)) \) is undefined when \( g(x) = -\frac{1}{2} \), which does not yield any valid \( x \).

Thus, the only restriction on the domain of \( f(g(x)) \) is from \( g(x) \):

\[ \text{Domain of } f(g(x)) = \mathbb{R} \setminus \left\{-\frac{5}{2}\right\} \]

### Final Answer

The domain of the function \( f(g(x)) \) is:

\[ \mathbb{R} \setminus \left\{-\frac{5}{2}\right\} \]