If `g (x) = x ^(2) -1 and gof (x) = x ^(2) + 4x + 3,` then `f ((1)/(2))` is equal `(f(x) gt 0 AA x in R)`

Options

A. `3/2`
B. 2
C. `5/2`
D. None of these

Explanation

To solve the problem, we need to find \( f\left(\frac{1}{2}\right) \) given the functions \( g(x) = x^2 - 1 \) and \( g(f(x)) = x^2 + 4x + 3 \).

### Step-by-step Solution:

1. **Understanding the Functions**:
We have two functions:
- \( g(x) = x^2 - 1 \)
- \( g(f(x)) = x^2 + 4x + 3 \)

2. **Substituting \( x = \frac{1}{2} \) into \( g(f(x)) \)**:
We need to evaluate \( g(f(\frac{1}{2})) \):
\[ g(f(\frac{1}{2})) = \left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right) + 3 \]
Calculating this:
\[ = \frac{1}{4} + 2 + 3 = \frac{1}{4} + \frac{8}{4} + \frac{12}{4} = \frac{21}{4} \]

3. **Setting Up the Equation**:
From the definition of \( g(f(x)) \):
\[ g(f(x)) = f(x)^2 - 1 \]
Therefore, we can set up the equation:
\[ f\left(\frac{1}{2}\right)^2 - 1 = \frac{21}{4} \]

4. **Solving for \( f\left(\frac{1}{2}\right) \)**:
Rearranging the equation:
\[ f\left(\frac{1}{2}\right)^2 = \frac{21}{4} + 1 = \frac{21}{4} + \frac{4}{4} = \frac{25}{4} \]
Taking the square root:
\[ f\left(\frac{1}{2}\right) = \pm \sqrt{\frac{25}{4}} = \pm \frac{5}{2} \]

5. **Considering the Condition**:
It is given that \( f(x) > 0 \) for all \( x \in \mathbb{R} \). Therefore, we take the positive value:
\[ f\left(\frac{1}{2}\right) = \frac{5}{2} \]

### Final Answer:
Thus, \( f\left(\frac{1}{2}\right) = \frac{5}{2} \).