If the functions f, g, h are defined from 'R' to 'R' by `f(x)=x^(2)-1, g(x)=sqrt(x^(2)+1), h(x)={{:(0",","if",xle0),(x",","if",xge0):}` then hofog is equal to :

Options

A. `x^(2)`
B. `sqrt(x^(4)-1)`
C. 0
D. none of these

Answer

To solve the problem, we need to find \( h(f(g(x))) \). Let's break it down step by step.

### Step 1: Find \( g(x) \)
The function \( g(x) \) is defined as:
\[ g(x) = \sqrt{x^2 + 1} \]

### Step 2: Find \( f(g(x)) \)
Now, we need to substitute \( g(x) \) into \( f(x) \). The function \( f(x) \) is defined as:
\[ f(x) = x^2 - 1 \]
Substituting \( g(x) \) into \( f \):
\[ f(g(x)) = f(\sqrt{x^2 + 1}) = (\sqrt{x^2 + 1})^2 - 1 \]
Calculating this gives:
\[ f(g(x)) = x^2 + 1 - 1 = x^2 \]

### Step 3: Find \( h(f(g(x))) \)
Now we need to find \( h(f(g(x))) \), which is \( h(x^2) \). The function \( h(x) \) is defined as:
\[ h(x) = \begin{cases} 0 & \text{if } x < 0 \\ x & \text{if } x \geq 0 \end{cases} \]
Since \( x^2 \) is always non-negative (it is either zero or positive), we will use the second case of \( h(x) \):
\[ h(x^2) = x^2 \quad \text{(since \( x^2 \geq 0 \))} \]

### Conclusion
Thus, we have:
\[ h(f(g(x))) = x^2 \]
The answer is \( x^2 \).