Let `f_(1):R to R, f_(2):[0,oo) to R, f_(3):R to R and f_(4):R to [0,oo)` be a defined by
`f_(1)(x)={{:(,|x|,"if "x lt 0),(,e^(x),"if "x gt 0):}:f_(2)(x)=x^(2),f_(3)(x)={{:(,sin x,"if x"lt 0),(,x,"if "x ge 0):}` and `f_(4)(x)={{:(,f_(2)(f_(1)(x)),"if "x lt 0),(,f_(2)(f_(1)(f_(1)(x)))-1,"if "x ge 0):}` then `f_(2)` of `f_(1)` is

Options

A. onto but not one-one
B. neither continuous nor one-one
C. differentiable but not one-one
D. continuous and one-one

Explanation

To solve the problem, we need to find \( f_2(f_1(x)) \) where the functions are defined as follows:

1. \( f_1(x) = \begin{cases} |x| & \text{if } x < 0 \\ e^x & \text{if } x \geq 0 \end{cases} \)

2. \( f_2(x) = x^2 \)

3. \( f_3(x) = \begin{cases} \sin x & \text{if } x < 0 \\ x & \text{if } x \geq 0 \end{cases} \)

4. \( f_4(x) = \begin{cases} f_2(f_1(x)) & \text{if } x < 0 \\ f_2(f_1(f_1(x))) - 1 & \text{if } x \geq 0 \end{cases} \)

### Step-by-Step Solution:

**Step 1: Determine \( f_1(x) \)**

For \( x < 0 \):
\[ f_1(x) = |x| = -x \]

For \( x \geq 0 \):
\[ f_1(x) = e^x \]

**Step 2: Substitute \( f_1(x) \) into \( f_2 \)**

Now we need to find \( f_2(f_1(x)) \).

**Case 1: When \( x < 0 \)**

Substituting \( f_1(x) \):
\[ f_2(f_1(x)) = f_2(-x) = (-x)^2 = x^2 \]

**Case 2: When \( x \geq 0 \)**

Substituting \( f_1(x) \):
\[ f_2(f_1(x)) = f_2(e^x) = (e^x)^2 = e^{2x} \]

**Step 3: Combine the results**

Thus, we can write:
\[ f_2(f_1(x)) = \begin{cases} x^2 & \text{if } x < 0 \\ e^{2x} & \text{if } x \geq 0 \end{cases} \]

### Final Result:
The function \( f_2(f_1(x)) \) is given by:
\[ f_2(f_1(x)) = \begin{cases} x^2 & \text{if } x < 0 \\ e^{2x} & \text{if } x \geq 0 \end{cases} \]