Let `f(x)={(2x+a",",x ge -1),(bx^(2)+3",",x lt -1):}`
and `g(x)={(x+4",",0 le x le 4),(-3x-2",",-2 lt x lt 0):}`
`g(f(x))` is not defined if

Options

A. `a in (10,oo), b in (5,oo)`
B. `a in (4,10), b in (5,oo)`
C. `a in (10,oo), b in (0,1)`
D. `a in (4,10), b in (1, 5)`

Explanation

To determine where \( g(f(x)) \) is not defined, we need to analyze the functions \( f(x) \) and \( g(x) \) given in the problem.

### Step 1: Analyze the Function \( f(x) \)

The function \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases} 2x + a & \text{if } x \ge -1 \\ bx^2 + 3 & \text{if } x < -1 \end{cases} \]

For \( g(f(x)) \) to be defined, \( f(x) \) must yield values that are within the domain of \( g(x) \).

### Step 2: Analyze the Function \( g(x) \)

The function \( g(x) \) is defined as:
\[ g(x) = \begin{cases} x + 4 & \text{if } 0 \le x \le 4 \\ -3x - 2 & \text{if } -2 < x < 0 \end{cases} \]

The domain of \( g(x) \) is \( -2 < x \le 4 \).

### Step 3: Determine the Range of \( f(x) \)

1. **For \( x \ge -1 \)**:
\[ f(x) = 2x + a \]
At \( x = -1 \):
\[ f(-1) = 2(-1) + a = -2 + a \]
As \( x \) increases, \( f(x) \) will also increase. Thus, the minimum value of \( f(x) \) when \( x \ge -1 \) is \( -2 + a \).

2. **For \( x < -1 \)**:
\[ f(x) = bx^2 + 3 \]
As \( x \) approaches \( -1 \) from the left, the value of \( f(x) \) approaches:
\[ f(-1) = b(-1)^2 + 3 = b + 3 \]
Since \( bx^2 \) is a parabola opening upwards (if \( b > 0 \)), the minimum value occurs at \( x = 0 \) (not applicable here since \( x < -1 \)). Thus, we consider values as \( x \) approaches \( -1 \).

### Step 4: Conditions for \( g(f(x)) \) to be Defined

For \( g(f(x)) \) to be defined, the output of \( f(x) \) must lie within the domain of \( g(x) \), which is \( -2 < x \le 4 \).

1. **From \( f(x) = 2x + a \)**:
\[ -2 < 2x + a \le 4 \]
- For the lower bound:
\[ -2 < 2x + a \implies 2x > -2 - a \implies x > \frac{-2 - a}{2} \]
- For the upper bound:
\[ 2x + a \le 4 \implies 2x \le 4 - a \implies x \le \frac{4 - a}{2} \]

2. **From \( f(x) = bx^2 + 3 \)**:
\[ -2 < bx^2 + 3 \implies bx^2 > -5 \implies x^2 > \frac{-5}{b} \quad (b > 0) \]
\[ bx^2 + 3 \le 4 \implies bx^2 \le 1 \implies x^2 \le \frac{1}{b} \]

### Step 5: Find Values of \( a \) and \( b \)

From the conditions derived, we can summarize:
- For \( 2x + a \):
- \( x > \frac{-2 - a}{2} \)
- \( x \le \frac{4 - a}{2} \)

- For \( bx^2 + 3 \):
- \( x^2 > \frac{-5}{b} \) (valid only if \( b > 0 \))
- \( x^2 \le \frac{1}{b} \)

### Conclusion

To find where \( g(f(x)) \) is not defined, we need to ensure that the values of \( a \) and \( b \) do not satisfy the conditions derived above.

Thus, \( g(f(x)) \) is not defined for:
- \( a \ge 10 \)
- \( b \le 5 \)