Let `g(x)=a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3) " and " f(x)=sqrt(g(x))`, f(x) has its non-zero local minimum and maximum values at -3 and 3, respectively. If `a_(3) in` the domain of the function
`h(x)=sin^(-1)((1+x^(2))/(2x))`.
f(10) is defined for

Options

A. `a_(0) gt 830`
B. `a_(0) lt 830`
C. `a_(0)=830`
D. None of these

Answer

To solve the problem step by step, we will analyze the given functions and conditions carefully.

### Step 1: Understanding the Functions
We have:
- \( g(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 \)
- \( f(x) = \sqrt{g(x)} \)

Given that \( f(x) \) has local minimum and maximum at \( x = -3 \) and \( x = 3 \) respectively, we know that the first derivative \( f'(x) \) must be zero at these points.

### Step 2: Finding Critical Points
The critical points occur where the derivative \( f'(x) = 0 \). Using the chain rule:
\[ f'(x) = \frac{1}{2\sqrt{g(x)}} g'(x) \]
Setting \( f'(-3) = 0 \) and \( f'(3) = 0 \) implies \( g'(-3) = 0 \) and \( g'(3) = 0 \).

### Step 3: Calculating \( g'(x) \)
The derivative of \( g(x) \) is:
\[ g'(x) = a_1 + 2a_2 x + 3a_3 x^2 \]
Setting \( g'(-3) = 0 \):
\[ a_1 - 6a_2 + 27a_3 = 0 \quad (1) \]
Setting \( g'(3) = 0 \):
\[ a_1 + 6a_2 + 27a_3 = 0 \quad (2) \]

### Step 4: Solving the System of Equations
Subtract equation (1) from equation (2):
\[ (a_1 + 6a_2 + 27a_3) - (a_1 - 6a_2 + 27a_3) = 0 \]
This simplifies to:
\[ 12a_2 = 0 \implies a_2 = 0 \]

### Step 5: Substituting \( a_2 \) Back
Substituting \( a_2 = 0 \) into either equation (1) or (2):
Using equation (1):
\[ a_1 + 27a_3 = 0 \implies a_1 = -27a_3 \]

### Step 6: Finding \( a_3 \)
We also know that \( a_3 \) is in the domain of:
\[ h(x) = \sin^{-1}\left(\frac{1+x^2}{2x}\right) \]
The argument of \( \sin^{-1} \) must lie between -1 and 1. Analyzing:
\[ -1 \leq \frac{1+x^2}{2x} \leq 1 \]
Cross-multiplying gives two inequalities:
1. \( 1 + x^2 \geq -2x \) (always true for \( x \neq 0 \))
2. \( 1 + x^2 \leq 2x \) leads to \( x^2 - 2x + 1 \leq 0 \) or \( (x-1)^2 \leq 0 \) which implies \( x = 1 \).

Thus, \( a_3 \) must be such that it does not violate the conditions of the function.

### Step 7: Finding \( f(10) \)
Substituting \( a_3 = 1 \) (as a possible value):
\[ a_1 = -27(1) = -27 \]
Thus, \( g(x) \) becomes:
\[ g(x) = a_0 - 27x + x^3 \]
Now, we need to find \( f(10) \):
\[ g(10) = a_0 - 27(10) + 10^3 = a_0 - 270 + 1000 = a_0 + 730 \]
Thus,
\[ f(10) = \sqrt{g(10)} = \sqrt{a_0 + 730} \]

### Step 8: Ensuring \( g(x) \) is Non-Negative
Since \( f(x) \) must be defined, we require:
\[ a_0 + 730 \geq 0 \implies a_0 \geq -730 \]

### Conclusion
To ensure \( f(10) \) is defined, we conclude that:
- \( a_0 \) must be greater than or equal to -730.