The complete set of values of a for which the function `f(x)=tan^(-1)(x^(2)-18x +a)gt 0 AA x in R` is

Options

A. `(81,oo)`
B. `[81,oo)`
C. `(-oo,81)`
D. `(-oo,81]`

Explanation

To solve the problem, we need to find the complete set of values of \( a \) for which the function \( f(x) = \tan^{-1}(x^2 - 18x + a) > 0 \) for all \( x \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) = \tan^{-1}(x^2 - 18x + a) \) is greater than 0 when the argument of the tangent inverse function, \( x^2 - 18x + a \), is greater than 0. This is because the range of \( \tan^{-1}(y) \) is positive when \( y > 0 \).

\[ x^2 - 18x + a > 0 \]

2. **Analyzing the Quadratic**:
The expression \( x^2 - 18x + a \) is a quadratic function in \( x \). For this quadratic to be greater than 0 for all \( x \in \mathbb{R} \), it must not have any real roots. This occurs when the discriminant \( D \) of the quadratic is less than 0.

3. **Finding the Discriminant**:
The discriminant \( D \) of the quadratic \( ax^2 + bx + c \) is given by the formula:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = -18 \), and \( c = a \). Thus, the discriminant is:
\[ D = (-18)^2 - 4(1)(a) = 324 - 4a \]

4. **Setting the Discriminant Condition**:
For the quadratic to be positive for all \( x \), we need:
\[ D < 0 \]
Therefore, we set up the inequality:
\[ 324 - 4a < 0 \]

5. **Solving the Inequality**:
Rearranging the inequality gives:
\[ 324 < 4a \]
Dividing both sides by 4:
\[ 81 < a \]
or equivalently:
\[ a > 81 \]

6. **Conclusion**:
The complete set of values for \( a \) for which \( f(x) > 0 \) for all \( x \in \mathbb{R} \) is:
\[ a \in (81, \infty) \]

### Final Answer:
The complete set of values of \( a \) is \( (81, \infty) \).