Let `f(x) = tan^(-1) (x^2-18x + a) gt 0 x in R`. Then the value of a lies in

Options

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

Explanation

To solve the problem, we need to analyze the function \( f(x) = \tan^{-1}(x^2 - 18x + a) \) and determine the conditions under which this function is greater than zero for all \( x \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Understanding the Function**:
We know that the function \( \tan^{-1}(y) > 0 \) when \( y > 0 \). Therefore, we need to ensure that:
\[ x^2 - 18x + a > 0 \quad \text{for all } x \in \mathbb{R} \]

2. **Analyzing the Quadratic**:
The expression \( x^2 - 18x + a \) is a quadratic function in \( x \). For this quadratic to be positive for all real numbers \( x \), its discriminant must be less than zero.

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

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

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

6. **Conclusion**:
Therefore, the value of \( a \) must be greater than 81. In interval notation, this can be expressed as:
\[ a \in (81, \infty) \]

### Final Answer:
The value of \( a \) lies in the interval \( (81, \infty) \).