The inequality `x^(2)-3x gt tan^(-1) x` is ture in

Options

A. `(4, oo)`
B. `(-oo, 0)`
C. `(0, 1)`
D. `(0, 2)`

Answer

To solve the inequality \( x^2 - 3x > \tan^{-1}(x) \), we will follow these steps:

### Step 1: Define the function
We start by defining a function \( f(x) \) based on the given inequality:
\[ f(x) = x^2 - 3x - \tan^{-1}(x) \]
We want to find where \( f(x) > 0 \).

### Step 2: Differentiate the function
Next, we differentiate \( f(x) \) to analyze its behavior:
\[ f'(x) = \frac{d}{dx}(x^2 - 3x) - \frac{d}{dx}(\tan^{-1}(x)) \]
Calculating the derivatives:
\[ f'(x) = 2x - 3 - \frac{1}{1+x^2} \]

### Step 3: Simplify the derivative
Now, we simplify \( f'(x) \):
\[ f'(x) = 2x - 3 - \frac{1}{1+x^2} \]
To combine these terms, we can write:
\[ f'(x) = \frac{(2x - 3)(1 + x^2) - 1}{1 + x^2} \]
Expanding the numerator:
\[ = \frac{2x + 2x^3 - 3 - 3x^2 - 1}{1 + x^2} = \frac{2x^3 - 3x^2 + 2x - 4}{1 + x^2} \]

### Step 4: Analyze the critical points
We need to find the critical points by setting \( f'(x) = 0 \):
\[ 2x^3 - 3x^2 + 2x - 4 = 0 \]
This is a cubic equation. We can use numerical methods or graphing to find the roots.

### Step 5: Evaluate the function at critical points
Let's evaluate \( f(x) \) at some critical points:
- \( f(0) = 0^2 - 3(0) - \tan^{-1}(0) = 0 - 0 = 0 \)
- \( f(1) = 1^2 - 3(1) - \tan^{-1}(1) = 1 - 3 - \frac{\pi}{4} < 0 \)
- \( f(2) = 2^2 - 3(2) - \tan^{-1}(2) = 4 - 6 - \tan^{-1}(2) < 0 \)

### Step 6: Determine the intervals
From our evaluations:
- \( f(0) = 0 \) indicates a boundary.
- \( f(1) < 0 \) and \( f(2) < 0 \) suggest that the function is negative in this interval.

### Step 7: Check behavior as \( x \to \infty \)
As \( x \to \infty \):
\[ f(x) \approx x^2 - 3x \to \infty \]
This indicates that \( f(x) > 0 \) for sufficiently large \( x \).

### Conclusion
From the analysis, we find that \( f(x) > 0 \) for \( x > 2 \). Thus, the solution to the inequality \( x^2 - 3x > \tan^{-1}(x) \) is:
\[ \text{The inequality is true for } x \in (2, \infty). \]