The largest integral value of x satisfying the inequality `(tan^(-1)(x))^(2)-4(tan^(-1)(x))+3gt0` is :

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the inequality \((\tan^{-1}(x))^2 - 4(\tan^{-1}(x)) + 3 > 0\), we will follow these steps:

### Step 1: Substitute \(k\) for \(\tan^{-1}(x)\)
Let \(k = \tan^{-1}(x)\). The inequality becomes:
\[ k^2 - 4k + 3 > 0 \]

### Step 2: Factor the quadratic expression
We need to factor the quadratic expression \(k^2 - 4k + 3\). We can rewrite it as:
\[ k^2 - 3k - k + 3 > 0 \]
Now, we can factor by grouping:
\[ k(k - 3) - 1(k - 3) > 0 \]
This simplifies to:
\[ (k - 1)(k - 3) > 0 \]

### Step 3: Determine the critical points
The critical points from the factors are \(k = 1\) and \(k = 3\).

### Step 4: Analyze the intervals
We will analyze the sign of the expression \((k - 1)(k - 3)\) in the intervals determined by the critical points:
- Interval 1: \( (-\infty, 1) \)
- Interval 2: \( (1, 3) \)
- Interval 3: \( (3, \infty) \)

1. **For \(k < 1\)** (e.g., \(k = 0\)):
\((0 - 1)(0 - 3) = (-1)(-3) = 3 > 0\) (positive)

2. **For \(1 < k < 3\)** (e.g., \(k = 2\)):
\((2 - 1)(2 - 3) = (1)(-1) = -1 < 0\) (negative)

3. **For \(k > 3\)** (e.g., \(k = 4\)):
\((4 - 1)(4 - 3) = (3)(1) = 3 > 0\) (positive)

### Step 5: Combine the intervals
The solution to the inequality \((k - 1)(k - 3) > 0\) is:
\[ k \in (-\infty, 1) \cup (3, \infty) \]

### Step 6: Convert back to \(x\)
Since \(k = \tan^{-1}(x)\), we need to find the corresponding \(x\) values:
1. For \(k < 1\):
\[ \tan^{-1}(x) < 1 \implies x < \tan(1) \approx 1.5574 \]

2. For \(k > 3\):
\[ \tan^{-1}(x) > 3 \implies x > \tan(3) \text{ (which is not relevant since } \tan^{-1}(x) \text{ cannot exceed } \frac{\pi}{2} \text{)} \]

### Step 7: Find the largest integral value of \(x\)
The largest integral value of \(x\) satisfying \(x < \tan(1)\) is:
\[ \lfloor 1.5574 \rfloor = 1 \]

### Final Answer
The largest integral value of \(x\) satisfying the inequality is:
\[ \boxed{1} \]