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 \(k = \tan^{-1}(x)\). Then, the inequality becomes:
\[ k^2 - 4k + 3 > 0 \]

### Step 2: Factor the quadratic expression
We can factor the quadratic expression:
\[ k^2 - 4k + 3 = (k - 3)(k - 1) \]
Thus, the inequality can be rewritten as:
\[ (k - 3)(k - 1) > 0 \]

### Step 3: Determine the critical points
The critical points from the factors are \(k = 3\) and \(k = 1\). We will analyze the sign of the product \((k - 3)(k - 1)\) in the intervals determined by these points.

### Step 4: Test the intervals
We have three intervals to test:
1. \(k < 1\)
2. \(1 < k < 3\)
3. \(k > 3\)

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

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

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

### Step 5: Write the solution in interval notation
The solution to the inequality \((k - 3)(k - 1) > 0\) is:
\[ k < 1 \quad \text{or} \quad k > 3 \]
In interval notation, this is:
\[ (-\infty, 1) \cup (3, \infty) \]

### Step 6: Convert back to \(x\)
Recall that \(k = \tan^{-1}(x)\). We need to find the corresponding values of \(x\):
- For \(k < 1\):
\(\tan^{-1}(x) < 1\) implies \(x < \tan(1)\).

- For \(k > 3\):
\(\tan^{-1}(x) > 3\) implies \(x > \tan(3)\).

### Step 7: Find the largest integral value of \(x\)
Calculating \(\tan(1)\) and \(\tan(3)\):
- \(\tan(1) \approx 1.557\)
- \(\tan(3) \approx -0.427\)

Thus, the intervals for \(x\) are:
\[ x < 1.557 \quad \text{or} \quad x > -0.427 \]
The largest integral value of \(x\) that satisfies \(x < 1.557\) is:
\[ \boxed{1} \]