The solution set of the inequality `(tan^(-1)x)^(2) le (tan^(-1)x) +6` is

Options

A. `[-tan2, tan3]`
B. `[tan2, - tan3]`
C. `[tan2, tan3]`
D. `(-oo, oo)`

Answer

To solve the inequality \( (\tan^{-1} x)^2 \leq \tan^{-1} x + 6 \), we can follow these steps:

### Step 1: Substitute \( t = \tan^{-1} x \)
We start by letting \( t = \tan^{-1} x \). This transforms our inequality into:
\[ t^2 \leq t + 6 \]

### Step 2: Rearrange the inequality
Rearranging the inequality gives us:
\[ t^2 - t - 6 \leq 0 \]

### Step 3: Factor the quadratic expression
Next, we can factor the quadratic expression:
\[ t^2 - t - 6 = (t - 3)(t + 2) \leq 0 \]

### Step 4: Find the critical points
The critical points of the inequality are found by setting the factors to zero:
\[ t - 3 = 0 \quad \Rightarrow \quad t = 3 \]
\[ t + 2 = 0 \quad \Rightarrow \quad t = -2 \]

### Step 5: Test intervals on a number line
Now we will test the intervals determined by the critical points \( t = -2 \) and \( t = 3 \):
- For \( t < -2 \): Choose \( t = -3 \):
\[ (-3 - 3)(-3 + 2) = (-6)(-1) = 6 \quad (\text{positive}) \]
- For \( -2 < t < 3 \): Choose \( t = 0 \):
\[ (0 - 3)(0 + 2) = (-3)(2) = -6 \quad (\text{negative}) \]
- For \( t > 3 \): Choose \( t = 4 \):
\[ (4 - 3)(4 + 2) = (1)(6) = 6 \quad (\text{positive}) \]

### Step 6: Determine the solution set
From the tests, we find that the product is less than or equal to zero in the interval:
\[ -2 \leq t \leq 3 \]

### Step 7: Convert back to \( x \)
Since \( t = \tan^{-1} x \), we need to convert this back to \( x \):
- For \( t = -2 \):
\[ x = \tan(-2) \]
- For \( t = 3 \):
\[ x = \tan(3) \]

Thus, the solution set for the inequality \( (\tan^{-1} x)^2 \leq \tan^{-1} x + 6 \) is:
\[ x \in [\tan(-2), \tan(3)] \]

### Final Answer
The solution set is:
\[ x \in [\tan^{-1}(-2), \tan^{-1}(3)] \]