If all values of `x in (a, b)` satisfy the inequality `tan x tan 3x lt -1, x in (0, (pi)/(2))`, then the maximum value (b, -a) is :

Options

A. `(pi)/(12)`
B. `(pi)/(3)`
C. `(pi)/(6)`
D. `(pi)/(4)`

Answer

To solve the inequality \( \tan x \tan 3x < -1 \) for \( x \in (0, \frac{\pi}{2}) \), we will follow these steps:

### Step 1: Rewrite the inequality
We start with the inequality:
\[ \tan x \tan 3x < -1 \]
Using the identity for \( \tan 3x \):
\[ \tan 3x = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} \]
Substituting this into the inequality gives:
\[ \tan x \left( \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} \right) < -1 \]

### Step 2: Substitute \( t = \tan x \)
Let \( t = \tan x \). The inequality becomes:
\[ t \left( \frac{3t - t^3}{1 - 3t^2} \right) < -1 \]
Multiplying both sides by \( 1 - 3t^2 \) (noting that this expression is positive for \( t < \frac{1}{\sqrt{3}} \)):
\[ t(3t - t^3) < - (1 - 3t^2) \]
This simplifies to:
\[ 3t^2 - t^4 + 1 - 3t^2 < 0 \]
which reduces to:
\[ -t^4 + 1 < 0 \quad \text{or} \quad t^4 > 1 \]

### Step 3: Solve the inequality \( t^4 > 1 \)
This implies:
\[ t > 1 \quad \text{or} \quad t < -1 \]
Since \( t = \tan x \) is positive in the interval \( (0, \frac{\pi}{2}) \), we only consider \( t > 1 \).

### Step 4: Find the corresponding \( x \)
The condition \( t > 1 \) translates to:
\[ \tan x > 1 \]
This occurs when:
\[ x > \frac{\pi}{4} \]

### Step 5: Determine the interval for \( x \)
Now we need to find the interval \( (a, b) \) where \( x \) satisfies the inequality. Since \( x \) must also be less than \( \frac{\pi}{2} \), we have:
\[ x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \]
Thus, we can identify:
- \( a = \frac{\pi}{4} \)
- \( b = \frac{\pi}{2} \)

### Step 6: Calculate \( b - a \)
Now we find the maximum value of \( b - a \):
\[ b - a = \frac{\pi}{2} - \frac{\pi}{4} = \frac{2\pi}{4} - \frac{\pi}{4} = \frac{\pi}{4} \]

### Step 7: Final answer
The maximum value of \( (b, -a) \) is:
\[ \frac{\pi}{4} \]