Considering only the principal values of inverse functions, the set `A={x ge 0 :tan^(-1)(2x)+tan^(-1)(3x)=(pi)/(4)}`

Options

A. Is an empty set
B. Contains two elements
C. Contains more than two elements
D. Is a singleton

Explanation

To solve the equation \( \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4} \) for \( x \geq 0 \), we can use the property of the inverse tangent function.

### Step-by-Step Solution:

1. **Use the Identity for Inverse Tangent:**
We know that:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
for \( ab < 1 \). Here, let \( a = 2x \) and \( b = 3x \).

2. **Apply the Identity:**
Substitute \( a \) and \( b \) into the identity:
\[ \tan^{-1}(2x) + \tan^{-1}(3x) = \tan^{-1}\left(\frac{2x + 3x}{1 - (2x)(3x)}\right) \]
This simplifies to:
\[ \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) \]

3. **Set the Equation:**
Now, we can set the equation:
\[ \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) = \frac{\pi}{4} \]

4. **Use the Property of Tangent:**
Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), we equate:
\[ \frac{5x}{1 - 6x^2} = 1 \]

5. **Cross-Multiply:**
Cross-multiplying gives:
\[ 5x = 1 - 6x^2 \]

6. **Rearrange the Equation:**
Rearranging the equation leads to:
\[ 6x^2 + 5x - 1 = 0 \]

7. **Use the Quadratic Formula:**
We can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 6, b = 5, c = -1 \). Therefore:
\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 6 \cdot (-1)}}{2 \cdot 6} \]
\[ x = \frac{-5 \pm \sqrt{25 + 24}}{12} \]
\[ x = \frac{-5 \pm \sqrt{49}}{12} \]
\[ x = \frac{-5 \pm 7}{12} \]

8. **Calculate the Roots:**
This gives us two potential solutions:
\[ x = \frac{2}{12} = \frac{1}{6} \quad \text{and} \quad x = \frac{-12}{12} = -1 \]

9. **Consider the Domain:**
Since we are considering \( x \geq 0 \), we discard \( x = -1 \). Thus, the only valid solution is:
\[ x = \frac{1}{6} \]

10. **Conclusion:**
Therefore, the set \( A \) is:
\[ A = \left\{ \frac{1}{6} \right\} \]
which is a singleton set.