Considering only the principal values of inverse trigonometric functions,the number of positive real values of "x" satisfying `tan^(-1)(x)+tan^(-1)(2x)=(pi)/(4)` is

Options

A. more than 2
B. 2
C. 0
D. 1

Answer

To solve the equation \( \tan^{-1}(x) + \tan^{-1}(2x) = \frac{\pi}{4} \), we can use the following steps:

### Step 1: Use the formula for the sum of inverse tangents
We know that:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
provided \( ab < 1 \).

In our case, let \( a = x \) and \( b = 2x \). Therefore, we can rewrite the equation:
\[ \tan^{-1}(x) + \tan^{-1}(2x) = \tan^{-1}\left(\frac{x + 2x}{1 - x \cdot 2x}\right) \]
This simplifies to:
\[ \tan^{-1}(3x) = \frac{\pi}{4} \]

### Step 2: Take the tangent of both sides
Taking the tangent of both sides gives us:
\[ 3x = \tan\left(\frac{\pi}{4}\right) \]
Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), we have:
\[ 3x = 1 \]

### Step 3: Solve for \( x \)
Now, solving for \( x \):
\[ x = \frac{1}{3} \]

### Step 4: Check for the number of positive real solutions
Since we derived \( x = \frac{1}{3} \) from the equation, we need to check if there are any other solutions. The function \( \tan^{-1}(x) + \tan^{-1}(2x) \) is strictly increasing for \( x > 0 \) because both \( \tan^{-1}(x) \) and \( \tan^{-1}(2x) \) are strictly increasing functions. Therefore, the equation \( \tan^{-1}(x) + \tan^{-1}(2x) = \frac{\pi}{4} \) can have at most one positive solution.

Since we found \( x = \frac{1}{3} \) as the only solution, we conclude that there is exactly one positive real value of \( x \).

### Final Answer
The number of positive real values of \( x \) satisfying the equation is:
\[ \boxed{1} \]