If `y = tan^(-1).(1)/(2) + tan^(-1) b, (0 b lt 1) and 0 lt y le (pi)/(4)`, then the maximum value of b is

Options

A. `1//2`
B. `1//3`
C. `1//4`
D. `2//3`

Explanation

To solve the problem step by step, we start with the given equation:

### Step 1: Write down the equation
Given:
\[ y = \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}(b) \]
where \( 0 < b < 1 \) and \( 0 < y \leq \frac{\pi}{4} \).

### Step 2: Set the maximum value of \( y \)
Since we want to find the maximum value of \( b \), we will consider the maximum value of \( y \):
\[ y = \frac{\pi}{4} \]

### Step 3: Rearrange the equation
Substituting \( y = \frac{\pi}{4} \) into the equation, we get:
\[ \frac{\pi}{4} = \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}(b) \]

### Step 4: Isolate \( \tan^{-1}(b) \)
Rearranging gives:
\[ \tan^{-1}(b) = \frac{\pi}{4} - \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 5: Use the identity for the difference of arctangents
Using the identity for the difference of arctangents:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
we can rewrite:
\[ \tan^{-1}(b) = \tan^{-1}\left(1\right) - \tan^{-1}\left(\frac{1}{2}\right) \]
This can be expressed as:
\[ b = \frac{1 - \frac{1}{2}}{1 + 1 \cdot \frac{1}{2}} \]

### Step 6: Simplify the expression for \( b \)
Calculating the right-hand side:
\[ b = \frac{\frac{1}{2}}{1 + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3} \]

### Conclusion
Thus, the maximum value of \( b \) is:
\[ \boxed{\frac{1}{3}} \]