Let `f (x) = tan ^(-1) ((1-x)/(1+x)).` Then difference of the greatest and least value of `f (x)` on`[0,1]` is:

Options

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

Answer

To find the difference between the greatest and least values of the function \( f(x) = \tan^{-1}\left(\frac{1-x}{1+x}\right) \) on the interval \([0, 1]\), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \tan^{-1}\left(\frac{1-x}{1+x}\right) \]
We can use the identity for the tangent of a difference:
\[ \frac{1 - \tan \theta}{1 + \tan \theta} = \tan\left(\frac{\pi}{4} - \theta\right) \]
By substituting \( x = \tan \theta \), we have:
\[ f(x) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) \]
This simplifies to:
\[ f(x) = \frac{\pi}{4} - \theta \]
where \( \theta = \tan^{-1}(x) \).

### Step 2: Express \( f(x) \) in terms of \( x \)
Since \( \theta = \tan^{-1}(x) \), we can write:
\[ f(x) = \frac{\pi}{4} - \tan^{-1}(x) \]

### Step 3: Evaluate \( f(x) \) at the endpoints of the interval
Now we evaluate \( f(x) \) at the endpoints of the interval \([0, 1]\).

1. **At \( x = 0 \)**:
\[ f(0) = \frac{\pi}{4} - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

2. **At \( x = 1 \)**:
\[ f(1) = \frac{\pi}{4} - \tan^{-1}(1) = \frac{\pi}{4} - \frac{\pi}{4} = 0 \]

### Step 4: Identify the maximum and minimum values
From our evaluations:
- The maximum value of \( f(x) \) on \([0, 1]\) is \( \frac{\pi}{4} \) (at \( x = 0 \)).
- The minimum value of \( f(x) \) on \([0, 1]\) is \( 0 \) (at \( x = 1 \)).

### Step 5: Calculate the difference
Now, we find the difference between the maximum and minimum values:
\[ \text{Difference} = \text{Maximum} - \text{Minimum} = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### Final Answer
Thus, the difference of the greatest and least value of \( f(x) \) on \([0, 1]\) is:
\[ \frac{\pi}{4} \]