Let `f(x) = (arc tan x)^(3) + (arc cot x)^(3) `.If the range of f (x) is `[a,b)`, then find the value of `b/(7a)`.

Answer

To solve the problem, we need to find the range of the function \( f(x) = (\tan^{-1} x)^3 + (\cot^{-1} x)^3 \) and then determine the value of \( \frac{b}{7a} \) where the range is given as \([a, b)\).

### Step-by-Step Solution:

1. **Understanding the function**:
We start with the function:
\[ f(x) = (\tan^{-1} x)^3 + (\cot^{-1} x)^3 \]
We know that:
\[ \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \]
Let \( t = \tan^{-1} x \). Then \( \cot^{-1} x = \frac{\pi}{2} - t \).

2. **Substituting in the function**:
We can rewrite \( f(x) \) in terms of \( t \):
\[ f(x) = t^3 + \left(\frac{\pi}{2} - t\right)^3 \]

3. **Expanding the cubic**:
Expanding \( \left(\frac{\pi}{2} - t\right)^3 \):
\[ \left(\frac{\pi}{2} - t\right)^3 = \frac{\pi^3}{8} - \frac{3\pi^2}{4}t + \frac{3\pi}{2}t^2 - t^3 \]
Therefore, we can combine the terms:
\[ f(x) = t^3 + \left(\frac{\pi^3}{8} - \frac{3\pi^2}{4}t + \frac{3\pi}{2}t^2 - t^3\right) \]
Simplifying gives:
\[ f(x) = \frac{\pi^3}{8} - \frac{3\pi^2}{4}t + \frac{3\pi}{2}t^2 \]

4. **Finding the range**:
The function \( f(x) \) is a quadratic in \( t \):
\[ f(t) = \frac{3\pi}{2}t^2 - \frac{3\pi^2}{4}t + \frac{\pi^3}{8} \]
The vertex of this parabola can be found using:
\[ t = -\frac{b}{2a} = -\frac{-\frac{3\pi^2}{4}}{2 \cdot \frac{3\pi}{2}} = \frac{\pi}{4} \]

5. **Calculating the maximum value**:
Substituting \( t = \frac{\pi}{4} \) into \( f(t) \):
\[ f\left(\frac{\pi}{4}\right) = \frac{3\pi}{2}\left(\frac{\pi}{4}\right)^2 - \frac{3\pi^2}{4}\left(\frac{\pi}{4}\right) + \frac{\pi^3}{8} \]
This simplifies to:
\[ = \frac{3\pi^3}{32} - \frac{3\pi^3}{16} + \frac{\pi^3}{8} \]
Finding a common denominator (32):
\[ = \frac{3\pi^3}{32} - \frac{6\pi^3}{32} + \frac{4\pi^3}{32} = \frac{3\pi^3}{32} \]

6. **Finding the endpoints**:
As \( t \) approaches \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \):
- At \( t = -\frac{\pi}{2} \):
\[ f\left(-\frac{\pi}{2}\right) = \frac{3\pi}{2}\left(-\frac{\pi}{2}\right)^2 - \frac{3\pi^2}{4}\left(-\frac{\pi}{2}\right) + \frac{\pi^3}{8} \]
This results in a value of \( \frac{7\pi^3}{32} \).

7. **Final range**:
The range of \( f(x) \) is:
\[ \left[\frac{3\pi^3}{32}, \frac{7\pi^3}{32}\right) \]
Hence, \( a = \frac{3\pi^3}{32} \) and \( b = \frac{7\pi^3}{32} \).

8. **Calculating \( \frac{b}{7a} \)**:
\[ \frac{b}{7a} = \frac{\frac{7\pi^3}{32}}{7 \cdot \frac{3\pi^3}{32}} = \frac{7}{21} = \frac{1}{3} \]

### Final Answer:
\[ \frac{b}{7a} = \frac{1}{3} \]