`(tan^-1x)^3+(cot^-1x)^3=k pi^3` then find the range of k

Options

A. `(1/32,9/8)`
B. `[1/32,7/8)`
C. `[1/32,9/8]`
D. `[1/32,1]`

Answer

To solve the equation \((\tan^{-1} x)^3 + (\cot^{-1} x)^3 = k \pi^3\) and find the range of \(k\), we can follow these steps:

### Step 1: Use the identity for \(\tan^{-1} x\) and \(\cot^{-1} x\)
We know that:
\[ \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \]
Let \(a = \tan^{-1} x\) and \(b = \cot^{-1} x\). Then we have:
\[ a + b = \frac{\pi}{2} \]

### Step 2: Apply the sum of cubes formula
Using the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
we can rewrite our equation as:
\[ a^3 + b^3 = \left(\frac{\pi}{2}\right) \left(a^2 - ab + b^2\right) \]

### Step 3: Express \(a^2 - ab + b^2\)
We can express \(a^2 + b^2\) in terms of \(a + b\) and \(ab\):
\[ a^2 + b^2 = (a + b)^2 - 2ab = \left(\frac{\pi}{2}\right)^2 - 2ab \]
Thus,
\[ a^2 - ab + b^2 = a^2 + b^2 - ab = \left(\frac{\pi^2}{4} - 2ab\right) - ab = \frac{\pi^2}{4} - 3ab \]

### Step 4: Substitute back into the equation
Now substituting back into our equation:
\[ a^3 + b^3 = \frac{\pi}{2} \left(\frac{\pi^2}{4} - 3ab\right) \]
This gives:
\[ a^3 + b^3 = \frac{\pi^3}{8} - \frac{3\pi}{2} ab \]

### Step 5: Set the equation equal to \(k \pi^3\)
We set this equal to \(k \pi^3\):
\[ \frac{\pi^3}{8} - \frac{3\pi}{2} ab = k \pi^3 \]
Dividing through by \(\pi^3\) (assuming \(\pi \neq 0\)):
\[ \frac{1}{8} - \frac{3}{2\pi^2} ab = k \]

### Step 6: Determine the range of \(ab\)
Now, we need to find the range of \(ab\). Since \(a = \tan^{-1} x\) and \(b = \cot^{-1} x\), we have:
\[ ab = \tan^{-1} x \cdot \cot^{-1} x \]
The maximum value of \(ab\) occurs when \(x = 1\):
\[ \tan^{-1}(1) = \frac{\pi}{4}, \quad \cot^{-1}(1) = \frac{\pi}{4} \implies ab = \left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{16} \]
The minimum value of \(ab\) occurs as \(x\) approaches 0 or \(\infty\), where \(ab\) approaches 0.

### Step 7: Substitute the bounds of \(ab\) into the equation for \(k\)
Substituting \(ab = 0\) into the equation for \(k\):
\[ k_{\text{min}} = \frac{1}{8} - 0 = \frac{1}{8} \]
Substituting \(ab = \frac{\pi^2}{16}\):
\[ k_{\text{max}} = \frac{1}{8} - \frac{3}{2\pi^2} \cdot \frac{\pi^2}{16} = \frac{1}{8} - \frac{3}{32} = \frac{4}{32} - \frac{3}{32} = \frac{1}{32} \]

### Final Step: Determine the range of \(k\)
Thus, the range of \(k\) is:
\[ \frac{1}{32} < k < \frac{1}{8} \]