Arithmetic mean of the non-zero solutions of the equation `tan^-1 (1/(2x + 1)) + tan^-1 (1/(4x + 1)) = tan^-1 (2/x^2)`

Options

A. 2
B. 3
C. 4
D. none of these

Explanation

To solve the equation \( \tan^{-1} \left( \frac{1}{2x + 1} \right) + \tan^{-1} \left( \frac{1}{4x + 1} \right) = \tan^{-1} \left( \frac{2}{x^2} \right) \), we will follow these steps:

### Step 1: Use the formula for the sum of arctangents
We can use the formula for the sum of two arctangents:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \quad \text{if } ab < 1 \]
Let \( a = \frac{1}{2x + 1} \) and \( b = \frac{1}{4x + 1} \).

### Step 2: Calculate \( a + b \) and \( ab \)
Calculating \( a + b \):
\[ a + b = \frac{1}{2x + 1} + \frac{1}{4x + 1} = \frac{(4x + 1) + (2x + 1)}{(2x + 1)(4x + 1)} = \frac{6x + 2}{(2x + 1)(4x + 1)} \]

Calculating \( ab \):
\[ ab = \frac{1}{(2x + 1)(4x + 1)} \]

### Step 3: Substitute into the formula
Now, substituting into the formula:
\[ \tan^{-1} \left( \frac{6x + 2}{(2x + 1)(4x + 1) - 1} \right) = \tan^{-1} \left( \frac{2}{x^2} \right) \]

### Step 4: Set the arguments equal
Since the arctangents are equal, we can set the arguments equal to each other:
\[ \frac{6x + 2}{(2x + 1)(4x + 1) - 1} = \frac{2}{x^2} \]

### Step 5: Cross-multiply
Cross-multiplying gives:
\[ (6x + 2)x^2 = 2((2x + 1)(4x + 1) - 1) \]

### Step 6: Simplify the equation
Expanding both sides:
- Left side: \( 6x^3 + 2x^2 \)
- Right side: \( 2(8x^2 + 6x + 1 - 1) = 16x^2 + 12x \)

Setting the equation:
\[ 6x^3 + 2x^2 = 16x^2 + 12x \]
Rearranging gives:
\[ 6x^3 - 14x^2 - 12x = 0 \]

### Step 7: Factor the polynomial
Factoring out \( 2x \):
\[ 2x(3x^2 - 7x - 6) = 0 \]
This gives us \( x = 0 \) or solving \( 3x^2 - 7x - 6 = 0 \).

### Step 8: Solve the quadratic equation
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 3 \cdot (-6)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ 49 + 72 = 121 \quad \Rightarrow \quad \sqrt{121} = 11 \]
Thus:
\[ x = \frac{7 \pm 11}{6} \]
Calculating the two possible values:
1. \( x = \frac{18}{6} = 3 \)
2. \( x = \frac{-4}{6} = -\frac{2}{3} \)

### Step 9: Identify non-zero solutions
The non-zero solutions are \( x = 3 \) and \( x = -\frac{2}{3} \).

### Step 10: Calculate the arithmetic mean
The arithmetic mean of the non-zero solutions:
\[ \text{Mean} = \frac{3 + \left(-\frac{2}{3}\right)}{2} = \frac{3 - \frac{2}{3}}{2} = \frac{\frac{9}{3} - \frac{2}{3}}{2} = \frac{\frac{7}{3}}{2} = \frac{7}{6} \]

Thus, the arithmetic mean of the non-zero solutions is \( \frac{7}{6} \).

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